Class 9 Maths

Coordinate Geometry Chapter 1 Distance Formulas

⇒ Coordinate Geometry: The branch of mathematics in which different problems of geometry are solved with the help of algebra is known as coordinate geometry.

⇒ Two types of co-ordinate geometry:

1. Two-dimensional or plane geometry.
2. Three-dimensional or solid geometry.

⇒  XOX’ and YOY’ are two perpendicular straight lines intersects at O.

⇒  They have divided the plane into four sections. Each of these sections is called a quadrant.

⇒  The section XOY, YOX’, X’OY’, and Y’OX’ are called the 1st, 2nd, 3rd, and 4th quadrants respectively.

⇒  The fixed point O is called the origin whose coordinate is (0, 0).

⇒  The straight lines together are called the coordinate axes.

⇒  \(\overleftrightarrow{\mathrm{XOX’}}\) is called the x-axis or abscissa and \(\overleftrightarrow{\mathrm{YOY’}}\) is called the y-axis or ordinate.

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Formulas at a glance:

1. Distance of a given point P(x, y) from the origin (0, 0) is OP = \(\sqrt{x^2+y^2}\)
2. The distance between two given points P (x₁, y₁) and Q (x₂, y₁) is \(\overline{P Q}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) units.

Coordinate Geometry Chapter 1 Distance Formulas True Or False

Example 1. The distance between the points A(a, 0) and B(b, 0) along the positive direction of the X-axis is (b – a) units (b > a).

Solution: The distance between the points A(a, 0)

and B (b, 0) is \(\sqrt{(b-a)^2+(0-0)^2}\) units

= \(\sqrt{(b-a)^2}\) units

= (b – a) units

∴ The statement is True.

Example 2. The distance between two points A(-p, 0) and (-q, 0) along the negative direction of X- axis (p – q) units [p < q].

Solution: The distance between A(- p, 0) and B(-q, 0) is

\(\sqrt{\{-q-(-p)\}^2+(0-0)^2}\) units

= \(\sqrt{(-q+p)^2}\) units

= \(\sqrt{{-(q-p)}^2}\) units

= \(\sqrt{(q-p)^2}\) units =(q – p) units

∴ The statement is False.

Example 3. The points (3, 4) and (-3, -4) are equidistant from origin.

Solution: The distance of point (3, 4) from origin (0, 0) is \(\sqrt{3^2+4^2}\) units = √25 units = 5 units

The distance between points (-3, -4) and (0, 0) is \(\sqrt{(-3)^2+(-4)^2}\) units

= \(\sqrt{9+16}\) units = √25 units = 5 units

∴ The statement is True.

Coordinate Geometry Chapter 1 Distance Formulas Fill In The Blanks

Example 1. The distance between the points A (0, m) and (0, – n) is _______ units.

Solution: The required distance is \(\sqrt{(0-0)^2+\{m-(-n)\}^2}\) units = \(=\sqrt{(m+n)^2}\) units = (m + n) units.

Example 2. The distance between the points (-7, 0) and (-2, 0) is ________ units.

Solution: The required distance is \(\sqrt{\{(-7)-(-2)\}^2+(0-0)^2}\) units

= \(\sqrt{(-7+2)^2+0} \text { units }\)

= \(\sqrt{(-5)^2} \text { units }\)

= √25 units = 5 units.

Example 3. If in a square (4, 4) and (-4, 4) are two adjacent vertices, then the perimeter of the square is ______ units.

Solution: The length of AD is \(\sqrt{(-4-4)^2+(4-4)^2}\) units

= \(\sqrt{(-8)^2+(0)^2}\) units

= √64 units = 8 units

Perimeter is (4 x 8) units= 32 units.

∴ The perimeter of the square is 32 units.

Coordinate Geometry Chapter 1 Distance Formulas Short Answer Type Questions

Example 1. Find the value of y if the distance of the point (-4, y) from origin is 5 units.

Solution: The distance of the point (-4, y) from the origin (0, 0) is \(\sqrt{(-4-0)^2+(y-0)^2}\) units

= \(\sqrt{16+y^2}\) units

According to question, \(\sqrt{16+y^2}\) = 5

⇒ 16 + y² = 25

⇒ y² = 25 – 16

⇒ y² = 9

⇒ y = ±√9

⇒ y = ±3

∴ The value of y is ±3.

Example 2. Find the coordinates of a point on the y-axis which is equidistant from two points (2, 3) and (-1, 2).

Solution: Let the coordinates of a point of the y-axis is (0, k)

The distance between the points (2, 3) and (0, k) is \(\sqrt{(2-0)^2+(3-k)^2}\) units

The distance between the points (1, 2) and (0, k) is \(\sqrt{(-1-0)^2+(2-k)^2}\) units

According to question, \(\sqrt{(2-0)^2+(3-k)^2=\sqrt{(-1-0)^2+(2-k)^2}}\)

⇒ 4 + (3 – k)² ⇒ 1 + (2 – k)² [squaring both sides]

⇒ 4 + 9 – 6k + k² = 1 + 4 – 4k + k²

⇒ -6k + k² + 4k – k² = 5 – 4 – 9

⇒ -2k = -8

⇒ k = 4.

∴ The coordinate of points on the Y-axis is (0, 4).

Example 3. Write the coordinates of two points on X-axis and Y-axis for which an isosceles right-angled triangle is formed with x-axis, y-axis, and straight line. joining two points.

Solution: As an isosceles right-angled triangle is formed with the x-axis, y-axis, and a straight line joining two points A and B;

∴ OA = OB

∴ The coordinates of two points on X-axis and Y-axis are (1, 0), (0, 1); (2, 0), (0, 2); (3, 0), (0, 3), etc.

Example 4. Write the coordinates of two points on opposite sides of x-axis which are equidistant from x-axis.

Solution: The coordinates of two points on opposite sides of x-axis are (2, 3), (2, -3); (-5, 6), (-5, -6) etc.

Example 5. Write the coordinates of two points on opposite sides of y-axis which are equidistant from y-axis.

Solution: The coordinates of two points on opposite sides of y-axis which are equidistant from y-axis are (2, 5), (-2, 3); (4, 6), (-4, 8) etc.

Example 6. Show that point (-2, -11) are equidistant from points (-3, 7) and (4, 6).

Solution: The distance between the points (-2, -11) and (- 3, 7) is \(\sqrt{\{-2-(-3)\}^2+(-11-7)^2} \text { units }\)

= \(\sqrt{(-2+3)^2+(-18)^2} \text { units }\)

= \(\sqrt{1+324} \text { units }=\sqrt{325} \text { units }\)

The distance between the points (-2, -11) and (4, 6) is \(\sqrt{(-2-4)^2+(-11-6)^2}\) units

= \(\sqrt{(-6)^2+(-17)^2} \text { units }\)

= \(\sqrt{36+289} \text { units }=\sqrt{325} \text { units }\)

∴ The point (-2,-11) is equidistant from points (-3, 7) and (4, 6).

Example 7. If the points A (2, -2), B (8, 4), C (5, 7), and D(-1, 1) are the vertices of a rectangle, then show that the lengths of diagonals AC and BD are equal.

Solution: The length of diagonal AC is \(\sqrt{(2-5)^2+(-2-7)^2} \text { units }\)

= \(\sqrt{(-3)^2+(-9)^2} \text { units }\)

= \(\sqrt{9+81} \text { units }=\sqrt{90} \text { units }\)

The length of diagonal BD is \(\sqrt{\{8-(-1)\}^2+(4-1)^2}\) units

= \(\sqrt{(9)^2+(3)^2}\) units

= \(\sqrt{81+9}\) units

= \(\sqrt{90}\) units

∴ AC = BD.

Example 8. Are the points A (0, 0), B (4, 3), and C (8, 6) co-linear? Verify the statement.

Solution: If A, B, and C are co-linear then AB + BC = AC

AB = \(\sqrt{(4-0)^2+(3-0)^2} \text { units }=\sqrt{25} \text { units }=5 \text { units }\)

BC = \(\sqrt{(8-4)^2+(6-3)^2} \text { units }\)

= \(\sqrt{16+9} \text { units }=\sqrt{25} \text { units }=5 \text { units. }\)

AC = \(\sqrt{(8-0)^2+(6-0)^2} \text { units }\)

= \(\sqrt{100} \text { units }=10 \text { units }\)

∴ AB+ BC= (5 + 5) units = 10 units

∴ AB + BC = AC

∴ The given points are co-linear.

Example 9. If the distance between the points (2, y) and (10,-9) is 10 units then find the value of y.

Solution: The distance between the points (2, y) and (10, -9) is \(\sqrt{(2-10)^2+\{y-(-9)\}^2} \text { units }\)

= \(\sqrt{64+(y+9)^2} \text { units }\)

According to question, \(\sqrt{64+(y+9)^2}\) = 10

⇒ 64 + y² + 18y + 81 = 100

⇒ y² + 18y + 45 = 0

⇒ y + 15y + 3y + 45 = 0

⇒ y (y + 15) + 3 (y + 15) = 0

⇒ (y + 15) (y + 3) = 0

either y + 15 = 0

⇒ y = – 15

or, y + 3 = 0

⇒ y = – 3

∴ The value of is – 3.

Example 10. Find the point on X-axis which are equidistant from points (3, 5) and (1, 3).

Solution: Let the point on X-axis is (h, 0)

The distance between (3, 5) and (h, 0) is \(\sqrt{(3-h)^2+(5-0)^2}\) units

The distance between (1, 3) and (h, 0) is \(\sqrt{(1-h)^2+(3-0)^2}\) units

According to question, \(\sqrt{(3-h)^2+(5-0)^2}=\sqrt{(1-h)^2+(3-0)^2}\)

⇒ 9 – 6h + h² + 25 = 1 – 2h + h² + 9

⇒ -6h + 2h + h² – h² = 1 + 9- 9 – 25

⇒ – 4h = – 24 ⇒ h = 6

∴ The required points is (6, 0).

Coordinate Geometry Chapter 3 Area Of Triangular Region

Area of formula:

1. The area of ΔABC where vertices are A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) respectively is \(\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\) sq. unit

or, = \(\frac{1}{2}\left|y_1\left(x_2-x_3\right)+y_2\left(x_3-x_1\right)+y_3\left(x_1-x_2\right)\right|\) sq. unit [Area is always taken positive]

2. The points (x₁, y₁), (x₂, y₂), (y₃, y₃) are collinear if x₁ (y₂ – y₃)+ x₂ (y₃ – y₁) + x₃ (y₁ – y₂) = 0

or, y₁ (x₂ – x₃) + y₂ (x₃ – x₁) + y₃(x₁ – x₂) = 0

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Area of the triangular region is

= \(\frac{1}{2}=\left|x_1 y_2+x_2 y_3+x_3 y_1-\left(y_1 x_2+y_2 x_3+y_3 x_1\right)\right|\) sq. unit

Similarly, area of quadrilateral region is

= \(\frac{1}{2}\left|\left(x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1\right)-\left(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1\right)\right|\) sq. unit

Coordinate Geometry Chapter 3 Area Of Triangular Region Fill In The Blanks

Example 1. Area of the triangle having vertices (2, 2), (4, 2), (1, 3) is _______ sq. unit.

Solution: 11

Example 2. (0, -2), (2, 4), (-1, -5) are ______ points.

Solution: Collinear.

Example 3. If (2, -1), (k, -1), (1, -1) are collinear points then k = ________

Solution: Any real value.

Example 4. If the points (1, 2), (2, 4), and (6, 6) are collinear then t = 

Solution: 3

Example 5. If the vertices of a triangle are (-1, 0), (0, 0), (0, 1) then its area is ______ sq. unit

Solution: \(\frac{1}{2}\)

Example 6. If the three points (0, 0), (2, -3), (x, y) are collinear then x = ______, y = _______

Solution: 4, -6.

Coordinate Geometry Chapter 3 Area Of Triangular Region True Or False

Example 1. Area of the triangle having vertices (-3, 1), (– 2, 2), and (3, 2) is 9\(\frac{1}{2}\) sq. unit.

Solution: The statement is true.

Example 2. If the vertices of a triangle are (a, b + c), (a, b – c), and (-a, c) then area is \(\frac{1}{2}\)/ab/sq.unit

Solution: The statement is false.

Example 3. (1, 4), (-1, 2), (-4, -1) are collinear points.

Solution: The statement is true.

Example 4. (a, b + c), (b, c + a), (c, a + b) are not collinear points.

Solution: The statement is false.

Example 5. Value of k for which (k, k), (-1, 5), (-7, 8) be collinear is 3.

Solution: The statement is true.

Example 6. If the points (a, c), (0, b), (\(\frac{1}{2}\), \(\frac{1}{2}\)) are collinear points then \(\frac{1}{a}+\frac{1}{b}=2\)

Solution: The statement is false.

Coordinate Geometry Chapter 3 Area Of Triangular Region Short Answer Type Questions

Example 1. If the three points (a, 0), (0, b), (1, 1) are collinear then show that \(\frac{1}{a}+\frac{1}{b}=1\)

Solution:

= \(\frac{1}{2}|a b+0+0-b-a|=0 \quad \text { or, } a+b=a b\)

∴ \(\frac{1}{a}+\frac{1}{h}=1\)

Example 2. Find the area of the triangular area region formed by the three points (1, 4), (-1, 2), and (-4, 1).

Solution:

⇒ Area = \(\frac{1}{2}|2-1-16+4+8-1| \text { sq.unit }=\frac{1}{2}|14-18| \text { sq.. unit }=2\)

The area of the triangular area region = \(\frac{1}{2}|2-1-16+4+8-1| \text { sq.unit }=\frac{1}{2}|14-18| \text { sq.. unit }=2\)

Example 3. If the points (0, -4), (-1, y), and (3, 2) are on the same straight line find y.

Solution: \(\frac{1}{2}|0(y-2)+(-1)\{2-(-4)\}+3(-4-4)|=0\)

⇒ or, -6 – 12 – 3y = 0, y = 6

Example 4. If the points A (3a, k), B (0, 3b), C (a, 2b) are collinear find k.

Solution: 3a (3b – 2b) + 0 (2b – k) + a (k – 3b) = 0

or, 3ab + ak – 3ab 0 or, ak = 0

∴ k = 0

Example 5. Show that (a, b + c), (b, c + a), and (c, a + b) are collinear.

Solution: \(\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\)

= \(\frac{1}{2}[a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)]\)

= \(\frac{1}{2}(a c-a b+a b-b c+b c-a c)=0\)

⇒ Here x₁ = a, y₁ = b + c

⇒ x₂ = b, y₂ = c + a.

⇒ x₃ = c, y₃ = a + b

Example 6. If A (x, 4), B (- 5, 7), C (-4, 5) are collinear, then show that 2x+y+3=0

Solution: \(\frac{1}{2}\)[x(7-5) + (-5)(5 − y) + (−4)(y −7)] = 0

⇒ or, \(\frac{1}{2}\)(2x+y+3)=0

∴  2x + y + 3 = 0

Mensuration Chapter 3 Area Of Circle

⇔ Area of circle = πr²

⇔ Area of semi-circle = \(\frac{1}{2}\) πr²

⇔ Area of circular ring = π (R² – r²) = π (R + r)(R -r)

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⇔ Sector of a circle = \(\pi r^2 \cdot \frac{\theta}{360}\)

⇔ Area of a square inscribed in a circle = \(\frac{(\text { diagonal })^2}{2}=\frac{(2 r)^2}{2}=2 r^2\)

⇔ Area of a square circumscribing the circle = (2r)² = 4r²

⇔ Area of an equilateral triangle, when the length of the radius of its in circle = r unit

⇔ Area of the triangle = \(\frac{\sqrt{3}}{4}(2 \sqrt{3} r)^2=3 \sqrt{3} r^2\)

⇔ Area of an equilateral triangle, when the length of the radius of its circumcircle is r unit

⇔ Area of the triangle = \(\frac{\sqrt{3}}{4}(4 \sqrt{3} r)^2=12 \sqrt{3} r^2\)

Mensuration Chapter 3 Area Of Circle Fill In The Blanks

Example 1. If the circumference of a circle be 22 cm, then its area will be ______ sq. cm

Solution: 38.5.

Example 2. The area of a sector making an angle 30° at the centre of a circle of radius 21 cm is ______ sq. cm.

Solution: 115.5.

Example 3. If the ratio of the circumference of two circles be 2: 3, the ratio of their areas will be ______

Solution: 4:9.

Example 4. If x be the area of a circle its circumference is ______

Solution: 2√πx

Example 5. The diameter of a circle inscribed in the square of area 49 sq cm is ______

Solution: 7 cm.

Mensuration Chapter 3 Area Of Circle True Or False

Example 1. If the circumference of a circle is A, then its diameter = \(\frac{A}{\pi}\).

Solution: The statement is true.

Example 2. The radius of a circle is 21 cm, its area is 616 sq. cm.

Solution: The statement is false.

Example 3. Area of a semi-circle of radius 7 cm is 77 cm.

Solution: The statement is true.

Example 4. When a square is inscribed in a circle, its diagonal will be equal to the diameter of the circle.

Solution: The statement is true.

Example 5. When a square circumscribed a circle, its diagonal will be equal to the diameter of the circle.

Solution: The statement is false.

Mensuration Chapter 3 Area Of Circle Short Answer Type Questions

Example 1. If the length of radius of a circular field was increased by 10%, let us write by calculating what % it increase the area.

Solution: Let radius = 100 r unit

Initial Area = 10000 πr² sq.unit

Increased Area = π(110)² sq.unit = 12100 πr² sq.unit

% increase = \(\frac{2100 \pi r^2}{10000 \pi r^2}\) x 100 sq.unit = 21 sq.unit

∴ 21% increased.

Example 2. If the perimeter of a circular field was decreased by 50%. Calculate what percent it decrease the area of circular field.

Solution: Let radius be r unit.

Initial perimeter and area = 2πr unit, πr²sq.unit

Reduced perimeter = \(\frac{2 \pi r}{2}\) unit = πr unit

Reduced radius = \(\frac{r}{2}\) unit

Area decrease = \(=\left\{\pi r^2-\pi\left(\frac{r}{2}\right)^2\right\} \text { sq. unit }=\frac{3 \pi r^2}{4} \text { sq. unit }\)

∴ % decrease = \(\frac{3 \frac{\pi r^2}{4}}{\pi r^2}\) x 100 sq.unit = 75%

Area is decreased by 75%.

Example 3. The length of radius of a circular field is r meter. If the area of other circle is x times of first circle, let us calculate how length of radius of other circle.

Solution: Area = x x πr²

πR² = πr²

∴ R= r√x unit.

Example 4. Calculate the area of a circular region circumscribe a triangle of which sides are 3 cm, 4 cm, and 5 cm.

Solution: 3²+ 4² = 5², It is a right-angled triangle.

Length of circumradius = \(\frac{5}{2}\) cm = 2.5 cm.

Area = \(\frac{22}{7}\) x 2.5 x 2.5 sq.cm = 19\(\frac{9}{14}\) sq.cm.

Example 5. Three circular plates were cut off from a tin plate with equal width if the ratio of the length of diameter of three circles is 3: 5: 7, calculate the ratio of their weight.

Solution: Let weight of per sq. unit tin = x gm

Let diameter be 3R, 5R, 7R units respectively (R > 0)

∴ Ratio weight = \(\pi\left(\frac{3 \mathrm{R}}{2}\right)^2 \cdot x: \pi\left(\frac{5 \mathrm{R}}{2}\right)^2 \cdot x: \pi\left(\frac{7 \mathrm{R}}{2}\right)^2 \cdot x\) = 9: 25: 49

Example 6. Find the area of the shadow region of the figure.

Solution: Area =\(\left\{\frac{90}{360} \pi \cdot(14)^2-\Delta A O B\right\} \mathrm{sq} \cdot \mathrm{cm}\)

= \(\left\{\frac{1}{4} \pi \cdot(14)^2-\frac{1}{2} \cdot 14 \cdot 14\right\} \mathrm{sq} \mathrm{cm}=56 \mathrm{sqcm}\)

Example 7. If the perimeter of a circle = perimeter of an equilateral triangle. Find the ratio of their areas.

Solution: 2πr = 3a  ⇒ r = \(\frac{3a}{2 \pi}\)

∴ Ratio = \(\pi\left(\frac{3 a}{2 \pi}\right)^2: \frac{\sqrt{ } 3}{4} a^2=63: 22 \sqrt{3}\)

Example 8. Find The area of the shadow region of

Solution: Area = Area of the square – Area of 4 sectors

= \(\left\{12^2-\frac{90^{\circ}}{360^{\circ}} \times \pi(6)^2 \times 4\right\} \mathrm{sq} \cdot \mathrm{cm}=30 \frac{6}{7} \mathrm{sq} . \mathrm{cm}\)

Example 9. Find the area of shadow region of

Solution: Area Of 4 setors

= \(\left\{\frac{22}{7} \times(3.5)^2\right\}-\left\{\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(3.5)^2\right\} \times 4 \text { sq.cm }=115.5 \text { sq.cm }\)

Example 10. Find the area of the shadow region of

Solution: Area = \(\left(\frac{1}{4} \times \frac{22}{7} \times 12^2-\frac{1}{2} \times 12 \times 12\right) \mathrm{sqcm}\)

= \(\frac{288}{7} \text { sq. }\)

= \(41 \frac{1}{7} \text { sq. } \mathrm{cm}\)

Mensuration Chapter 2 Circumference Of Circle

Some Important Facts:

In any circle:

\(\frac{Circumference}{Diameter}\) = constant (denoted by π and π = \(\frac{22}{7}\))

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∴ circumference = π x diameter = π x 2 x radius

= πd = 2πr (d = length of the diameter, r = length of the radius)

Circular ring:

Let length of the inter radius and length of outer radius = R

∴ The width of the ring = R – r.

Mensuration Chapter 2 Circumference Of Circle True Or False

Example 1. If the circumference of a circle is A cm, then its radius is \(\frac{A}{\pi}\) cm.

Solution: 2πr = A

⇒ r = \(\frac{A}{2 \pi}\)

∴ The statement is False.

Example 2. The difference of the circumference and the radius is 3.7 cm. Length of its diameter is 1.4 cm.

Solution: 2πr – R = 3.7

⇒ R(2π – 1) = 3.7

⇒ R = \(\frac{3.7 \times 7}{37}\) =0.7, 2R = 1.4 cm

∴ The statement is True.

Example 3. The width of a circular ring is 7 cm. The difference between the outer and inner circumference of the ring is 4.4 cm.

Solution: R – r = 7

∴ 2π (R – r) = 7 x 2 x \(\frac{22}{7}\) = 44

∴ The statement is False.

Example 4. Perimeter of a semi-circular ring is π (R + r) + 2(R – r) when R and r are the length of outer and inner radius.

Solution: Perimeter Outer semi-circumference + inner circumference + 2 (width)

= πR + πr + 2 (R – r)

∴ The statement is True.

Example 5. The outer and inner circumference of a ring-shaped circular plate are x cm and y cm respectively. The width of the ring-shaped plate is \(\frac{x-y}{\pi}\) cm.

Solution: 2πR – 2πr = x – y

or, R – r = \(\frac{x-y}{2 \pi}\)

∴ The statement is False.

Mensuration Chapter 2 Circumference Of Circle Fill In The Blanks

Example 1. If the circumference of a circle be 44 cm, then its length of the diameter is ________ cm.

Solution: 14.

2πr = 44 ⇒ 2r = \(\frac{44 \times 7}{22}\) = 14

Example 2. In radius of an equilateral triangle is 7 cm, so the measure of its circumradius is _______ cm.

Solution: 14 cm.

∴ \(\frac{\sqrt{3}}{2} \times \frac{2}{3}=21 \times \frac{2}{3} \mathrm{~cm}=14 \mathrm{~cm}\)

Example 3. The circumference of a circular wheel is 250 dm. The no. of revolution to move 1 km is _______

Solution: 30.

No. of revolution \(\frac{10000}{250}=30\)

Example 4. A circular wheel revolves 80 times to 8088 m. Circumference is _________

Solution: 110 cm.

Circumference = \(\frac{88}{80}\) m = 1.1 m = 110 cm.

Example 5. The diameter of a circular wheel is 3.5 mt. The distance which is carved by 1400 revolution _______

Solution: 15.4 km.

Distance covered = 1400 circumference = 1400 x \(\frac{22}{7}\) x 3.5 mt = 15.4 km

Mensuration Chapter 2 Circumference Of Circle Short Answer Type Questions

Example 1. If the perimeter of a semicircle is 180 m. Find its diameter.

Solution: If the length of the radius is r m then πr + 2r = 180

or, \(r\left(\frac{22}{7}+2\right)=180\)

or, r = \(\frac{180 \times 7}{36} \times 35 \mathrm{~m}\)

or, 2r = 70m

∴ Its diameter = 70 m.

Example 2. The length of a minute’s hand is 7 cm. How much length will Minute’s hand go to rotate 90°?

Solution: Length = \(\frac{\text { angle of centre }}{360^{\circ}}\) x circumference

= \(\frac{90^{\circ}}{360^{\circ}}\) x 2π x 7 cm = 11 cm

∴ 11 cm much length will minute hand go to rotate.

Example 3. What is the ratio of radii of the inscribed and circumscribed circle of a square?

Solution: Let length of the side of the square be a unit

Length of the circumradius = \(\frac{1}{2}\) diagonal of the square

= \(\frac{\sqrt{2}}{2}\) a unit

∴ Length of the inradius = \(\frac{1}{2}\) length of the side = \(\frac{a}{2}\)

ratio = \(\frac{a \sqrt{ } 2}{2}: \frac{a}{2}\) = √2:1

∴ The ratio of radii of the inscribed and circumscribed circle of a square.

Example 4. The minute’s hand of a clock is 7 cm. How length does the minute’s hand move is 15 minutes?

Solution: circumference = 2 x \(\frac{22}{7}\) x 7 cm = 44 cm

Now minute’s hand covers 44 cm in 60 minutes.

∴ In 15 minutes it covers \(\frac{44}{60}\) x 15 = 11 cm.

Example 5. What is the ratio of perimeter of a square and perimeter of a circle when the length of diameter of circle is equal to the length of the side of the square.

Solution: d = a (d is the length of the diameter and a is the length of the side of the square)

∴ Ratio = 4a: πd

= 4а: πа (d = a)

= 4 : π = 4 x 7: 22 = 14: 11

∴ The ratio of perimeter of a square and perimeter of a circle is 14: 11.

Example 6. A wire of length 36 cm is made a semi-circle. Find its length of the radius.

Solution: R (π + 2) = 36

(Length of the radius is R cm)

or, \(\frac{36 R}{7}\) = 36

⇒ R = 7 cm.

∴ The length of the radius is 7 cm.

Example 7. Circumference of a wheel is 2 m 5 dem. If the speed is 15 km/hr find the revolution of the wheel per minute.

Solution: Circumference = 25 dcm.

In 60 minute wheel covers 15 km = 150000 dcm

In 1 minute wheel covers \(\frac{150,000}{60}\)dcm = 2500 dcm

∴ No. of revolution per minute = \(\frac{2500}{25}\) = 100 dcm

∴ The revolution of the wheel per minute is 100 dcm.

Example 8. Ratio of the circumference of two circles is 2: 3 and difference of the length of radii is 2 cm. Find the smaller radius.

Solution: Let length of the smaller radius r cm.

∴ Length of bigger radius = (r + 2) cm

∴ \(\frac{2 \pi r}{2 \pi(r+2)}=\frac{2}{3}\)

⇒ 3r = 2r + 14

∴ r = 4 cm

∴ The smaller radius is 4 cm.

Example 9. If the circumference is 2π² unit, then find its diameter.

Solution: 2πr = 2π² ⇒ 2r = 2π

∴ Length of the diameter = 2π unit.

Example 10. The circumference of a circles is 22 cm. Then find the length of the diagonal of a square inscribed in that circle.

Solution: Let length of the radius be r cm circumference 2πr = 22 cm

∴ 2 x \(\frac{22}{7}\) x r = 22

or, r = \(\frac{7}{2}\)

Length of the side of the square = Diameter of the circle = \(\frac{7}{2}\) x 2 cm = 7 cm

∴ Length of the diagonal of the square = √2 x 7 cm = 7√2 cm.

Mensuration Chapter 1 Area And Perimeter Of Triangle And Quadrilateral Useful Information And Formulae

Rectangle:

Let the length of a rectangle be a units and its breath be is units.

1. Area of the rectangle = lengths x breadth = ab sq units. ∴ length = \(\frac{\text { area }}{\text { breadth }}\); breadth = \(\frac{\text { area }}{\text { length }}\)
2. Perimeter of the rectangle 2 (length + breadth) = 2(a + b) units ∴ length breadth = semi perimeter.
3. Diagonal of the rectangle = \(\sqrt{(\text { length })^2+(\text { breadth })^2}=\sqrt{a^2+b^2} \text { units. }\)

Square:

Let each side of a square be a units

1. Area of the square = a² sq. units.
2. Perimeter of the square = 4a units ∴ side = \(\frac{\text { perimeter }}{4}\)
3. Diagonal of the square = a√2 units ∴ area of the square = \(\frac{1}{2}\) (diagonal)².

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Rhombus:

1. Area of the rhombus = \(\frac{1}{2}\) x product of the diagonals base x height.
2. Perimeter of the rhombus = 4 x side.

Trapezium and parallelogram:

1. Area of the trapezium = \(\frac{1}{2}\) x sum of the distance of the parallel side between them.
2. Area of the parallelogram = base × height.
3. Area of the four walls of a rectangular room = Perimeter of the floor x height of the room = 2 (lengths + breadth) x height.

Triangle:

Let the sides of a triangle be a units, b units and c units.

1. Area of the triangle = \(\frac{1}{2}\) x base x height.
2. Area of the triangle = \(\sqrt{\mathrm{S}(s-a)(s-b)(s-c)}\) sq units. where S semi-perimeter of the triangle = \(\frac{1}{2}\) (a+b+c) units.
3. Height of the equilateral triangle = \(\frac{\sqrt{3}}{2}\) x side
4. Median of the equilateral triangle = height of that triangle.
5. Circum radius of the equilateral triangle = \(\frac{2}{3}\) x median.
6. In radius of the equilateral triangle = \(\frac{1}{3}\) x median.
7. Area of the equilateral triangle = \(\frac{\sqrt{3}}{4}\) x (side)².
8. If the base of an isosceles triangle be a units and each of the two equal sides of b units, then height of that isosceles triangle \(\sqrt{b^2-\frac{a^2}{4}}\) units and area of that isosceles triangle = \(\frac{1}{2} \times a \times \sqrt{b^2-\frac{a^2}{4}}\) units.

Mensuration Chapter 1 Area And Perimeter Of Triangle And Quadrilateral Useful Information And Formulae Fill In The Blanks

Example 1. The length of the hypotenuse of an isoscels triangle is 3√2 cm, the area is _______

Solution: 4\(\frac{1}{2}\)

Example 2. When each side of an equilateral triangle is increased by 10%, its altitude increased by ________%.

Solution: 10%.

Example 3. When each side of an equilateral triangle is increased by 10%, its area increased by _________%.

Solution: 19%.

Example 4. The length of a median of an equilateral triangle of area 4√3 sq cm is ________ cm

Solution: 2√3

Example 5. In an isosceles right-angled triangle, the length of hypotenuse is √2 cm then the length of each equal sides is _______ cm.

Solution: 1.

Example 6. The length of two sides holding the right angle in a right-angled triangle are 3 cm and 4 cm. The length of the perpendicular drawn from the right angle to the hypotenuse is _______ cm.

Solution: 2 \(\frac{2}{5}\)cm.

Example 7. If each side of a square is increased by 10%, then the area is increased by ________ %.

Solution: 21%.

Example 8. The perimeter of a square is m metre. It is area is _______ sq m.

Solution: \(\frac{m^2}{16}\)

Example 9. The area of a square is 8k metre, the length of its diagonal is ________ m.

Solution: 4√k

Example 10. The perimeter of the square whose area is equal to sum of the areas of two square of sides 8 metre and 6 metre respectively is _______ metre.

Solution: 40.

Mensuration Chapter 1 Area And Perimeter Of Triangle And Quadrilateral Useful Information And Formulae True Or False

Example 1. The two diagonals of a rhombus are 24 m, 10 m. It is area is 120 sq cm.

Solution: The statement is True.

Example 2. If the three sides of an equilateral triangle be (x + 3) cm, (y + 4) cm, (x + y) cm, then the triangle with the sides (2xy + 1) cm, (6x + 1) cm and (8y+ 1) cm is also an equilateral triangle.

Solution: The statement is True.

Example 3. The length of a rectangle is 3 cm longer than its breadth. If its breadth be x cm, then the area is (x² + x) sq cm.

Solution: The statement is False.

Example 4. The length of a rectangle is 3 times that of its breadth. If the perimeter of the rectangle is 32 cm, its area is 4.8 sq cm.

Solution: The statement is False.

Example 5. The area of an equilateral triangle is 9√3 sq cm. The length of the median is 3√3 cm.

Solution: The statement is True.

Example 6. The centriod of an equilateral triangle ΔABC is G. If AB = 6 m, then AG = 2√3 cm.

Solution: The statement is True.

Example 7. The measure of each side of two equal sides of an isosceles triangle is 10 cm and measure of its base is 16 cm. The area is 16 sq cm.

Solution: The statement is False.

Example 8. A square and an equilateral triangle stand on same base. Area of the triangle = √3/2 x area of the square.

Solution: The statement is False.

Example 9. If number of metres of the perimeter of an equilateral triangle is equal to the number of sq metres of the area of the triangle. Measure of each side of the triangle is 4√3 metres.

Solution: The statement is True.

Example 10. D, E, and F are the midpoints of the sides BC, CA, and AB of a triangle ABC respectively. If the area of the triangle ABC = 24 sq em then ΔDEF = 6 sq cm.

Solution: The statement is True.

Mensuration Chapter 1 Area And Perimeter Of Triangle And Quadrilateral Useful Information And Formulae Short Answer Type Questions

Example 1. If the length of a square increased by 10%, then what % of the area will be increased?

Solution: Let length of the side be a unit.

∴ Area = a² sq units

increased side = \(\frac{11a}{10}\) unit

increased area = \(\frac{121 a^2}{1 a}\)

area % increased = \(=\frac{\frac{121 a^2}{1 a}-a^2}{a^2} \times 100=21\)

∴ 21% of the area will be increased.

Example 2. If the length is increased by 10% and breadth is decreased by 10% of a rectangle, then what % of area will be increased or decreased?

Solution: Let length and breadth be a and b units area = ab sq units.

Increased length = \(\frac{11a}{10}\) unit

decreased breadth = \(\frac{9a}{10}\)

New area = \(\frac{99 a^2}{100}\) sq units.

area % decreased = \(\frac{\frac{a^2-99 a^2}{100}}{a^2} \times 100=1\)

∴ 1% of area will be decreased.

Example 3. The length of rectangle is 5 cm. The length of the perpendicular on a breadth of the rectangle from an intersecting point between two diagonals is 2 cm. What is the length and breadth?

Solution: OP = 5 cm.

BC = 2 x 2 cm = 4 cm (from mid point theorem)

Let DC = x cm

∴ 4² + x² = 5²

x = 3

∴ The length and breadth 3 cm.

Example 4. The perimeter of a rectangle is 34 cm and area is 60 sq cm. Find the length of each diagonal.

Solution: If length and breadth are x cm, y cm then 2 (x + y) = 34

or, x + y = 17

xy = 60

(x + y)² = 17²

or, x² + y² = 17² – 2 x 60

= 289 – 120 = 169

∴ \(\sqrt{x^2+y^2}\) = 13

∴ Length of each diagonal = 13 cm.

Example 5. The numerical values of area and height of an equilateral triangle are cuqal. Find the length of each side.

Solution: \(\frac{\sqrt{3}}{4}\) x (side)²

= \(\frac{\sqrt{3}}{2}\) x (side)

∴ The length of each side = 2 units.

Example 6. The length of each side of an equilateral triangle is doubled. What % of area will be increased?

Solution: Let each side be a unit, area = \(\frac{\sqrt{3}}{4}\) a² sq unit

New area = \(\frac{\sqrt{3}}{4}\) (2a)² sq. unit = √3a²

% area increased = \(=\frac{\sqrt{3} a^2-\frac{\sqrt{3}}{4} a^2}{\frac{\sqrt{3}}{4} a^2} \times 100=\frac{3 \sqrt{3}}{4} 4^2 \times \frac{4}{\sqrt{3} q^2} \times 1 a=300\)

∴ 300% of area will be increase.

Example 7. The length of sides of a right-angled triangle are (x – 2) cm, x cm and (x + 2) cm. Find the hypotenuse?

Solution: (x + 2)² = x² + (x – 2)²

or, (x + 2)² – (x – 2)² = x²

or, 4.x.2 = x² or, x = 8

∴ Length of the hypotenuse 10 cm.

Example 8. A square drawn on height of equilateral triangle. Find ratio of areas of triangle and square.

Solution: Let length of each side of the equilateral triangle be x cm

∴ height = \(\frac{\sqrt{3}}{2}\)x cm

Required ratio = \(\frac{\sqrt{3}}{4} \dot{x}^2:\left(\frac{\sqrt{3}}{2} x\right)^2\)

= \(\frac{\sqrt{3}}{4} x^2: \frac{3 x^2}{4}=1: \sqrt{3}\)

∴ Ratio of areas of triangle and square = 1: √3.

Example 9. In parallelogram ABCD, AB = 4 cm, BC 6 cm, ∠ABC 30°, find the area.

Solution: We draw ⊥ from A to BC which intersects AC at P.

Extend AP to Q such that AP = PQ

From ΔAPB and ΔQPB,

AP = QP, ∠APB = ∠QPB (= 90°)

BP in common.

∴ ΔAPB = ΔQΡΒ

∴ ∠ABP = ∠QBP = 30°

∠ABQ = 30° + 30° = 60°

∠BAP = 180° – ∠APB – ∠ABP

= 180° – 90° – 30° = 60°

∴ ∠BQP = ∠BAP 60°

In ΔABQ, ∠ABQ = ∠BAQ = ∠BQA = 60°

∴ ΔABQ is equilateral of side equal to 4 cm.

AP = \(\frac{1}{2}\) = AQ = 2 cm  (AP = QP)

Required area = BC x AP = 6 x 2 sq cm = 12 sq cm.

Example 10. Length of height of a rhombus is 14 cm and length of side is 5 cm. Find the area.

Solution: Area = base x height

= 14 x 5 sq cm = 70 cm.

∴ Area of the rhombus is 70 cm.

Geometry Chapter 4 Theorems On Concurrence

⇔ Concurrent lines: If two or more different straight lines having a common point is said to be concurrent lines.

⇒ AB, CD, EF and GH are concurrent lines.

⇔ Circumcentre of a triangle: The point where the three perpendicular bisectors of sides of a triangle intersect is called the circumcentre of the triangle.

⇒ In ΔABC, the three perpendicular bisectors of AB, BC and CA meet at the point O; the point O is called the circumcentre and OA or OB or OC is the circumradius of ΔABC.

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⇒ [OA = OB = OC]

⇒ A circle is drawn passing through the points A, B and C is called circum circle.

⇔ Incentre of a triangle: The point where the three internal bisectors of angles of a triangle intersect is called the incentre of the triangle.

In ΔABC, the internal bisectors of angles ∠A, ∠B and ∠C intersect each other at the point O.

I drawn OD ⊥ BC;

A circle is drawn with centre O and length of equal radius of OD is said to be in circle of ΔABC.

The circle touch AC and AB at E and F respectively. Centre of the circle is called incentre.

⇔ Centroid of a triangle: The point where the three medians of a triangle intersect is said to be centroid of a triangle.

⇒ Three medians AD, BE and CF of a triangle ABC, intersect at the point G.

⇒ The point G is called the centroid of the ΔABC.

⇒ The centroid divides any median from the vertex in the ratio 2: 1.

⇔ Orthocentre of a triangle: The point where perpendiculars on the opposite sides from the three vertices of any triangle is called the orthocentre of the triangle.

⇒ In ΔABC, AD ⊥ BC, BE ⊥ CA and CF ⊥ AB;

⇒ AD, BE and CF meets at point O.

⇒ The point O is the orthocentre of ΔABC.

⇒ The triangle DEF obtained by joining the three points D, E, and F of ΔABC, is called a pedal triangle.

⇔ External centre: The point where the external bisectors of two angles and one internal bisector of an angle of a triangle intersect is called external centre.

⇒ In ΔABC, external bisectors of ∠ABC and ∠ACB and internal bisector of ∠BAC intersect at point O.

So O is the external centre.

OD, OE, OF are said to be external radius and the circle passes through the points D, E and F are called external circles of the triangle ABC.

Theorems:

1. The perpendicular bisectors of sides of triangle are concurrent.
2. The perpendiculars from vertices of triangle on the opposite sides are concurrent.
3. The internal bisectors of angles of triangle are concurrent.
4. The three medians of a triangle are concurrent.

Geometry Chapter 4 Theorems On Concurrence True Or False

Example 1. The sum of lengths of three medians of a triangle is greater than three-fourth of its perimeter.

Solution: In ΔABC, the medians AD, BE and CF intersects at G (centroid).

∴ \(\frac{A G}{G D}=\frac{2}{1}\)

[centroid of a triangle divides any median from the vertex in the ratio 2: 1]

⇒ \(\frac{G D}{A G}=\frac{1}{2}\)

⇒ \(\frac{G D}{A G}+1=\frac{1}{2}+1\)

⇒ \(\frac{G D+A G}{A G}=\frac{3}{2}\)

⇒ \(\text { i.e. } \frac{A D}{A G}=\frac{3}{2}\)

⇒ \(A G=\frac{2}{3} A D\)

Similarly, BG = \(\frac{2}{3}\)  BD and CG = \(\frac{2}{3}\) CF

In ΔABG, AG + BG > AB…….(1)

In ΔBCG, BG + CG > BC……..(2)

In, ΔACG, CG + AG > AC……….(3)

[The sum of lengths of two sides of a triangle is greater than the length of third side]

(1) + (2) + (3) we get

2(AG + BG + CG) > AB + BC + AC

2(\(\frac{2}{3}\) AD + \(\frac{2}{3}\) BE + \(\frac{2}{3}\) CF) > AB + BC + CA

⇒ \(\frac{4}{3}\)(AD + BE + CF) > AB + BC + CA

⇒ AD + BE + CF > \(\frac{3}{4}\) (AB + BC + CA)

So the statement is true.

Example 2. The orthocentre of a triangle is a point equidistance from its three sides.

Solution: In the adjoining figure, if O is the incentre of the triangle ABC, then OD = OE = OF [Inradius]

i.e. the point equidistance from three side of a triangle is incentre.

So the statement is false.

Example 3. In ΔABC, the internal bisectors ∠B and ∠C are meets at point O; if ∠BOC = 112°, then the value of ∠BAC is 44°.

Solution: In ΔABC, OB and OC are bisectors of ∠B and ∠C.

∴ ∠OBC = \(\frac{1}{2}\) ∠ABC and ∠OCB = \(\frac{1}{2}\) ∠ACB

∠OBC + ∠OCB = \(\frac{1}{2}\) (∠ABC + ∠ACB)

⇒ 180° – ∠BOC = \(\frac{1}{2}\) (180° – ∠BAC)

⇒ 180° – ∠BOC = 90° – \(\frac{1}{2}\) ∠BAC

⇒ 180° – 112° = 90° – \(\frac{1}{2}\) ∠BAC

⇒ 68° = 90° – \(\frac{1}{2}\) ∠BAC

⇒ \(\frac{1}{2}\) ∠BAC = 90° – 68° = 22°

⇒ BAC = 44°

∴ The statement is true.

Geometry Chapter 4 Theorems On Concurrence Fill In The Blanks

Example 1. The length of circumradius of a right-angled triangle is ________ of hypotenuse.

Solution: Half

[OA = OB = OC]

Example 2. The two medians of triangle are together than the third median.

Solution: Greater.

Example 3. If in ΔABC, three medians AD, BE and CF meets at point G then area of ΔABC area of ΔAGE is __________

Solution: ΔAGE = \(\frac{1}{6}\) ΔABC

⇒ \(\frac{\triangle \mathrm{ABC}}{\triangle \mathrm{AGE}}=\frac{6}{1}\)

⇒ ΔABC: ΔAGE = 6 : 1

Geometry Chapter 4 Theorems On Concurrence Short Answer Type Questions

Example 1. If the lengths of sides of triangle are 6 cm, 8 cm and 10 cm, then write where the circumcentre of this triangle lies.

Solution: 6²+ 8² = 100 = 10²

⇒ So the triangle is a right-angled triangle.

⇒ So the circumcentre of the triangle lies on the midpoint of hypotenuse i.e. lies on the midpoint of side with 10 cm in length.

Example 2. AD is the median and G is the centroid of an equilateral triangle. If the length of side 3√3 cm, then find the length of AG.

Solution: The medians and heights of any equilateral triangle are equal in length.

∴ The length of median AD.

= \(\frac{\sqrt{3}}{2}\) x length of side

= \(\left(\frac{\sqrt{3}}{2} \times 3 \sqrt{3}\right) \mathrm{cm}\)

= \(\frac{9}{2} \mathrm{~cm}=4.5 \mathrm{~cm}\)

centroid of a triangle divides any median from the vertex in the ratio 2: 1

∴ \(\frac{AB}{GD}\) = \(\frac{1}{2}\)

let AG = 2x cm

and GD x cm [x is common multiple and x > 0]

AG + GD = (2x + x) cm = 3x cm

3x = 4.5

⇒ x = 1.5

∴ AG = (2 x 1.5) cm = 3 cm.

Example 3. DEF is a pedal triangle of an equilateral triangle ABC. Find the value of ∠FDA.

Solution: In ΔABD and ΔACD,

∠ADB = ∠ADC = 90°  [AD ⊥ BC]

⇒ hypotenuse AB = hypotenuse AC  [ΔABC is an equilateral]

⇒ and AD = AD [common side]

∴ ΔABD ≅ ΔACD [by R-H-S criterion of congruency]

∴ BD = CD i.e. D is the mid point of BC

and ∠BAD = ∠CAD = \(\frac{60^{\circ}}{2}\) = 30°

Similarly, F and E are the mid points of AB and AC respectively.

∴ FE || BC and FE = \(\frac{1}{2}\) BC

⇒ Similarly, DE = \(\frac{1}{2}\) AB and FD = \(\frac{1}{2}\) AC

As AB = BC = CA

∴ DE = FE = FD

∴ ΔDEF is an equilateral triangle.

∴ ∠ZDFE = 60°

⇒ As FE || BC and AB is intersection

∴ ∠AFE corresponding ∠ABC = 60°

∴ ∠AFD = ∠AFE + ∠DEF

= 60° + 60° = 120°

In ΔAFD, ∠FDA + ∠AFD + ∠FAD = 180°

∠FDA + 120° + 30 ° = 180°

⇒ ∠FDA = 30°

Example 4. ABC is an isosceles triangle in which ∠ABC = ∠ACB and median AD = \(\frac{1}{2}\) BC. If AB = √2 cm, then find the length of the circumradius of this triangle.

Solution: In ΔABC, ∠ABC = ∠ACB

∴ AC = AB

⇒ In ΔABD and ΔACD,

⇒ AB = AC, AD = AD  [common side]

⇒ and BD = CD  [D is the mid point of BC]

∴ ΔABD ≅ ΔACD [By S-S-S criterion of congruency]

∴ ∠ADB = ∠ADC

∠ADB + ∠ADC = 180°

∴ ∠ADB + ∠ADB = 180°

⇒ 2 ∠ADB = 180°

⇒ ∠ADB = 90°

∴∠ADC = 90°

⇒ Again, AD = \(\frac{1}{2}\) BC = BD = CD

∴ BD or CD or AD is the circumradius of ΔABC.

⇒ In right-angled triangle ABD, ∠ADB = 90°

∴ AD²+ BD² = AB² [By Pythagorus theorem]

BD² + BD² = (√2)² cm²

⇒ 2BD² = 2 cm²

⇒ BD² = 1 cm²

⇒ BD = √1 cm = 1 cm

⇒ The length of circumradius of ΔABC is 1 cm.

Example 5. In ΔABC, two medians AD and BE are perpendicular each other at point G. If BC = 8 cm and AC = 6 cm, then find the length of AB.

Solution: In ΔABC, the medians AD and BE intersects at G.

⇒ So G is the centroid of the ΔABC.

⇒ ∠AGB = ∠AGE = ∠DGB = 90° [AD ⊥ BE]

⇒ AG: GD = 2 : 1 and BG: GE = 2:1

⇒ Let, AG = 2x cm and GD = x cm

⇒ BG= 2y cm and GE = y cm

⇒ BG = 2y cm and GE = y cm [x and y are common multiples and x > 0, y > 0]

∴ BD = \(\frac{1}{2}\) BC = (\(\frac{1}{2}\) x 8) cm = 4 cm and AE = \(\frac{1}{2}\) AC = (\(\frac{1}{2}\) x 6) cm = 3 cm

⇒ In ΔAGE, ∠AGE = 90°

∴ AG² + GE² = AE² [By Pythagorus theorem]

⇒ (2x)² + (y)² = (3)²

⇒ 4x² + y² = 9………(1)

⇒ In ΔBGD, ∠BGD = 90°

∴ BG² + GD² = BD²

⇒ (2y)² + x² = 4²

⇒ 4y² + x² = 16 …….(2)

⇒ (1) + (2), we get,

⇒ 4x² + y² + 4y² + x² = 9 + 16

⇒ 5x² + 5y² = 25  ⇒ x² + y² = 5

⇒ In ΔABG, ∠AGB = 90°

∴ AB² = AG² + BG²

= ((2x)² + (2y)²) cm²

= 4(x² + y²) cm²

= 4 x 5 cm² = 20 cm²

⇒ AB = √20 cm

= √4×5 cm = 2√5 cm

⇒ The length of AB is 2√5 cm.

Example 6. O is the circumcentre of triangle ABC. If ∠OBC = 30° then find the value of ∠BAC.

Solution: I join A, O and AO is extended at P.

⇒ In ΔBOC, OB = OC [circumradius]

∴ ∠OCB = ∠OBC = 30°

⇒ ∠BOC = 180° (30° + 30°) = 120°

⇒ In ΔAOB, OA = OB

∴ ∠OAB = ∠OAB

The exterior ∠BOP = ∠OAB + ∠OBA

= ∠OAB + ∠OAB = 2 ∠OAB

Similarly, ∠COP = 2 ∠OAC

∠BAC = ∠OAB + ∠OAC

= \(\frac{1}{2}\)(∠BOP + ∠COP)

= \(\frac{1}{2}\) ∠BOC = \(\frac{1}{2}\) x 120° = 60°

Example 7. If the point O is the orthocentre of ΔABC then find the relation between ∠BOC and ∠BAC.

Solution: In ΔABC,

⇒ AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.

⇒ AD, BE and CF intersects at point O.

∴ ∠AEB = ∠AFC = 90°

⇒ In quadrilateral AEOF,

⇒ ∠EAF + ∠AFO + ∠EOF + ∠AEO = 360°

⇒ ∠EAF + 90° + ∠EOF + 90° = 360°

⇒ ∠EAF + ∠BOC = 360° – 180°  [∠EOF = vertically opposite ∠BOC]

⇒ i.e. ∠BAC + ∠BOC = 180° [required relation]

Example 8. In ΔABC, D, E and F are midpoints of side, BC, CA and AB respectively. If AB = 5 cm, BC = 6 cm and CA = 7 cm, then find the perimeter of ΔDEF.

Solution: In ΔABC, F and E are midpoints of side AB and AC respectively,

∴ FE = \(\frac{1}{2}\) BC = (\(\frac{1}{2}\) x 6) cm = 3 cm

⇒ Similarly, DE = \(\frac{1}{2}\) = AB = (\(\frac{1}{2}\) x 5) cm = 2.5 cm

⇒ and FD = \(\frac{1}{2}\) AC = (\(\frac{1}{2}\) x 7) cm = 3.5 cm

∴ Perimeter of ΔDEF = (3 + 2.5+ 3.5) cm = 9 cm.

Geometry Chapter 3 Theorems On Area

Theorem:

1. The area of parallelograms which stand on same base and between same parallels are equal.
2. When a triangle and any parallelogram are on the same base and between the same parallels, the area of a triangle is half the area of the parallelogram.
3. The area of triangular regions being on the same base and between same parallels is equal.
4. Triangular regions of equal area standing on same base and being on the same side of it, they will be between the same parallels.

Geometry Chapter 3 Theorems On Area True Or False

Example 1. The heights of each parallelogram between the same parallel lines are same.

Solution: The perpendicular distance between two parallel lines are same.

⇒ The parallelogram ABCD and parallelogram PQRS are between the same parallels.

⇒ Therefore, the statement is True.

Example 2. The area of parallelogram ABCD is 32 sq. cm. O is any point on diagonal BD. If the area of ΔAOD is 5 sq. cm then the area of ΔAOP is 4 sq. cm.

Solution: I join A, C.

⇒ The diagonal AC intersects BD at point P;

⇒ So AC and BD bisects each other at point P.

ΔADC = \(\frac{1}{2}\) parallelogram ABCD

= (\(\frac{1}{2}\) x 32) sq. cm = 16 sq. cm

⇒ DP is a median of ΔADC

∴ ΔADP = \(\frac{1}{2}\) ΔADC

= (\(\frac{1}{2}\) x 16) sq. cm = 8 sq. cm

⇒ ΔAOD = 5 sq. cm

⇒ ΔΑΟΡ = ΔADP – ΔAOD

= (8 – 5) sq. cm = 3 sq. cm

⇒ So the statement is false.

Geometry Chapter 3 Theorems On Area Fill In The Blanks

Example 1. The _______ of a triangle divides the triangular region into two equal parts of triangular regions.

Solution: median.

⇒ [ΔABD = ΔACD]

Example 2. If a triangle and a parallelogram are on the same base and between the same parallels, the ratio of areas between a parallelogram and a triangle is _______

Solution: 2 1.

⇒ ΔABC and parallelogram BCDE are on same base BC and between the same parallels.

∴ ΔABC = parallelogram BCDE

⇒ or, \(\frac{\text { parallelogram } \mathrm{BCDE}}{\triangle \mathrm{ABC}}=\frac{2}{1}\)

as, parallelogram BCDE: ΔABC= 2:1

Example 3. ΔABC and rhombus BCDE are on the same base and between the same parallels BC and FD. If ΔABC = 18 sq. cm and BC = 8 em then height of rhombus BCDE is _______

Solution: 4.5 cm.

⇒ Area of rhombus BCDE = 2 ΔABC = (2 x 18) sq.cm = 36 sq. cm

⇒ BC = 8 cm

∴ Height = \(\frac{36}{8}\) 36 cm = 4.5 cm.

Geometry Chapter 3 Theorems On Area Short Answer Type Questions

Example 1. DE is perpendicular on side AB from point D of parallelogram ABCD and BF is perpendicular on side AD from the point B ; If AB = 10 cm, AD = 8 cm, and DE 6 cm; Find the length of BF.

Solution: Area of parallelogram = base x height DE = AD × BF

⇒ Area of parallelogram ABCD = AB x DE = AD x BF

⇒ 10 cm = 6 cm = 8 cm x BF

⇒ or, BF = \(\frac{60}{8}\) cm = \(\frac{15}{2}\) = 7.5 cm

∴ The length of BF is 7.5 cm.

Example 2. The area of the parallelogram-shaped region ABCD is 100 sq. units. P is the midpoint of side BC; Find the area of triangular region ABP.

Solution: I join A and C.

⇒ AC is a diagonal of parallelogram ABCD.

∴ ΔABC = \(\frac{1}{2}\) parallelogram ABCD

= (\(\frac{1}{2}\) x 100)sq. unit = 50 sq. unit

AP is a median of ΔABC

∴ΔABP = \(\frac{1}{2}\) ΔABC = (\(\frac{1}{2}\) x 50) sq. unit

= 25 sq. unit.

∴ The area of the triangular region of ΔABP is 25 sq. unit.

Example 3. AD is the median of triangle ABC and P is any point on side AC in such a way that area of ΔADP: area of ΔABD = 2: 3. Find the area of ΔPDC: area of ΔABC.

Solution: AD is a median of ΔABC

∴ ΔABD = ΔADC = \(\frac{1}{2}\) ABC

\(\frac{\triangle \mathrm{ADP}}{\triangle \mathrm{ABD}}=\frac{2}{3}\) [given]

∴ \(\frac{\Delta \mathrm{ADP}}{\Delta \mathrm{ADC}}=\frac{2}{3}\)

[ΔABD = ΔADC]

⇒ or, \(\triangle \mathrm{ADP}=\frac{2}{3} \triangle \mathrm{ADC}\)

⇒ as, \(\frac{\Delta \mathrm{PDC}}{\Delta \mathrm{ABC}}=\frac{\Delta \mathrm{ADC}-\Delta \mathrm{ADP}}{\Delta \mathrm{ABC}}\)

= \(\frac{\Delta \mathrm{ADC}-\frac{2}{3} \Delta \mathrm{ADC}}{\Delta \mathrm{ABC}}=\frac{\frac{1}{3} \Delta \mathrm{ADC}}{\Delta \mathrm{ABC}}\)

= \(\frac{\frac{1}{3} \times \frac{1}{2} \Delta \mathrm{ABC}}{\Delta \mathrm{ABC}}=\frac{1}{6}\)

∴ ΔPDC: ΔABC = 1: 6.

Example 4. BDE is a parallelogram. F is midpoint of side ED. If area of triangular region ABD is 20 sq. unit, then find the area of triangular region AEF.

Solution: AD is a diagonal of parallelogram ABDE

⇒ ΔADE = ΔABD = 20 sq. cm

⇒ AE is an median of ΔADE

∴ ΔAEF = \(\frac{1}{2}\) ΔADE

= (\(\frac{1}{2}\) x 20) sq cm = 10 sq. cm

∴ The area of triangular is 10 sq. cm.

Example 5. PQRS is a parallelogram. X and Y are the midpoints of side PQ and SR respectively. Join diagonal SQ, Find the area of the parallelogram shaped region XQRY: area of triangular region QSR.

Solution: I join Q, Y.

⇒ In quadrilateral XQRY,

⇒ QX || RY  [PQ || SR]

and QX = RY  [\(\frac{1}{2}\) PQ = \(\frac{1}{2}\) SR]

∴ XQRY is a parallelogram.

⇒ QY is a diagonal of parallelogram XQRY.

∴ ΔQRY = \(\frac{1}{2}\) parallelogram XQRY

⇒ or, Parallelogram XQRY = 2 ΔQRY……..(1)

⇒ QY is a median of ΔQSR.

∴ ΔQRY = \(\frac{1}{2}\) ΔQSR

⇒ or, ΔQSR = 2 ΔQSR

⇒ or, ΔQSR = 2 ΔQRY……(2)

⇒ From (1) and (2), parallelogram XQRY = ΔQSR

∴ Parallelogram XQRY: ΔQSR = 1: 1.

Example 6. If the area of parallelogram PQRS is 15 sq. cm, then find the area of parallelogram PMNO.

Solution: I join O, Q.

⇒ ΔPOQ and parallelogram PQRS are on same base and between the same parallels.

∴ ΔPOQ = \(\frac{1}{2}\) parallelogram PQRS

= (\(\frac{1}{2}\)x 15) sq. cm

= 7.5 sq. cm

⇒ Similarly, ΔPOQ = \(\frac{1}{2}\) = parallelogram PMNO

⇒ or, parallelogram PMNO = 2 ΔPOQ

= (2 x 7.5) sq. cm = 15 sq. cm

∴ The area of parallelogram PMNO is 15 sq. cm.

Example 7. In ΔABC D is the midpoint of side AB and DBCE is a parallelogram; If the area of ΔABC is 40 sq. cm, then find the area of parallelogram DBCE.

Solution: DE intersects AC at point F.

⇒ In ΔABC, D is the midpoint of AB and DE || BC;

∴ F is the midpoint of AC.

In ΔADF and ΔCEF,

∠AFD = ∠CFE [vertically opposite angles]

∠DAF = alternate ∠ECF

[AB || CE and AC is intersection]

∴ AF = CF

∴ ΔADF ≅ ΔCEF [By A-A-S criterion of B congruences]

∴ ΔADF = ΔCEF

Area of parallelogram DBCE

= Area of ΔCEF + area of quadrilateral ΔCFD

= Area of ΔADF+ area of quadrilateral BCFD = Area of ΔABC= 40 sq. cm.

∴ The area of parallelogram DBCE is 40 sq. cm.

Example 8. In trapezium, ABCD, AD || BC, and P and Q are midpoints of side AB and DC 18 em then find the ratio of area of quadrilateral APQD respectively. If AD = 12 cm and BC and area of quadrilateral PBCQ.

Solution: P and Q are the midpoints of side AB and DC respectively.

∴ PQ || AD || BC

and PQ = \(\frac{1}{2}\) (AD + BC) = \(\frac{1}{2}\) (12 + 18) cm = 12 cm

AS, AD || PQ  ∴ APQD is a trapezium

Similarly, PBCQ is a trapezium

As P and Q are midpoints of side AB and DC respectively

∴ Height of trapezium APQD and PBCQ are same.

Let, this height be h cm.

∴ Area of trapezium APQD = \(\frac{1}{2}\) (AD + PQ) x h

= \(\frac{1}{2}\) (12 + 15) x h sq. cm = \(\frac{27h}{2}\) sq. cm.

Area of trapezium PBCQ = \(\frac{1}{2}\) (PQ + BC) x h

= \(\frac{1}{2}\) (15 + 18) x h sq. cm. = \(\frac{33h}{2}\) sq. cm

∴ Ratio of areas between two trapezium is \(\frac{27h}{2}\): \(\frac{33h}{2}\)

= \(\frac{27h}{2}\): \(\frac{33h}{2}\) = 9: 11

∴ The ratio of area of quadrilateral APQD and area of quadrilateral PBCQ is 9: 11.

Example 9. In ΔABC, D and E are midpoints of side AB and BC respectively; F is the midpoint of side AD. If the area of ΔABC is 24 sq. cm then find the area of ΔCEF.

Solution: I join C, D. CD is a median of ΔABC

∴ ΔADC = \(\frac{1}{2}\) ΔABC

Again, CF is a median of ΔADC

∴ ΔACF = \(\frac{1}{2}\) ΔADC

= \(\frac{1}{2}\) x \(\frac{1}{2}\) ΔABC = \(\frac{1}{2}\) ΔABC

FE is a median of ΔFBC

∴ ΔCEF = \(\frac{1}{4}\) ΔFBC

= \(\frac{1}{2}\) (ΔABC – ΔACF)

= \(\frac{1}{2}\) (ΔABC – \(\frac{1}{4}\) ΔABC)

= \(\frac{1}{2}\) x \(\frac{3}{4}\) ΔABC

= (\(\frac{3}{8}\) x 24) sq. cm = 9 sq. cm

∴ The area of ACEF is 9 sq. cm.

Example 10. In parallelogram ABCD, the points P and Q on sides AB and DC such that BP = 2AP and DQ = 2CQ; If the area of parallelogram ABCD is 36 sq. cm then find the area of quadrilateral APCQ.

Solution: BP = 2AP

as, AB – AP = 2AP

or, AB = 3AP

or, AP = \(\frac{1}{3}\) AB

Similarly, CQ = \(\frac{1}{3}\) DC

Again AB = DC

∴ 3AP = 3CQ

or, AP = CQ

In quadrilateral APCQ, AP = CQ and AP || CQ

∴ APCQ is a parallelogram

As parallelogram APCQ and parallelogram ABCD are between the same parallels AB and DC

∴ Their heights are same. Let height is h cm

∴ Area of parallelogram ABCQ = AP x h

= \(\frac{1}{3}\) AB x h = \(\frac{1}{3}\) x area of parallelogram ABCD

= (\(\frac{1}{3}\) x 36) sq. cm = 12 sq. cm

∴ The area of quadrilateral APCQ is 12 sq. cm.

Geometry Chapter 2 Transversal And Mid Point Theorem

Theorem:

1. The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
2. In any triangle, through the midpoint of any side, if a line segment is drawn parallel to the second side, then it will bisect the third side and the length of the line segment intersected by the two sides of the triangle is equal to the length of the half of the second side.
3. If three or more parallel straight lines make equal intercepts from a transversal, then they will make equal intercepts from another transversal.

[Proof of the Theorem (3) is not for evaluation]

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Geometry Chapter 2 Transversal And Mid Point Theorem True Or False

Example 1. In ΔABC, D, and E are the midpoints of AB and AC respectively. If the length of DE is 5 cm, then the length of BC is 10 cm.

Solution: In ΔABC, D and E are the midpoints of AB and AC respectively

∴ DE = \(\frac{1}{2}\) BC

⇒ BC = 2 DE (2 × 5) cm = 10 cm

⇒ So the statement is true.

Example 2. In the equilateral triangle ABC, D, E, and F are midpoints of side AB, BC, and CA respectively. If the perimeter of ΔDEF is 8 cm.

Solution: AB = BC = CA = \(\frac{12}{3}\) cm = 4 cm

⇒ D and F are midpoints of AB and AC.

∴ DF = \(\frac{1}{2}\) BC = (\(\frac{1}{2}\) x 4) cm = 2 cm

⇒ Similarly, EF = DE = 2 cm

∴ The perimeter of ΔDEF is (2 × 3) cm = 6 cm

∴ The statement is false.

Geometry Chapter 2 Transversal And Mid Point Theorem Fill In The Blanks

Example 1. If three or more _______ straight lines make equal intercepts from a transversal, then they will make equal intercepts from another transversal.

Solution: Parallel.

⇒ [If AB || CD || EF and MN = NO then GH = HI]

Example 2. The triangle formed by joining midpoints of consecutive sides of an equilateral triangle is a _______ triangle.

Solution: Equilateral.

Geometry Chapter 2 Transversal And Mid Point Theorem Short Answer Type Questions

Example 1. In the triangle, ABC, AD, and BE are two medians and DF is parallel to BE, meets AC at the point F. If the length of the side AC is 8 cm. Find the length of the side CF.

Solution: In ΔBCE, D is the midpoint of BC and DF || BE;

∴ F is the midpoint of EC,

∴ CF = \(\frac{1}{2}\) EC = \(\frac{1}{2}\) x \(\frac{1}{2}\) AC  [Midpoint of AC is E]

= \(\frac{1}{4}\) AC = (\(\frac{1}{4}\) x 8) cm

∴ CF = 2 cm.

∴ The length of the side CF is 2 cm.

Example 2. In the triangle, ABC, the midpoints of BC, CA, and AB are P, Q, and R respectively, if AC = 21 cm, BC = 29 cm, and AB = 30 cm perimeter of the quadrilateral ΔRPQ.

Solution: In ΔABC, P and Q are the midpoints of BC and AC respectively.

∴PQ = \(\frac{1}{2}\)AB = (\(\frac{1}{2}\)x 30)cm = 15 cm

⇒ Similarly, RP = \(\frac{1}{2}\) AC = \(\frac{1}{2}\)(21) cm = 10.5 cm

⇒ Again AR = \(\frac{1}{2}\) AB = (\(\frac{1}{2}\) x 30) cm = 15 cm

⇒ AQ = \(\frac{1}{2}\)AC = (\(\frac{1}{2}\) x 21)cm = 10.5 cm

∴ Perimeter of quadrilateral ARPQ is AR+ RP+ PQ+ AQ

= (15+ 10.5 + 15 + 10.5)cm = 51 cm

Example 3. In the triangle, ABC, D is any point on the side AC. The midpoints of AB, BC, AD, and DC are P, Q, X, and Y respectively. If PX = 5 cm, then write the length of the side QY.

Solution: XY = DX + DY

= \(\frac{1}{2}\) AD + \(\frac{1}{2}\) DC

= \(\frac{1}{2}\)(AD + DC)

= \(\frac{1}{2}\) AC

In ΔABC, P and Q are the midpoints of AB and BC

∴ PQ = \(\frac{1}{2}\) AC and PQ || AC

∴ XY = PQ and PQ || XY  [PQ || AC]

∴ PQYX is a parallelogram

∴ QY= PX = 5 cm.

∴ The length of the side QY is 5 cm.

Example 4. In the triangle ABC, the medians BE and CF intersects at the point G. The midpoints of BG and CG are P and Q respectively. If PQ = 3 cm, then find the length of BC.

Solution: In ΔGBC, the midpoints of BG and CG are P and Q respectively.

∴ PQ = \(\frac{1}{2}\) BC

⇒ BC = 2 PQ = (2 x 3) cm = 6 cm.

∴ The length of BC is 6 cm.

Example 5. In the triangle ABC, the midpoints of BC, CA, and AB are D, E, and F respectively, EF intersects AD at the point O. If AD = 6 em, then write the length of AO.

Solution: In ΔABC, F, and E are the midpoints of AB and AC respectively.

= FE || BC

⇒ In ΔABD, F is the midpoint of AB and FO || BD [FE || BC]

∴ O is the midpoint of AD

∴ AO = \(\frac{1}{2}\) AD = (\(\frac{1}{2}\) x 6) cm = 3 cm

∴ The length of AO is 3 cm.

Example 6. In the parallelogram ABCD, BC is extended to point P such that BC CP. Join A, P; AP is intersect CD and BD at the points Q and R respectively. If DR = 3 cm then find the length of BR.

Solution: Through point C a line segment parallel to BD is drawn which intersects AP at the point S.

⇒ In parallelogram ABCD, AD = BC

⇒ Again, BC = CP ∴ AD = CP.

⇒ In ΔADQ and ΔCPQ,

⇒ ∠AQD = ∠CQP [vertically opposite angles]

⇒ ∠DAQ alternate ∠CPQ [AD || BP and AP is intersection]

and AD = CP

∴ ΔADQ ≅ ΔCPQ  ∴ DQ = CQ

⇒ In ΔDQR and ΔCQS

⇒ ∠DQR = ∠CQS [vertically opposite angles]

⇒ ∠RDQ = alternate ∠SCQ and DQ = CQ

∴ ∠DQR ≅ ΔCQS,  ∴ DR = CS

⇒ In ΔBPR, C is the midpoint of BP and CS || BR

∴ CS = \(\frac{1}{2}\) BR

⇒ or, BR = 2 CS = 2 DR (2 x 3) cm = 6 cm.

∴ The length of BR is 6 cm.

Example 7. In the triangle, ABC, D, and E are the midpoints of AB and BC respectively. BC is extended to the point F such EC CF; Join D, F. If AC = 8 cm, then find the length of CG.

Solution: In ΔABC, D, and E are the midpoints of AB and BC respectively.

∴ DE || AC and DE = \(\frac{1}{2}\) AC

= (\(\frac{1}{2}\)x 8) cm = 4 cm

⇒ In ΔDEF, C is the mid point of EF and GC || DE [DE || AC]

⇒ In CG = \(\frac{1}{2}\) DE = (\(\frac{1}{2}\) x 4) cm = 2 cm.

‍ ∴ The length of CG is 2 cm.

Example 8. The area of a square is x² sq. cm. Find the perimeter of the quadrilateral formed by joining midpoints of consecutive sides of that square.

Solution: Let, P, Q, R and S are the midpoints of the sides AB, BC, CD, and DA respectively of a square ABCD.

I join P, Q; Q, R; R, S; S, P; A, C and B, D.

⇒ Area of the square ABCD is x² sq. cm.

∴ The length of each side is √x² cm or x cm and length of the each diagonal is √2x cm.

∴ AB = BC = CD = DA = x cm and AC = BD = √2x cm

⇒ In ΔABD, P and S are the midpoints of side AB and AD respectively.

⇒ PS || BD and PS = \(\frac{1}{2}\) BD \(\frac{\sqrt{2} x}{2}\) cm

∴ In the quadrilateral PQRS, PS || QR and PS = QR

∴ PQRS is a parallelogram.

⇒ In ΔABC, P and Q are the midpoints of AB and BC respectively

∴ PQ = \(\frac{1}{2}\) AC = \(\frac{\sqrt{2} x}{2}\) cm

∴ PQ = QR = \(\frac{\sqrt{2} x}{2}\) cm

⇒ In ΔBPQ, BP = BQ and ∠PBQ = 90°

∴ ∠BPQ = ∠BQP = \(\frac{180^{\circ}-90^{\circ}}{2}=45^{\circ}\)

⇒ Similarly, ∠APS = 45°

∴ ∠SPQ = 180° – ∠APS – ∠BPQ

= 180° – 45° – 45° = 90°.

⇒ Similarly, ∠PQR = 90°

⇒ As, in parallelogram PQRS, PQ = QR and ∠PQR = 90°

∴ PQRS is a square whose length of each sides is \(\frac{\sqrt{2} x}{2}\) cm

∴ Perimeter = (4 x \(\frac{\sqrt{2} x}{2}\)) cm = 2√2x cm.

∴ The perimeter of the quadrilateral form by joining midpoints of consecutive sides of that square is 2√2x cm.

Example 9. In the ΔABC, D, E, and F are midpoints of sides AB, BC, and CA respectively. In the ΔDEF, P, Q, and R are the midpoints of the sides DE, EF, and FD respectively. If PR = 3 cm then the length of AB.

Solution: In the ΔDEF, P and Q are midpoints of DE and DF respectively.

∴ PR = \(\frac{1}{2}\) EF

⇒ In the ΔABC, E and F are midpoints of BC and AC

∴ EF = \(\frac{1}{2}\) AB

⇒ or, AB = 2 EF = 2 x 2PR (4 x 3) cm = 12 cm.

∴ The length of AB is 12 cm.

Example 10. In the ΔABC, AB = x cm, BC = y cm, and ∠ABC = 90°. If D and E are midpoints of AB and BC, then find the perimeter of ΔBDE.

Solution: BD = \(\frac{1}{2}\) AB = \(\frac{x}{2}\) and BE = \(\frac{1}{2}\) BC = \(\frac{y}{2}\) cm

⇒ In ΔABC, ∠ABC = 90°

∴ AB² + BC² = AC² [By Pythagorus theorem]

⇒ AC = \(\sqrt{A B^2+B C^2}\)

= \(\sqrt{x^2+y^2} \mathrm{~cm}\)

⇒ In ΔABC, D, and E are the midpoints of AB and BC

∴ DE = \(\frac{1}{2}\) AC = \(\frac{\sqrt{x^2+y^2}}{2}\) cm

⇒ Perimeter of ΔBDE = BD + BE + DE

= \(\left(\frac{x}{2}+\frac{y}{2}+\frac{\sqrt{x^2+y^2}}{2}\right)\) cm

The perimeter of ΔBDE = \(\left(\frac{x}{2}+\frac{y}{2}+\frac{\sqrt{x^2+y^2}}{2}\right)\) cm

Algebra Chapter 2 Graph

⇒ The region within angle XOY is called 1st quadrant.

⇒ The region within angle YOX’ is called the 2nd quadrant.

⇒ The region within angle X’OY’ is called the 3rd quadrant and the region YOX is called the 4th quadrant.

⇒ We get the distance from Y-axis is X coordinate and the distance from X-axis is the Y coordinate.

⇒ O is called the origin. The X coordinate is called the abscissa and the Y-co-ordinate is called the ordinate.

⇒ The co-ordinate of origin O is (0, 0).

⇒ Signs of abscissa and ordinate in the different quadrants is shown.

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Algebra Chapter 2 Graph Fill In The Blanks

Example 1. Co-ordinate of origin is ______

Solution: (0, 0)

Example 2. Equation of X axis is ______

Solution: y = 0

Example 3. Equation of Y axis is ______

Solution: y = 0

Example 4. X + 8 = 0 is parallel to _______ axis.

Solution: Y axis

Example 5. (0, 8) point lies on ______ axis.

Solution: Y axis

Example 6. (-6, 0) lies on ______ axis.

Solution: X axis

Example 7. (-3, +5) lies in the ________ quadrant.

Solution: 2nd

Example 8. Distance of A (6, 8) from Y axis is ________ units.

Solution: 6

Example 9. 2x + 3y = 12 intersects X axis at ________

Solution: (6, 0)

Example 10. 2x + 3y 12 intersects Y axis at _________

Solution: (0, 4)

Algebra Chapter 2 Graph True Or False

Example 1. The equation of the straight line parallel to X axis is x = c (c is a constant)

Solution: The statement is False.

Example 2. (-8, 5) lies in the 3rd quadrant.

Solution: The statement is False.

Example 3. (6, 0) lies in X axis.

Solution: The statement is True.

Example 4. \(\frac{X}{2}-\frac{Y}{2}\) = 1 intersects Y axis at (0, 2).

Solution: The statement is True.

Example 5. The distance of the point (-a, -b) from X axis is a units. (where a, b > 0).

Solution: The statement is True.

Example 6. The distance of the point (8, 6) from the origin is 10 units.

Solution: The statement is True.

Example 7. y = -x is passing through origin.

Solution: The statement is True.

Example 8. 2x + 3y = 5 passes through origin.

Solution: The statement is False.

Example 9. The angle between the lines y = constant and x = constant is 90°.

Solution: The statement is True.

Example 10. The line joining the points (a, b) and (-a, -b) passes through the origin.

Solution: The statement is True.

Algebra Chapter 2 Graph Short Answer Type Questions

Example 1. Let us write the coordinates the point of intersection of the graph of equation 2x + 3y = 12 and the X axis.

Solution: At X axis ordinate = 0

∴ 2x + 0 = 12 or, x = 6

⇒ The point of intersection is (6, 0).

Example 2. Let us write the coordinates of the point of intersection of the graph of the equation 2x – 3y = 12 and the Y axis.

Solution: At Y axis, abscissa = 0

∴ 0 – 3y = 12 or, y = -4

⇒ The point of intersection is (0, -4).

Example 3. Let us write the distance of the point (6,-8) from X axis and Y axis.

Solution: Distance from X axis= 8 units.

⇒ Distance from Y axis = 6 units.

Example 4. Let us write the angle derived from the equation XY from the positive direction of X axis.

Solution: y = x straight lines make 45° angle with the positive side of X axis.

Example 5. Answer the following

1. ΔAPC =?
2. ΔBPD =?

Solution:

1. AC (54) unit = 1 unit
   height ΔAPC 3 units
   ΔAPC = \(\frac{1}{2}\).1.3 sq units
   = 1\(\frac{1}{2}\) sq units.
2. BD = (6 – 5) unit = 1 unit
   height of ΔBPD = 2 units
   ΔBPD = \(\frac{1}{2}\) x 1 x 2 sq units = 1 sq unit

Example 6. 2x + 3y = 11 intersects X axis at (α, β) then find α, β.

Solution: At X axis y = 0

∴ 2x + 3 x 0 = 11

⇒ or, x = +\(\frac{11}{2}\)

∴ The intersection point is (\(\frac{11}{2}\),0)

∴ α = \(\frac{11}{2}\), β = 0

Example 7. Find the point of intersection of the straight lines y = x and 3x – 2y = 0

Solution: By putting y = -x in 3x – 2y = 0, 3x – 2(-x) = 0

⇒ or, 5x = 0

⇒ or, x = 0 and y = 0

∴ The point of intersection is (0, 0)

Example 8. Find the point of intersection of x = 2, y = 3.

Solution: The straight line x = 2 is parallel to Y axis and the straight

⇒ line y = 3 is parallel to X axis.

∴ The point of intersection is (2, 3).

Example 9. Find the area of the triangle formed by the coordinate axes and the straight lines 2x + 3y = 6.

Solution: At X axis, y = 0

∴ 2x + 0 = 6, x = 3

⇒ The straight line intersects X axis at (3, 0)

⇒ Similarly at Y axis, Y = 0

∴ 2.0 + 3y = 6, y = 2

⇒ The straight line intersects Y axis at (0, 2)

⇒ D = \(\frac{1}{2}\) x 3 x 2 sq. units = 3 sq. units.

Arithmetic Chapter 2 Profit And Loss

⇒ Cost price: The amount paid to purchase an article is known as its cost price.

⇒  Selling price: The price at which an article is sold is known as its selling price.

⇒  The cost price and selling price are abbreviated as C. P. and S. P. respectively.

⇒  Profit: If S.P. > C. P. then the difference between S. P. and C. P. is called profit.

⇒ Loss: If C. P. > S. P. then the difference between C. P. and S. P. is called loss.

∴ Profit = Selling price (S. P.) – Cost price (C. P.) and Loss Cost price (C. P.) – Selling price (S. P.)

⇒ Marked price: While buying goods we have seen that on every article there is a price marked. This price is known as the marked price of the article.

⇒ Discount = Marked price x Rate of discount

⇒ S. P. = Marked price – Discount

⇒ Profit percentage = \(=\frac{\text { Total profit }}{\text { Cost price }}\) x 100

⇒ Loss percentage = \(=\frac{\text { Total loss }}{\text { Cost price }}\) x 100

⇒ Equivalent discount: On a particular principal the equivalent discount is equal to more than one successive discount on that principal.

⇒ The discount equivalent to successive discounts of a% and b% is (a + b – \(\frac{ab}{100}\))%

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Arithmetic Chapter 2 Profit And Loss True Or False

Example 1. If selling price of an article is smaller than the cost price, then there will be profit.

Solution: The statement is False.

Example 2. If S.P. and C.P. of an article are equal then there will be no profit or loss.

Solution: The statement is True.

Example 3. Ram purchased a pen for ₹50 and sold it to Shyam at a loss of 20% then the selling price of that pen was ₹40.

Solution: C.P. = ₹50 and S.P. = ₹40

Loss = ₹(50 – 40) = ₹10

Loss percentage = \(\frac{10}{50}\) x 100 = 20

∴ The statement is True.

Arithmetic Chapter 2 Profit And Loss Fill In The Blanks

Example 1. Profit percentage x ________ = Total profit x 100.

Solution: cost price (C. P.).

Example 2. Cost price = \(=\frac{100 \times}{100-\text { Loss percentage }}\)

Solution: Selling price.

Example 3. There is a ________ relation between cost price and selling price

Solution: Direct.

Arithmetic Chapter 2 Profit And Loss Short Answer Type Questions

Example 1. If 20% profit is on cost price, what is profit percentage on selling price?

Solution: If cost price is ₹100 then profit is ₹20

∴ Selling price = ₹(100+ 20) = ₹120

⇒ If selling price is ₹120 then profit is ₹20

⇒ If selling price is ₹1, then profit is ₹\(\frac{20}{120}\)

⇒ If selling price is ₹100, then profit is ₹\(\frac{20}{120}\) x 100

= ₹\(\frac{50}{3}\) = ₹16\(\frac{2}{3}\)

∴ Profit is 16\(\frac{2}{3}\)% on selling price.

Example 2. If 20% profit is on selling price, what is the profit percentage on cost price?

Solution: If S.P. is ₹100 then profit is ₹20

∴ C.P. = ₹(100 – 20) = ₹80

⇒ If C.P. is ₹80, then profit is ₹20

⇒ If C.P. is ₹1, then profit is ₹\(\frac{20}{80}\)

⇒ If C.P. is ₹100, then profit is ₹(\(\frac{20}{80}\) x 100) = ₹25

∴ 25% profit on cost price.

Example 3. By selling 110 mangoes, if the cost price of 120 mangoes has been got, what will be the profit percentage?

Solution: Let S.P. of 110 mango is ₹x

⇒ S.P. of 1 mango is ₹\(\frac{x}{110}\)

⇒ According to condition, C.P. of 120 mangoes is ₹x [x > 0]

∴ C.P. of 1 mango is ₹\(\frac{x}{120}\)

⇒ Profit = \(₹\left(\frac{x}{110}-\frac{x}{120}\right)\)

= \(₹\left(\frac{12 x-11 x}{1320}\right)=₹ \frac{x}{1320}\)

⇒ If C.P. is \(₹ \frac{x}{120}\) then profit is \(₹ \frac{x}{1320}\)

⇒ If C.P. is 1, then profit is \(₹\left(\frac{x}{1320} \times \frac{120}{x}\right)\)

⇒ If C.P. is 100, then profit is \(₹\left(\frac{120 \times 100}{1320}\right)\) = \(\frac{100}{11}\) = 9\(\frac{1}{11}\)

∴ Profit is 9\(\frac{1}{11}\).

Example 4. To submit electricity bill in due time, 15% discount can be obtained. Sumanbabu has got 54 as discount for submission of electricity bill in due time. How much was his electricity bill?

Solution: Let, the electricity bill of Sumanbabu was ₹x

⇒ Discount = \(₹\left(x \times \frac{15}{100}\right)=₹ \frac{3 x}{20}\)

⇒ According to question, \(₹ \frac{3 x}{20}\) = 54

⇒ x = \(\frac{54 \times 20}{3}\)

⇒ x = 360

∴ The electricity bill was ₹360

Example 5. A commodity is sold at ₹480 with a loss of 20% on selling price, what is the cost price of the commodity?

Solution: If selling price of a commodity is ₹100 then loss is ₹20.

⇒ C.P. = ₹(100+20) = ₹120

⇒ If S.P. is ₹100, then C.P. is ₹120

⇒ If S.P. is ₹1 then C.P. is \(₹ \frac{120}{100}\)

⇒ If S.P is ₹480 then C.P. is ₹\(\frac{120 \times 480}{100}\) = ₹576

∴ Cost price of the commodity is ₹576.

Example 6. If a commodity is sold with successive discounts of 20% and 10%, what will be the equivalent discount?

Solution: Let the marked price of the commodity is ₹100

⇒ Then first discount is ₹20

⇒ The net price after 1st discount = ₹(100 – 20) = ₹80

⇒ Second discount = 10% of ₹80 = \(₹ \left(80 \times \frac{10}{100}\right)\) = ₹8

⇒ Total discount = ₹(20 + 8) = ₹28

∴ The equivalent discount is 28%.

Example 7. By selling a clock for 180, Rohit loses 10%, for what amount should be sell it as to gain 10%. [By proportion]

Solution: The loss is 10%

If C.P. of the clock is ₹100, then S.P. will be ₹(100 – 10) = ₹90

In mathematical language, the problem is,

⇒ S.P.(₹)
90
180

⇒ C.P.(₹)
100
?

⇒ The relation between S.P. and C.P. is direct.

∴ The direct proportion is, 90: 180 100: ? (The required C.P.)

∴ The required cost price = \(₹ \frac{180 \times 100}{90}=₹ 200\)

⇒ Rohit wants to make 10% profit

⇒ In mathematical language, the problem is

C.P(₹)
100
200

S.P. (₹)
100+ 10 = 110
?

⇒ The relation between S.P. and C.P. is direct

∴ The direct proportion is, 100: 200 : : 100:? (The required C.P.)

∴ The required selling price = \(₹ \frac{200 \times 110}{100}\) = ₹220

∴ To get 10% profit, Rohit has to sell the clock at ₹220

Example 8. Some toffees are bought at 15 for a rupee and the same number at 10 a rupee. Find the gain or loss percent.

Solution: C.P. of 15 toffees is ₹1

⇒ C.P. of 1 toffee is ₹\(\frac{1}{5}\)

⇒ S.P. of 10 toffees is ₹1

⇒ S.P. of 1 toffee is ₹\(\frac{1}{5}\)

⇒ Profit = \(₹\left(\frac{1}{10}-\frac{1}{15}\right)=₹ \frac{1}{30}\)

⇒ Profit % = \(\frac{\frac{1}{30}}{\frac{1}{15}} \times 100=\frac{15}{30} \times 100=50\)

∴ The percentage profit of toffees is 50.

Example 9. Akash sells a shirt at a loss of 20%. Had he sold the shirt for 200 more, he would have earned a profit of 5%. Determine the cost price of the shirt.

Solution: Let, the C.P. of the shirt is ₹x [x > 0]

⇒ Loss = 20% of ₹x = \(₹\left(x \times \frac{20}{100}\right)=₹ \frac{x}{5}\)

⇒ S.P = \(₹\left(x-\frac{x}{5}\right)=₹ \frac{4 x}{5}\)

⇒ Had Akash sold the shirt for ₹200 more, i.e. \(₹\left(\frac{4 x}{5}+200\right)\), he would have earned a profit of 5%

∴ x + x x \(\frac{5}{100}\) = \(\frac{4x}{5}\) + 200

⇒ x + \(\frac{x}{20}\) – \(\frac{4x}{5}\) = 200

⇒ \(\frac{20 x+x-16 x}{20}=200\)

⇒ 5x = 200 x 20

⇒ x = \(\frac{200 \times 20}{5}\)

⇒ x = 800

∴ Cost price of the shirt is ₹800

Example 10. A dishonest businessman defrauds by false balance, to the extent of 10% both in buying and in selling his goods. Find the actual gain percent of the businessman.

Solution: Since the businessman defrauds to the extent of 10% in buying.

⇒ So he takes goods of ₹110 in exchange of ₹100

⇒ Again, the businessman sells goods of ₹100 at ₹110

⇒ the businessman sells goods of ₹1 at ₹\(\frac{110}{100}\)

⇒ the businessman sells goods of ₹110 at \(₹ \frac{110 \times 110}{100}\) = ₹121

∴ The businessman finally gets ₹121 in exchange of ₹100

∴ The actual gain percent = (121 – 100) = 21.