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Geometry Chapter 6 Parallel Lines And Transversal

Question 1. What are parallel lines?
Solution:

Parallel lines: If two straight lines on the same Plane do not intersect when produced in any direction the two straight lines are said to be parallel to one another.

In the Below Figure the straight line, AB, CD, and EF are parallel to each other i.e., AB || CD || EF

Question 2. what are vertically opposite Angles?
Solution:

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Vertically Opposite Angles: If two straight lines intersect at a point, the angles formed on the opposite sides of the common point (vertex) are called vertically opposite angles. Here are two straight lines AB, CD interest at ‘O’. ∠AOC, ∠BOD are two vertically opposite angles.

Also LAOD, ∠Boc are two vertically opposite angles.

WBBSE Class 7 Maths Geometry Chapter 6

Question 3. what are corresponding angles?
Solution:

Corresponding Angles: Two angles lying on the same side of the transversal are known as corresponding angles if both lie either above are below the two given lines.

There are four pairs of corresponding angles.

(∠1, ∠5), (∠2,∠6), (∠8,∠4), and (∠7, ∠3).

Question 4. what is alternate Angles?
Solution:

Alternate Angles: The pair of interior angles on the opposite side of the transversal are called alternate angles.

There are two pairs of alternate angles (∠4,∠6), (∠3,∠5)

Class 7 Geometry Chapter 6 Exercise Solutions

Question 5. From the Figure write down the pairs of corresponding angles, alternate angles, and vertically opposite angles.

Corresponding Angles:- (∠1,∠5), (∠2, ∠6) (∠8,∠4), and (∠7, ∠3).

Alternate Angles: (∠4, ∠6) and (∠3, ∠5).

Vertically opposite Angles: (∠AGE, ∠OHF) (∠EGB, ∠CHF)

Question 6. In the adjacent Figure AB II CD Find the value of (∠BGH + ∠GHD)
Solution:

∴ The Sum of the measurement of two interior angles on the same side of the transversal is 180°

∴ ∠BGH + ∠GHD = 180°

Class 7 Maths Chapter 6 Geometry PDF

Question 7. In the adjacent Figure, ABIICD, IF ∠EGB= 40° then find the value of ∠AGH, ∠AGE, ∠BGH, ∠GHD, ∠GHC, ∠CHF, ∠DHF
Solution:

∠EGB = 40°

∠AGH = 40° Transversal the corresponding angles are equal.

∠AGE = 40° is an external angle to ∠EGB. exterior angle theorem ∠AGE = ∠EGB

∠BGH = 40° LBGH is the corresponding angle to ∠EGB.∠BGH = ∠EGB = 40°

∠GHD = 140° The corresponding angles on the same side ∠GHD+ ∠BGH = 180° ∠GHD = 180 – 40 = 140°

∠GHC = 40° LGHC corresponds to ∠AGH. ∠GHC = ∠AGH =40°

∠CHF = 140°

∠DHF = 140° ∠DHF Corresponds to ∠GHD.

∠DHF = ∠GHD = 140°

Geometry Chapter 4 Construction Of Triangles

Question 1. The length of the three sides of the triangle is given. Identify the cases where triangles can be Constructed or not Give reasons for the cases. 3.5cm, 4.5cm, 8cm

Solution:

Triangle inequality Theorem

This theorem states that the sum of the lengths Of any two sides of a triangle must be greater than the length of the third side.

Triangle having sides a, b, c

1. a+b> c
2. a+c>b
3. b+c>a

1. 3.5cm, 4.5cm, 8cm

⇒ a = 3.5cm,

b = 45cm,

C = 8cm.

b + c > a

4.5+8 >3.5

12.5>3.5

∴ with these sides of a triangle Can be constructed.

Class 7 Geometry Chapter 4 Solutions

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2. 3cm, 4cm, 5cm.

Let a = 3cm, (From Triangle inequality Theorem)

a+b > c

3+4 > 5

7>5

b = 4cm,

a+c>b

3+5 > 4

8>4

c=5cm

a+c > a

4+53

9>3

∴ with the sides of a, b, c lan constructed a triangle. is possible.

3. 4cm, 5cm, 10cm

Let a = 4cm, b=5cm, c = 10cm (from Triangle inequality Theorem)

a+b > c ⇒ 4+5>10 → False

a+c > b ⇒ 4+10>5 → True

b+c > a ⇒ 5+10>4 → True

∴ with this three side, we can construct a triangle in two ways.

4. 6cm, 7cm, 8cm.

Let a = 6cm, b= 7cm, c=8cm (From Triangle inequality Theorem)

a+b > c ⇒ 6 +7 > 8 → False

a+c > b ⇒ 6 7 8 > 7 → True

b+c > a ⇒ 7+876 → True.

∴ with these three sides, we can construct a triangle in two ways.

5. 6·4cm, 5.5cm, 7cm

Let a = 6·4cm, b = 5.5cm, c = 7cm (Triangle inequality Theorety.

a+b > c ⇒ 6·4 + 5.5 > 7 → True

a+c > b ⇒ 6·4+7 > 5.5 -> True

b+c>a ⇒ 5.5 +7 > 6·4 → True

∴ with the three given sides, we can draw the triangle.

Class 7 Maths Geometry Important Questions

Question 2. Draw a triangle ABC in which AB = 4.5cm, BC= 6cm, and CA = 7.5cm. Also, measure its three angles with a protractor and write their measure.

Solution:

Given:

AB = 4.5cm

BC= 6cm

CA = 7.5m

∠ABC = 90°

∠BCA = 35°

∠САВ = 55°

1. Draw a line segment BC of length 6cm with the help of Scale and pencil
2. Take the compass with a measurement of 4.5cm, and Draw an Arc From point B.
3. Next, take measurement of 7.5cm. Draw an Arc From point ‘C’
4. Mark the Joining points of the Arc and connect the Sides of a triangle with the point.
5. Measuring the angle of a triangle with a protractor and noting down

Question 3. Construct an equilateral triangle having each side =5.5cm. Also, measure its angles.

Solution:

1. Draw a Base of the length of 5.5cm with the help of a Scale and pencil
2. Next, take the compass with a pencil take the measurement of 5.5cm and make an Arc From both sides of the line segment of the triangle Base
3. Mark the Joining of Arcs
4. Connect the lines from the Arc to both sides of the base of a triangle.
5. All the sides of a triangle have a measurement of 5.5cm
6. Measuring that all angles of a triangle are equal ie 60°
7. Construction of the equilateral triangle is completed

Class 7 Maths Chapter 4 Geometry PDF

Question 4. Draw an isosceles triangle, the base of which is 4.8cm and the sum of the base angles of which is 105°
Solution:

sum of Base Angle = \(\frac{105}{2}\) = 52.5

1. Draw a baseline of 4.8cm with the help of Scale and pencil.
2. Given that the sum of base angles is 105° divided into 2 parts each angle has 52.5°
3. Take a protractor and make the two angles of the baseline Draw a line through the angles until they intersect at a point:
4. Now the Isosceles triangle with a base length of 4.8cm and a base angle of 52.5cm is formed.

Question 5. Draw a triangle PQR in which QR = 7cm, ∠PQR = 30°, and ∠PRG=60° measure the third angle.
Solution:

1. Draw a line segment QR with a measurement of 7cm.
2. Make the 30° Angle from the side and mark a point.
3. Next, make the 60° Angle from the R side and mark a point.
4. Now extend or draw a line with the particular angles made at a point where both lines will intersect each other and Note down that point as ‘p’
5. Now take the protractor and measure the angle third angle it is 90°

Question 6. Draw a right angle triangle ABC in which AB=3cm, BC=4CM and ∠ABC = 90° Measure the length of the hypotenuse
Solution:

1. Draw a line segment of length 4C and mark it as BC, with the help of scale and pencil
2. Now make a 90° angle from point ‘B’ Draw a line Connect the angle.
3. Next cut off the line with 3cm length and mark the point as ‘A’
4. Joining the ‘Ac’ to form a right-angle triangle
5. Now measure the length of ‘Ac’ and Note down the length of the hypotenuse is 5cm

WBBSE Class 7 Geometry Answers

Question 7. Draw an Isosceles right-angled triangle PQR in which the length of hypotenuse PR is 10cm
Solution:

1. Draw the hypotenuse PR of 10cm with the help of a scale and pencil.
2. Next, make a bisector of PR by taking more than half of its length and making or drawing an arc on both sides.
3. Connect the bisector. and take the compass and adjust to the midpoint PR where the bisector intersects and with the radius make the semi-circle as shown.
4. Make a point where the Bisector and semicircle meets Note it and Join the the lines from PQ, and RQ. Isosceles right-angled triangle is formed.

Question 8. Draw a right-angled triangle ABC such that ∠ACB=90°. AB=7.5cm and ∠ABC = 30°
Solution:

Given AB = 7.5cm

∠ACB 90°, ∠ABC = 30°

Let the third angle be ‘x’

A sum of angles in a Δle = 180°

90°+30°+2° = 180°

х° = 180°-120°

x = 60°

∴ ∠BAC = 60°

1. Draw AB with 7.5 cm
2. Take the compass with some radius and draw an arc from A and Bas shown
3. From the meeting point of the arc on AB draw an area on the first area which will make 600 Angles. Same as on the other side.
4. Next, Draw the two Arcs from the intersection of two arcs on the B side. and the intersection of the Arc on AB. which makes the intersection” of the Points make 30° Angle.

5. Now take the midpoint of AB as the radius and draw a semicircle.

6. Now extend or draw the lines from A and B with the given. angle. which meet at a semicircle which makes a 90° angle

∴ The right-angled triangle is formed.

WBBSE Class 7 Geometry Answers

Question 9. Draw an isosceles triangle ABC in which AB = AC=6cm and BC=4.8cm. Draw AD⊥BC and measure the length of A.D.
Solution:

1. Draw a line segment with a length of 6cm and name it AB with the help of scale and pencil.
2. Take a compass with a measurement of 6cm and draw an Arc From point ‘A’
3. Take the measurement of 4.8cm and draw an Arc From point ‘B’ and these two Arc intersect at a point mark the point as ‘C’
4. Now Joining the other sides of the triangle to point ‘C’ the triangle ABC is formed.
5. Draw a perpendicular to BC and Name it ‘D’ Next find the length of ‘AD’ with the help of scale.
6. The length of AD is 5.5cm.

Question 10. Draw an isosceles triangle ABC in which the length of base BC is 4.8cm and the length of perpendicular AD.
Solution:

1. Draw a baseline of length BC = 4.8cm with the help of Scale and pencil.
2. Take the with any radius and draw an Arc from point B and point as shown.
3. Next, take the compass and put the compass at the intersection of the baseline Arc which makes the angle of 60°. Do it on Both sides of the Intersection.
4. Next, extend the lines through the Intersection of the two arcs. The extended lines form a triangle at point ‘A’
5. Draw a perpendicular to BC and Name it ‘AD’

Arithmetic Chapter 4 Approximation Of Values

Question 1. Determine the approximate value of 5378 up to tens, hundreds, and thousands of places.

Solution:

5378

5-Thousand

3-Hundred

7-Tens

8-units

The number at the tens place is 7 and on its immediate right side is 8 which is between 5 to 9.

So up to tens place the approximate value is (537+1)ten

= 5380

The digit on the hundreds place is 3 and on its immediate right side is 7 So the approximate value upto hundreds place is 5300

The digit on the thousands place is 5 and on its. right side we have 3, so the approximate value. up to a thousand places is 6000

5378 ≈ 5380 (upto tens place).

5378 ≈ 5400 (upto Hundreds place)

5378 ≈ 5000 (upto Thousands place).

Read and Learn More Class 7 Maths Solutions

Class 7 Maths Arithmetic Chapter 4 Solutions

Question 2. Find the approximate value of 6.748564 up to whole number and 1, 2, 3, 4, and 5 decimal places

Solution:

6.748564 ≈ The number is up to 6th place is 4

6.748564 16.748571 [5th place after the decimal is 6 Hence 1 is added to 5th decimal place value 6, 6+1=7]

6.748564 ≈ 6.7486 [ 4th place after decimal is 6 Hence 1 is added to 3rd place value 5, 5+1=6

6.748564 ≈ 6.749[3rd place after decimal is 5 Hence 1 is added to 2nd place value 8, 8+1=9]

6.748564 ≈ 6.75 [2nd place after decimal is 4 Hence 1 is added to 1 decimal place value 44+1=5]

6.748564 ≈ 6.8

6.748564 ≈ 6.74856

≈ 6.7486

≈ 6.749

≈ 6.75

≈ 6.8

≈ 7

Question 3. Find the approximate value of \(5 \frac{7}{37}\) upto 2,3,4 and 5 decimal place
Solution:

⇒ \(5 \frac{7}{37} \Rightarrow \frac{192}{37}\)

⇒ \(5 \frac{7}{37}\) = 5.18 [TWO decimal place]

⇒ \(5 \frac{7}{37}\)= 5.189 [3 decimal place]

⇒ \(5 \frac{7}{37}\) = 5.1891 [4 decimal place]

⇒ \(5 \frac{7}{37}\) = 5.18918 [5 decimal place]

Class 7 Arithmetic Chapter 4 Questions

Question 4. Divide ₹37 among 5 boys and 3 girls equally. Find how much each would get. (approximated upto 2 places of decimal) Also, find the total money received by 5 boys and 3 girls and how much this total amount is more or less than ₹37

Solution:

The total number of boys and girls is (5+3) or 8

₹37 is divided among 8$ boys and girls

Each gets = ₹\(\frac{37}{8}\)

= 4.625 paise

The total money received by 5 boys is

= (4.625×5) paise

= 23.125 paise

The total money received by 3 girls is

= (4.625×3) Paise

= 13.875 Paise

Total money received by 5 boys and 3 girls is (23.125+13-87)

=₹37

This amount is equal to ₹37

WBBSE Class 7 Maths Solutions

Question 5. Simplify the following up to 3 decimal places. 21-3574 + 5.72 + 3.602

Solution:

21.3574+ 5.727272… +3.602602.

⇒ 30.687202.

Rounded the decimal upto

3 digits i.e. 30.687

⇒ \(\begin{array}{r}
21.3574 \\
+5.7272 \\
+3.602602 \\
\hline 30.687201 \\
\hline
\end{array}\)

Question 6. Find the difference between 47.286 and 28.6 upto 2 places of decimal

Solution:

Given:

First number = 47.286

Second number = 28.6

The difference = (First number) – (second number)

= 47.286 28.6

= 18.686

Rounded to two decimals places the difference is 18.69

Class 7 Arithmetic Textbook Solutions

Question 7. If 1 inch = 2.54 cm then find the value of 1cm into an inch upto 3 decimal places.

Solution:

Given:

1 inch = 2.54cm.

So to find how many inches are in 1 centimeter

we’ll divide by 2:54

⇒ \(\frac{1}{2.54} \approx 0.3937 .\)

The Rounded to three decimals places, 1 centimeter is approximately equal to 0.394 inches.

Question 8. Write the approximate value of the following numbers upto lacs thousands and hundreds of places.

Solution:

Question 9. Find the values of the following correct to 3 places of decimals.

1. \(7 \frac{2}{7}\)
2. \(3 \frac{8}{45}\)
3. 8.0645

Solution:

1. \(7 \frac{2}{7}\)

⇒ \(\frac{51}{7}\)

⇒ 7.285714….

Rounded to three decimals places it’s approximately 7.286

2. \(3 \frac{8}{45}\)

⇒ \(3 \frac{8}{45}\)

⇒ \(\frac{143}{45}\)

= 3.17778 ≈ 3.178

3. 8.0645

Rounded to three decimal places It’s approximately 8.064

Class 7 Maths Arithmetic Problems

Question 10. Simplify and Find the approximate value upto 1 decimal place.

⇒ \(7 \frac{2}{3} \times \frac{4}{5} \div \frac{3}{4} \text { of } \frac{2}{3}\)

Solution:

⇒ \(7 \frac{2}{3} \times \frac{4}{5} \div \frac{3}{4} \text { of } \frac{2}{3}\)

⇒ \(7 \frac{2}{3} \times \frac{4}{5} \div \frac{3}{4} \times \frac{2}{3}\)

⇒ \(\frac{23}{3} \times \frac{4}{5} \div \frac{3}{4} \times \frac{2}{3}\)

⇒ \(\frac{92}{15} \div \frac{6}{4}\)

⇒ \(\frac{\frac{92}{15}}{\frac{6}{4}}\)

⇒ \(\frac{92}{15} \times \frac{4}{6}=\frac{368}{90}=4.0\)

⇒ \(7 \frac{2}{3} \times \frac{4}{5} \div \frac{3}{4} \text { of } \frac{2}{3} \text { is } 4.0\)

Class 7 Maths Solutions For Arithmetic Chapter 5 Square Root Of Fractions

Question 1. Choose the correct answers.

The least positive integer which will multiply \(\frac{144}{605}\) to make it perfect square is

1. 3
2. 5
3. 7
4. 11

⇒ \(\frac{144}{605}=\frac{2 \times 2 \times 6 \times 6}{5 \times 11 \times 11}\)

⇒ \(\frac{2^2 \times 6^2}{11^2 \times 5}\)

So \(\frac{144}{605}\) is to be divided by the least positive integer

5 so as to make it a perfect square fraction.

So the correct answer is (2) 5

Class 7 Arithmetic Chapter 5 Exercise Solutions 

Question 2. Write true or False

1.5128 is a perfect decimal Square → False

2. The value of \(\sqrt{0.0256}+\sqrt{2.56}+\sqrt{0.000256}\) is 1.776.

⇒ \(\sqrt{0.0256}+\sqrt{2.56}+\sqrt{0.000256}\)

⇒ 0.16 + 1·6+0-016

⇒ 1.776 → True

3. The square root of \(3 \frac{109}{225} \text { is } 1 \frac{13}{15}\)

⇒ \(3 \frac{109}{225}\)

⇒ \(\frac{784}{225}\)

⇒ \(\sqrt{\frac{784}{225}}\)

⇒ \(\frac{28}{15}\)

⇒ \(1 \frac{13}{15}\) → True

Question 3. Fill In the blanks:

1. If \(\sqrt{\frac{x}{225}}=\sqrt{\frac{9}{25}}\) then the value of x is 81

⇒ \(\sqrt{\frac{x}{225}}=\sqrt{\frac{9}{25}}\)

⇒ \(\frac{\sqrt{x}}{\sqrt{225}}=\frac{\sqrt{9}}{\sqrt{25}}\)

⇒ \(\frac{\sqrt{x}}{15}=\frac{3}{5}\)

⇒ \(5 \sqrt{x}=3 \times 15\)

⇒ \(\sqrt{x}=\frac{45}{5}\)

⇒ \(\sqrt{x}=9\)

squaring on Both sides.

⇒ \((\sqrt{x})^2=(9)^2\)

x =81

2. If \(\sqrt{1024}\) = 32 then the value of \(\sqrt{10.24}+\sqrt{0.1024}+\sqrt{0.001024}\) is 3.552

⇒ \(\sqrt{10.24}+\sqrt{0.1024}+\sqrt{0.001024}\)

Given \(\sqrt{1024}=32\)

⇒ \(10.24=10.24 \times 1=(32 \times 0.32)=\left(\frac{32}{10}\right)^2\)

⇒\(\sqrt{10.24}=\sqrt{\left(\frac{32}{10}\right)^2}=\frac{32}{10}=3.2\)

⇒ \(0.1024=0.1024 \times 1=\left(\frac{32}{100}\right)^2=\left(\frac{32}{10^2}\right)^2\)

⇒ \(\sqrt{0.1024}=\sqrt{\left(\frac{32}{100}\right)^2}=\frac{32}{100}=0.32\)

⇒ \(0.001024=0.001024 \times 1=\left(\frac{32}{1000}\right)^2=\left(\frac{32}{10^3}\right)^2\)

⇒\(\sqrt{0.001024}=\sqrt{\left(\frac{32}{1000}\right)^2}=\frac{32}{1000}=0.032\)

⇒\(\sqrt{10.24}+\sqrt{0.1024}+\sqrt{0.001024}\)

⇒ 3.2 + 0.32 + 0.032

⇒ 3.552

Class 7 Arithmetic Chapter 5 Important Questions

Question 4. Find the value of \(\sqrt{\sqrt{\frac{1 \cdot 44}{0.25}}+\sqrt{\frac{0.64}{0.25}}}\)

Solution.

Given

⇒ \(\sqrt{\sqrt{\frac{1 \cdot 44}{0.25}}+\sqrt{\frac{0.64}{0.25}}}\)

⇒ \(\sqrt{\frac{\sqrt{1.44}}{\sqrt{0.25}}+\frac{\sqrt{0.64}}{\sqrt{0.25}}}\)

⇒ \(\sqrt{\frac{1.2}{0.5}+\frac{0.8}{0.5}}\)

⇒ \(\sqrt{2 \cdot 4+1 \cdot 6}\)

⇒ \(\sqrt{4}\)

= 2

⇒ \(\sqrt{\sqrt{\frac{1 \cdot 44}{0.25}}+\sqrt{\frac{0.64}{0.25}}}\) = 2

Question 5. Find the length of each side of a square whose area is equal to the sum of the two squares having the lengths of each side 1.2m and 0.5m respectively

Solution:

Given:

Length of First Square side = 1.2m

Length of second square side = 0.5m

The area of the First square

Area 1 = (1.2m)² = 1044 m²

The area of the second square

Arear = (0.5m)² = 0.25 m²

Total Area = Area 1 + Area2 = 1.44m² +0.25 m² = 1.69m²

Side length = \(\sqrt{\text { Total Area }}\)

⇒ \(\sqrt{1.69 \mathrm{~m}^2}\)

= 1.3m

∴ The length of each side of the new square is 1.3 meters.

Class 7 Maths Arithmetic Chapter 5 Notes

Question 6. If product of two positive number is \(\frac{12}{25}\) and their quotient is \(1 \frac{26}{49}\) then find the numbers.

Solution:

Let the two positive numbers be ‘x’ and ‘y’.

According to Question xxy = \(\frac{12}{25}\) and \(\frac{x}{y}\) = \(1 \frac{26}{49}\)

⇒ \(x \times y \times \frac{x}{y}=\frac{12}{25} \times \frac{75}{49}\)

⇒ \(x^2=\frac{900}{1225}\)

⇒ \(x=\sqrt{\frac{900}{1225}}\)

⇒ \(x=\frac{30}{35}=\frac{6}{7}\)

⇒ \(x=\frac{6}{7}\)

⇒ xxy = \(\frac{12}{25}\)

⇒ \(\frac{6}{7} x y=\frac{12}{25}\)

⇒ \(y=\frac{12}{25} \times \frac{7}{6}\)

⇒ \(y=\frac{14}{25}\)

The two numbers are $ and \(\frac{6}{7}\) and \(\frac{14}{25}\)

Question 7. Find the approximate value of the following Upto 3 decimal places:-

1. \(\sqrt{3}\)
2. \(\sqrt{5}\)
3. \(\sqrt{18}\)

1. \(\sqrt{3}\)

∴ \(\sqrt{3} \simeq 1.732\)

2. \(\sqrt{5}\)

∴ \(\sqrt{5}=2.236\)

3. \(\sqrt{18}\)

∴ \(\sqrt{18}=4.242\)

Class 7 Maths Chapter 5 Arithmetic PDF

Question 8. Find the square root of the following.

Solution:

1. \(4 \frac{220}{729}\)

⇒ \(\frac{3136}{729}\)

⇒ \(\sqrt{\frac{3136}{729}}\)

⇒ \(\frac{56}{27}\)

= 2.0740

∴ \(\sqrt{4 \frac{220}{729}}=\frac{56}{27}\)

2. 206.094736

⇒ \(\sqrt{206.094736}\) = 14.357

3. 10732.96

⇒ \(\sqrt{10732.96}\) = 103.6

4. \(3 \frac{1321}{2025}\)

⇒ \(\frac{7396}{2025}\)

⇒ \(\sqrt{\frac{7396}{2025}}\)

⇒ \(\frac{86}{45}\)

⇒ \(1 \frac{41}{45}\)

∴ \(\sqrt{3 \frac{1321}{2025}}=1 \frac{41}{45}\)

Question 9. What is the least number that must be subtracted From 0.001528?

Solution:

⇒ \(0.001528 \approx \sqrt{\frac{1528}{1000000}} \approx \frac{\sqrt{1528}}{1000} \approx \frac{39.08}{1000} \approx 0.03908\)

⇒ \((0.039)^2=0.039^2=0.001521\)

⇒ \((0.040)^2=0.040^2=0.001600\)\

The closest perfect square less than 0.001528 is 0.001521

Now subtract this value from 0·001528:

0.001528 – 0.001521 = 0·000007.

∴ The least number that must be subtracted from 0.001528 to make it a perfect Square number is 0.000007.

Class 7 Maths Chapter 5 Arithmetic PDF

Question 10. Find the value of \(\sqrt{1-(0.05)^2}\) upto 3 decimal places.

Solution:

⇒ \(\sqrt{1-(0.05)^2}\)

⇒ \(\sqrt{1-0.0025}\)

⇒ \(\sqrt{0.9975}\)

= 0·998749

Rounded upto 3 decimals = 0.998

Question 11. By which Fraction should \(\frac{35}{91}\) be multiplied so that the square root of the product is 3?

Solution:

Let the Fraction be \(\frac{a}{b}\)

⇒ \(\sqrt{\left(\frac{35}{91} \times \frac{a}{b}\right)}=3\)

⇒ \(\frac{35}{91} \times \frac{a}{b}=3^2\)

⇒ \(\frac{35}{91} \times \frac{a}{b}=9\)

⇒ \(\frac{35}{91}=\frac{5 \times 7}{7 \times 13}=\frac{5}{13}\)

Substitute \(\frac{5}{13} \text { for } \frac{35}{91}\)

⇒ \(\frac{a}{b}=9 \times \frac{13}{5}=\frac{117}{5}\)

⇒ \(\frac{35}{91} \times \frac{117}{5}=\frac{35 \times 117}{91 \times 5}=\frac{5 \times 7 \times 117}{7 \times 13 \times 5}=\frac{117}{13}=9\)

⇒ \(\sqrt{9}=3\)

Hence the Fraction \(\frac{117}{5}\) is correct.

Thus the Fraction by which \(\frac{35}{91}\) should be multiplied so that the square root of the product is 3 is

⇒ \(\frac{117}{5} \Rightarrow 23 \frac{2}{5}\)

Geometry Chapter 1 Revision Of Old Lesson

Question 1. Choose the correct answer

1. The acute angle is

1. 30°
2. 90°
3. 105°
4. 270°

The acute Angle lies between 0° to 90°

∴ The option (1) 30° is the correct answer

Class 7 Maths Geometry Chapter 1 Solutions

2. The surface of the pyramid is.

1. Rectangle
2. Square
3. Triangle
4. None of these

∴ The option (3) triangle is the correct answer.

Read and Learn More Class 7 Maths Solutions

3. The number of vertices of a cube is

1. 4
2. 6
3. 8
4. 12

∴ The option (3) 8 is the correct answer

Question 2. write true or false

1. Only one curved line can be drawn through two points. → False

2. Two straight lines can be drawn through two Fixed points → False

3. 120° is an obtuse angle → True

Class 7 Geometry Chapter 1 Questions

Question 3. Fill in the blanks

1. The greatest chord of a circle is known as the Diameter
2. 180° is called a straight angle
3. Many straight lines can be drawn through one fixed point.

Question 4. write the definition with a Figure of the Following

1. Adjacent angle

Adjacent angles: are two angles that share a common side and a common vertex, and do not overlap.

In the given diagram ∠ABD and ∠CBD are adjacent angles.

They share the common arm or common side BD and a common vertex B.

2. Vertically opposite angle

vertically opposite angles are angles that are opposite one another at a specific vertex and are created by two straight intersecting lines.

vertically opposite angles are equal to each other.

WBBSE Class 7 Maths Geometry Solutions

3. Reflex Angle:

A reflex Angle is an angle that is more than 180° and less than 360°

4. Straight Angle

A straight angle measures 180° and looks like a straight line

Mathematically

It is an angle whose arms lie in opposite directions from the vertex and they join together to form 180°.

5. Obtuse Angle

An angle that is larger than go but smaller than 180° is called an obtuse angle.

Question 5. Draw a line segment of length 10cm with a scale and divide it into four equal parts. Measure
Solution:

Line segment AB = 10cm

It is divided into ‘4’ equal parts.

Each part is divided into 2.5CM

AC = 2.5cm,

CD = 2.5 cm,

DE = 2.5 CM,

EB = 2.5 cm

∴ AB = AC+ CD +DE+EB

= 2·5+2·5+2·5+2·5

AB = 10cm

∴ The parts are equal AC = CD= DE = EB.

Class 7 Geometry Textbook Solutions 

Question 6. Draw an angle of 120° with the help of a protractor and divide it into four equal parts with the help of a compass.
Solution:

∠BOA = 120°

∠OAE = 30°

∠AOD = 60°

∠AOC = 90°

Question 7. Draw a triangle ABC and bisect each of the three angles and find if the angle bisectors are concurrent.
Solution:

Question 8. Draw a triangle ABC and bisect the sides BC, AC, and AB perpendicularly. If the three perpendicular bisectors are concurrent then name it’ Now with. center o and radius equal to the line segment AO draw a circle.
Solution:

Class 7 Maths Geometry Problems

Question 9. Draw a triangle PQR and from the three vertices P, Q, and R of the triangle, perpendicular PA, BQ, and RC are drawn on the sides QR, RP, and PQ respectively If the line segment PA, QB, and RC are concurrent.
Solution:

Question 10. A line segment AB of length 6cm is drawn. A point p is taken on the line segment AB. With the help of a scale and pencil compass draw a perpendicular PQ on AB at p.
Solution:

1. Draw a line segment AB of length 6cm
2. Draw a point ‘P’ on a line segment
3. Draw a perpendicular on ‘AB’.

Geometry Chapter 2 Drawing Angles With Compass

Question 1. Draw the following angles using a scale, pencil 1350, and compass.

1. 1.5°
2. \(22 \frac{1}{2}^{\circ}\)
3. 30°
4. 45°
5. 60°
6. \(7 \frac{1}{2}^{\circ}\)
7. 75°
8. 90°
9. 105°
10. 120°
11. 135°
12. 150°

Solution:

∠AOB = \(7 \frac{1}{2}^{\circ}\)

∠AOC = 15°

Read and Learn More Class 7 Maths Solutions

∠AOD = \(22 \frac{1}{2}^{\circ}\)

∠AOE = 30°

∠AOF = 45°

∠AOG = 60°

∠AOH = 75°

∠LAOI = 90°

∠AOJ = 105°

∠AOK = 120°

∠AOL = 135°

∠AOM = 150°

Class 7 Geometry Chapter 2 Questions

Question 2. Draw an angle ABC where ∠ABC = 45° Then CB is produced to any point D. Measure the LABD using a protractor. Bisects the ∠ABC and LABD wring compass and measures the angle between the bisectors of ∠ABC and ∠ABD.

Solution:

∠ABC= 45°

∠ABD = 15°

Angle between the bisectors of ∠ABC and ∠ABD = \(37\frac{1}{2}\)

WBBSE Class 7 Maths Geometry Solutions

Question 3. Draw an angle ∠PQR without a protractor, where ∠PQR = 120° Then divide ∠PQR into four equal parts.

Solution:

∠PQA = 30°

∠PQB = 60°

∠PQC = 90°

∠PQR = 120°

Geometry Chapter 3 Properties Of Triangle

Question 1. Choose the correct answer

1. If the measurement of the angle of a triangle is 100° then the triangle is

1. Right-angled triangle
2. Acute angled triangle
3. Obtuse angled triangle
4. None of these

∴ The option (3), an obtuse-angled triangle ≤ 90° to 2120° is an obtuse angle.

2. If the measurement of one angle is equal to the sum of measurements of other two angles then the triangle is

1. Acute angled
2. Right-angled
3. Obtuse angled
4. None of these

Read and Learn More Class 7 Maths Solutions

Class 7 Geometry Chapter 3 Questions

∴ The option (2) Right angled triangle is the correct answer

Greater than 90° to equal to 120° the angle is called obtuse Angle

Hypotenus = side+ Adjacent

If the measurement of two angles is x and y then the measurement of the third angle is

1. (X-Y)°
2. (x+y)°
3. 180-(2+1)°
4. {180 -(x-x)}

∴ The option (3) is the correct answer

A sum of Angles in a triangle is 180°

Here we have two angles given x° and y°

∴ x° + y° + third angle = 180°

third angle = 180° – (x°+ y°)

WBBSE Class 7 Maths Geometry Solutions

Question 2. write the true or false.

1. Obtuse triangle has two obtuse angle

→ False

∵ The obtuse angle lies between 90° to 120°

2. If the length of the three sides of a triangle 3 cm, 4cm, and 5cm then the triangle is a right-angled triangle

→ True

3. The measurement of the three angles of an isosceles right angled triangle is 45°, 45° and 90°

→ True

In an Isosceles right angled triangle two sides (or) two angles are equal then the triangle is Isosceles and having 90° is called an Isosceles right isosceles-angled triangle.

Class 7 Geometry Textbook Solutions

Question 3. Fill in the blanks

Question 1. The median of a triangle is ______
Answer: Concurrent.

Question 2. The point of intersection of perpendicular bisectors of a triangle is called ______
Answer: Orthocentre

Question 3. The height and median of a ______ are equal
Answer: Equilateral triangle

Question 4. If the ratio of measurement angles of a twangle is 3:4:5 then write the name of the triangle.
Solution:

The Sum of angles in a triangle is 180°

Angles of a triangle are in a ratio of 3:4:5

Let’s denote the angles are 3x,48,5x

3x+4x+5x=180°

12x = 180°

x = \(\frac{180}{12}\)

x = 15

∴ 3x = 3 ×15 ⇒ 45°
་
4x = 4 ×15 ⇒ 60°

5x = 5 ×15 ⇒ 75°

By observing that all three angles lie between 0° to 9°

∴ The triangled formed by the angles are Acute angle triangle

Class 7 Maths Geometry Problems

Question 5. The length of the base of a triangle is 12cm and its height is 10cm Find its area.
Solution:

Given:-

Base = 12cm

Height = 10cm

Area of triangle = \(\frac{1}{2}\) [Base x Height]

⇒ \(\frac{1}{2}\)[12×10]

⇒ \(\frac{1}{2}\)[120]

Are of triangle = 60 sqcm

Question 6. If the area of a triangle is 100sqm and length. of its base is 20cm find its altitude.
Solution:

Given:

Area of triangle = 100 sq cm

Length of base = 20cm

Altitude =?

Area of triangle = \(\frac{1}{2}\)[Base × Height]

100 = [\(\frac{20}{2}\) × Height ]

100 × \(\frac{2}{1}\) = [\(\frac{20}{2}\) × Height]

200 = 10 × Height

⇒ \(\frac{200}{20}\) = Height

200 = Height

Height Altitude = 20cm ⇒ 20cm

∴ The Altitude of a triangle is 20cm

Class 7 Geometry Formulas

Question 7. If the length of hypotenuse use and the length of one side are 20cm and 16cm respectively. Find the length of the third side.
Solution:

Given

Length of hypotenuse = 200m

Length of one side = 16cm

From Pythagoras Theorem

(Hypotenuse)² = (Side)² + (side)²

(20)² = (16)²+(Side)²

(20)² – (16)² = (Side)²

(20+16) (20-16) =(Side)²

36 × 4 = (Side)²

144 = (Side)²

side = √144

Side = 12

∴ The length of the third side is 12cm

Class 7 Geometry Chapter 3 Explanation

Question 8. If the length of three sides of a triangle is (m²-n²) unit, 2mn unit, and (m²+n²) unit. then Write the name of the triangle.
Solution:

Given:-

The first side of the triangle = (m²-n²) unit.

The second side of the triangle = 2mn unit

The third Side of the triangle = m²+n² unit

m²+n² is the largest side.

m²+n² will always be greater than the other two sides, given that m and n are positive integers.

Apply Pythagoras theorem

(m² + n²)² = (m²-n²)² + (2mn)² + (m²+n²)²

m⁴+2m²n²+n⁴ = m⁴ – 2m²n² + n⁴ + 4m²n²

m⁴+2m²n²+n⁴ = m⁴ +n⁴+2m²n²

(m²+n²)² = (m²+n²)²

∵ The sum of the squares of the two smaller sides equals the square of the largest side, of a triangle. satisfies the Pythagoreus theorem.

A triangle with sides (m²-n²) unit, 2mm unit, and (m²+n²) is a right angled triangle

Arithmetic Chapter 3 Proportion

Question 1. Find if the pair of ratios given below are equal and also Find if Four members are in proportion.
Solution:

1. 4:5 and 20:25

4:5 = 4:5

20:25 = 4:5

∴ 4:5 and 20:25 are equal

Hence 4,5, 20, 25 are in proportion.

2. 2ac: 3ac and 14 ab: 21ab. [a ≠ 0; b ≠ 0. c ≠ 0]

2ac:3ac ⇒ \(\frac{2 a c}{3 a c} \Rightarrow \frac{2}{3}\) ⇒ 2:3

14ab: 21ab ⇒ \(\frac{14 a b}{21 a b} \Rightarrow \frac{2}{3}\) ⇒ 2:3

∴ 2ac : 3ac and 14ab:21ab are equal.

Hence 2ac, 3ac and 14ab, 21ab are in proportion.

3. 5:7 and 15:18

5:7 = 5:7

15:18 = 5:6

∴ 5:7 = 5:6

Hence 5:7, 15 and 18 are not in proportion.

Class 7 Arithmetic Problems With Solutions

Question 2. Fill in the blank squares

1. 3: 7:: 21: ▢
2. ▢: 6:: 20: 24
3. 9: 8:: ▢: 56
4. 18: ▢:: 30: 45

1. 3: 7:: 21: ▢

Let the square be ‘X’

∴ \(\frac{3}{7}=\frac{21}{x}\)

3x = 21×7

3x = 147

x = \(\frac{147}{3}\)

x = 49

∴ 3:7::21:49

2. ▢: 6:: 20: 24

Let the square be ‘x’

∴ \(\frac{x}{6}=\frac{20}{24}\)

24x = 120

x = \(\frac{120}{24}\)

x = 5

∴ 5: 6:: 20: 24

3. 9: 8:: ▢: 56[/latex]

Let the square be ‘x’

∴\(\frac{9}{8}=\frac{x}{56}\)

8x = 9×56

x = \(\frac{504}{8}\)

x = 63

∴ 9:8::63:56

4. 18:▢:: 30: 45

Let the square be ‘x’.

⇒ \(\frac{18}{x}=\frac{30}{45}\)

30x = 18×45

x = \(\frac{810}{30}\)

x = 27

∴ 18:27:30:45

WBBSE Class 7 Arithmetic Chapter 3

Question 3. Verify whether the following numbers are in proportion.

1. 2, 3, 4, 6
2. 10,8, 5,4
3. 6,2,5,9
4. 16, 24, 6, 9

1. 2,3,4,6

Here product of extreme = 2×6 = 12 and product of mean = 3×4 =12

∴ Product of extreme = product of means

Hence the four members are in proportion.

2. 10, 8, 5,4

Here product of extreme = 10×4 = 40 and product of mean = 8×5 = 40·

∴ Product of extreme = product of mean

Hence the four members are in proportion.

3. 6,2,5,9

Here product of extreme = 6×9= 54 and product of mean = 2×5=10

∴ Product of extreme product of mean

Hence the four members are not in proportion

4. 16, 24,6,9

Here product of extreme = 16×9 = 144 product of mean = 24×6 = 144

∴ Product of extreme = product of mean

Hence the Four members are in proportion.

1. 2,3,4,6

⇒ 2:3 and 4:6

⇒ 2:3 = 2:3

4:6 = 2:3

2:3 and 4:6 are equal Hence 2, 3, 4, 6 are in proportion.

2. 10,8,5, 4

⇒ 10:8 and 5:4

⇒ 10:8 = 5:4

5:4 = 5:4

10:8 and 5:4 are equal.

Hence 10,8,5,4 are in proportion.

Alternating method

3. 6, 2,5,9

⇒ 6:2 and 5:9

⇒ 6:2 = 3:1

5:9 = 5:9

∴ 6:2 and 5:9 are not equal.

∴ 6, 2, 5, 9 are not in proportion.

4. 16, 24, 6, 9

⇒ 16:24 and 6:9

⇒ 16:24 = 2:3

6:9 = 2:3

16:24 and 6:9 are equal

Hence 16, 24, 6,9 are in proportion.

Class 7 Maths Arithmetic Solutions WBBSE

Question 4. Form different proportions with the following numbers.

1. 7,5, 14, 10,
2. \(\frac{1}{3}, \frac{1}{2}, \frac{1}{6}, \frac{1}{4}\)
3. 8,7,16,14

1. 7,5, 14, 10,

2. \(\frac{1}{3}, \frac{1}{2}, \frac{1}{6}, \frac{1}{4}\)

3. 8,7, 16, 14

WBBSE Class 7 Arithmetic Exercise Solutions

Question 5. Find whether the following sets of numbers are in continued proportion and write the proportionality:

1. 5,25, 125
2. 7, 21,63,
3. 12, 48, 192,
4. 6, 10, 18.

1. 5, 25, 125

5×125 = 625 = (25)²

i.e, 1st and × 3rd term = (mean)²

so, 5, 25 and 15 are in continued proportion

The proportionality is 5:25:25:125

2. 7, 21, 69

7×63 = 441=(21)²

i·e, 1st × 3rd term = (mean)²

so, 7, 21, 63 are in continued proportion.

The proportionality is 7:21:21:63

3. 12, 48, 192

12×192 = 2304 =(48)²

i.e; 1st and 3rd term = (mean)²

So 12, 48, 192 are in continued proportion.

The proportionality is 12:48:48:192

4. 6, 10, 18

6×18 = 108 ≠ (10)²

i.e, 1st term x 3rd term 7 (mean)²

so 6, 10, 18 are not in continued proportion.

WBBSE Maths Study Material Class 7 

Question 6. The three numbers in continued proportion 1st and 2nd numbers are 18 and 3 respectively. Find the third number of this proportion.
Solution:

Givent

The first number is 18

the second number is 3

18:3::3:X

⇒ \(\frac{18}{3}=\frac{3}{x}\)

18x = 9

x = \(\frac{9}{18}\)

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

∴ The third number in this proportion = \(\frac{1}{2}\)

Verify

18, 3, \(\frac{1}{2}\)

18 × 1/2 = 9 = (3)² = (mean)²

∴ The three numbers are in continued proportion.

The proportion will be 18:3:3:\(\frac{1}{2}\)

Question 7. Find the mean of the following two numbers.

1. 15,135
2. \(\frac{1}{2}, \frac{1}{8}\)

Let positive mean proportion of 15 and 135 is ‘X’

∴ 15:x :: 2:135

⇒ \(\frac{15}{x}=\frac{x}{135}\)

x² = 15×135

x= \(\sqrt{2025}\)

x= 45

∴ Mean proportion 15, 135, 45.

2. \(\frac{1}{2}, \frac{1}{8}\)

Let the positive mean proportion of \(\frac{1}{2} {and}\frac{1}{8}\) is ‘X’.

∴ \(\frac{1}{2}: x:: x: \frac{1}{8}\)

⇒ \(\frac{\frac{1}{2}}{x}=\frac{x}{\frac{1}{8}}\)

⇒ \(\frac{1}{2} \times \frac{1}{8}=x^2\)

⇒ \(\frac{1}{16}=x^2\)

x= \(\sqrt{\frac{1}{16}}\)

x= \(\frac{1}{4}\)

∴ Mean proportion \(\frac{1}{2}\) and \(\frac{1}{8}\) is \(\frac{1}{4}\)

WBBSE Maths Study Material Class 7 

Question 8. If \(x: \frac{27}{64}:: \frac{3}{4}: x\) (x≠0) then find the value of x.
Solution:

⇒ \(x: \frac{27}{64}:: \frac{3}{4}: x\)

⇒ \(\frac{x}{\left(\frac{27}{64}\right)}=\frac{\left(\frac{3}{4}\right)}{x}\)

Cross-multiply

⇒ \(x^2=\frac{3}{4} \times \frac{27}{64}\)

⇒ \(\frac{81}{256}\)

x = \(\sqrt{\frac{81}{256}}\)

x = \(\frac{9}{16}\)

∴ The proportion will be \(\frac{9}{16}: \frac{27}{64}:: \frac{3}{4}: \frac{9}{16}\)

Question 9. The ratio of the price of two books is 5:7. If the price of the Second book is 63, Find the price of 1st book.
Solution:

Given:

The ratio of the price of two books is 5:7

Let the price of the first book is ‘x’

Let the price of the second book be ‘y’

∴ х:y = 5:7

The price of the second book is (y) = 63

∴ \(\frac{x}{63}=\frac{5}{7}\)

7x = 63×5

7x = 315

x = \(\frac{315}{7}\)

x = 45

∴ The price of the first book is ₹45

Question 10. If 10 men can do a piece of work in 12 days. Find in how many days will 15 men finish the same work.
Solution:

In the mathematical language.

For a particular work, if the number of persons increases less time is required to finish the work.

⇒ \(\frac{10}{15} \times 12\)

⇒ \(\frac{2}{3} \times12\)

⇒ 8

∴ In 8 days 15 men finish the Same work.

Class 7 Arithmetic Problems With Solutions

Question 11. In a relief camp, there is a provision of food For 200 people for 60 days. After 15 days 50 people went away elsewhere. Find how many days the remaining food lasts for the remaining people in this camp.
Solution:

Let the required number of days be ‘x’

In mathematical language the problem is.

200-50 = 150 Remaining.

150: 200: 45:x

⇒ \(\frac{150}{200}=\frac{45}{x}\)

150x = 200×45

x = \(\frac{9000}{150}\)

x = 60

∴ The food will go for Bodays for 150 people.

Question 12. If 8 men can do a piece of work in 12 days. Find in how many days amen can do the same work?
Solution:

In the mathematical language, the problem is.

For a particular work, if the number of persons increases, less time is required to finish the work.

⇒ \(\frac{3}{9} \times 12\)

⇒ \(\frac{1}{3} \times 12\)

⇒ 1×4

⇒ 4

∴ In 4 days 9men can do the same work.

Class 7 Arithmetic Problems With Solutions

Question 13. The ratio of boys and girls in a school with 612 students is 5:4 which will be the new ratio of 12 new boys admitted.
Solution:

Total number of students = 612

The ratio of boys and girls is 5:4

5x+4x = 612

9x = 612

x = \(\frac{612}{9}\)

x = 68

∴ 5x = 5×68

5x = 340

4x = 4×68

4x = 272

with 12 new boys, the number of boys becomes.

340+12= 352

∴ The new ratio = \(\frac{352}{272}\)

The GCD OF 352 and 272 is 16.

So we divide both by 16.

⇒ \(\frac{352 \div 16}{272 \div 16}=\frac{22}{17}\)

∴ The new ratio of boys and girls is 22:17

Question 14. The Sum of the ages of A and B is 95 years. 5 years ago the ratio of their ages were 10:7. Find the ratio of their ages 5 years hence.
Solution:

Let’s denote the current ages of A and B as ‘A’ and ‘B’ respectively.

A+B = 95 ⇒ A = 95-B

Five years ago their ages were A-5, B-5

∴ \(\frac{A-5}{B-5}=10: 7\)

⇒ \(\frac{(95-B)-5}{B-5}=\frac{10}{7}\)

⇒ \(\frac{95-B-5}{B-5}=\frac{10}{7}\)

⇒ \(\frac{90-B}{B-5}=\frac{10}{7}\)

7×90-7×B = 10B-50

630-7B = 10B-50

630+50 = 10B+7B

680 = 17B

⇒ \(\frac{680}{17}\)

B = 40

5 years Hence the ages will be A+ 5 = 55+5=60,

B+5= 40 +5=45

The ratio of their ages:

⇒ \(\frac{A+5}{B+5}=\frac{60}{45}=\frac{4}{3}\)

The ratio of the ages 5year Hence will be 4:3

Class 7 Arithmetic Problems With Solutions

Question 15. There are 5 litres of Syrup in 15 litres of a Soft drink Find how much syrup will be required to make 21 litres of soft drink.
Solution:

Given:

Syrup Soft drink.

5     → 15

?    ← 21

⇒ \(\frac{21}{15} \times 5\)

⇒ 7 litres

∴ 7 litres of Syrup is required to make 21 litres of Soft drink.

Arithmetic Chapter 7 Areas Of Rectangles And Square

Question 1. Choose the correct answer:

1.  If the perimeter of a square is xom then its area is.

1. 4x² sq.m.
2. \(\frac{x^2}{16}\) sqcm
3. \(\frac{x^2}{4}\) sqm.
4. \(\frac{x^2}{2}\) sq.cm.

The perimeter of the square = 4 × length of each side.

x = 4 × length of each side

length of each side = \(\frac{x}{4}\)

Area of square = (length of side)²

⇒ \(\left(\frac{x}{4}\right)^2\)

Area of square = \(\frac{x^2}{16}\) sq. cm.

∴ The option (2) \(\frac{x^2}{16}\) 22 sq,cm is correct answer.

Class 7 Arithmetic Chapter 7 Questions

Read and Learn More Class 7 Maths Solutions

2. 36000 sq. mm =?

1. 3.65qcm.
2. 36 sq.cm.
3. 360 sq.cm.
4. 3600sq.cm.

1cm = 10mm

? = 36000

⇒ \(\frac{36000}{10} \times 1\) = 3600 sqcm

1.sq mm = 0.01sq centimeter

⇒ \(\frac{36000}{1} \times 0.01\)

= 3605qcm.

∴ The option (3) 360 sq cm is the correct answer

3. How many tiles each \(1 \frac{1}{3}\) meter square will cover a floor 20 metres square?

1. 200
2. 225
3. 15
4. None of these.

Area of the Flood = 20m x 20m = 400m²

Each tile has dimensions of \(\frac{4}{3} m \times \frac{4}{3} m \text {. }\)

Area of one tile = \(\frac{4}{3} m \times \frac{4}{3} m=\frac{4 \times 4}{3 \times 3}=\frac{16}{9} m^2\)

Number of tiles \(=\frac{\text { Area of the floor }}{\text { Area of one tile }}\)

⇒ \(\frac{400 m^2}{\frac{16}{9} m^2}\)

⇒ \(400 \times \frac{9}{16}\)

⇒ 400 × 0.5625

Number of tiles = 225

∴ The option (2) 225 is the correct answer

WBBSE Class 7 Maths Solutions

Question 2. write true or False:

1. 1Arc = 10 sq.m. → False

2. The length. of a rectangle is 4 times its breadth

If the breadth is 6.5 cm then its area is 169 sq. cm

length of a rectangle is l = 4xb

If breadth is 6.5

The area is 169 sq. cm

l = 4×6·5

l = 26cm

Area = l × b

=26×6.5

= 165 sq. cm.

→ True

3. If the perimeter of a square is (4a+16) com then the length of each side is (a +4) cm.

The perimeter of the square = 4 × length of each side.

4a+16 = 4(a+4)

4a+16

→ True.

Question 3. Fill in the blanks:

1. 1Sq·Hm = 10000 Sq.m

2. If the area of the square is (x²+4x+4) Cm then its perimeter is

Area of Square = (length of side)²

⇒ \(\sqrt{x^2+4 x+4}\) = length of side.

⇒ \(\sqrt{(x)^2+2 \cdot 2 x+(2)^2}\) = length of side.

⇒ \(\sqrt{(x+2)^2}\) = length of side.

x+2 = length of side.

perimeter of a square = 4x length of each side

= 4x(x+2)

perimeter of a square = 4x+8 cm.

3. The ratio of length and breadth of a rectangle is 3:2 and its area is 600 sq.m. The perimeter is _______m

l:b= 3:2 ⇒ l = 3x, b = 2x Area = 600sq.m

Area = 600 Sqm

lxb = 600 Sq.m.

3x × 2х =600

6x² = 600

x² = \(\frac{600}{6}\)

x² = 100

x = 10m

4. l = 32, b = 2x

l=3×10 , b = 2×10

l= 30m, b=20m

perimeter of a rectangle is 2(l+b)

⇒ 2(30+20)

⇒2(50)

⇒ 100m

Class 7 Arithmetic Textbook Solutions

Question 4. The area of a rectangle is (x²-64) sq m (x>8) and the length is (x+8)m then find its breadth.
Solution:

Given:

The area of the rectangle is x²-64 square meters.

The length of the rectangle is x+ 8 meters

The area of the Rectangle is

Area = Length x Breadth

Let ‘b’ be the breadth of the rectangle.

x²-64 = (x+8)×b

b= \(\frac{x^2-64}{x+8} \Rightarrow \frac{(x)^2-(8)^2}{x+8}\)

a²-b² = (a+b)(a-b)

b= \(\frac{(x+8)(x-8)}{x+8}\)

Since x>8, x+8=0 and we can cancel x+8 in the numerator and denominator

b=2-8

The breadth of the rectangle is x-8 meters.

Question 5. The cost of cultivation of a 45m long piece of land is ₹200. If the breadth of land were less the cost would have been ₹150 calculate the breadth of the land.
Solution:

Let’s denote the breadth of the piece of land as ‘b’ meters

The length of the land is 45 meters.

The area of the land with breadth b is 45 × b sq.m.

The cost of cultivation is ₹200

If the breadth is reduced by 9m

breadth b-9 meters

The new Area is 45×(b-9)

The cost of cultivation is ₹150.

Cost per square meter = \(\frac{200}{45 b}\)

Reduced Breadth

Cost per square meter = \(\frac{150}{45(b-9)}\)

⇒ \(\frac{200}{45 b}=\frac{150}{45(b-9)}\)

⇒ \(\frac{200}{b}=\frac{150}{b-q}\)

⇒ 200(b-9) = b(150)

⇒ 2006-1800 = 150b.

Subtract 150b from both sides.

50b -1800 = 0

506 =1800

b = \(\frac{1800}{50}\)

b = 36

The breadth of the piece of land is 36 meters.

Class 7 Maths Arithmetic Problems

Question 6. A square plot of land of side 26m. A path 2m wide runs all around it from outside. Find the area of the path.
Solution:

Given

The Square Plot land side is 26m.

1. Calculate the Area of the large square.

The path runs all around it and is 2 meters wide.

The side length of the larger square is

26m+ 2m+2m = 30m

The area of the larger square is:

Area_{large sequence} = 30mx30m = 900m²

2. Calculate the Area of the original square. Path

Area_{Origina sequence} = 26m × 26m = 676m²

3. Calculate the Area of the path.

Area_Path = Area_{large sequence} – Area_{Original sequence}

= 900m²-676m² = 224m²

∴ The area of the path around the square plot of land is 224 Sq.m

Question 7. The length of the rectangle floor of a room is thrice of its breadth. It cost ₹4761 to cover the whole floor with a carpet. If one sq.m of carpet costs ₹3 then find the perimeter of the room.
Solution:

Given:

The cost to cover the whole floor ₹4761

Cost per square meter = ₹3m2

Area of the Floor = \(=\frac{\text { Total cost }}{\text { cost per square meter }}\)

= \(\frac{4761}{3}\)

= 1587Sqm²

Let the breadth of the room be ‘b’ meters.

Since the length is thrice the breadth, the length ‘l’ is 3b meters.

Area = l×b

= 3b×b

= 3b²

The area of the Floor is 1587Sq.m²

3b² = 1587

b² = \(\)

= 529

Class 7 Maths Chapter 7 Notes

Question 8. \(b^2=\frac{1587}{3}=529\)

b = \(\sqrt{529}\) = 23m

b = 23m

l is three times the breadth:-

l = 3b

= 3×23

1=69m

The perimeter p of a rectangle is given by

P = 2 × (l+b)

P = 2 × (69+23)

= 2 × 92

= 184m

∴ The perimeter of the room is 184 meters.

Question 9. A room 24 meters long and 12 meters high, costs 48960 For whitewashing its walls at 700 per square meter. Find the breadth of the room.
Solution:

Let the breadth of the room be m

Length of the room = 24m, and 12m high

The area of the room is (24×x) sq. m ⇒ 24x sq.m

Cost per square meter = ₹60

Total cost = ₹48960

Total Area of walls = Total cost/cost per square meter

Total Area of walls = \(\frac{48960}{60}\)
་
=816 sq.m.

Two longer walls each have an area of 24m × 12m

Two Shorter walls each have an area of bm × 12m

Total Area of the four walls is = 2× (24×12)+2×(b×12)

= 2×288+2×12b

576+246

The total area of the walls whitewashed is 816 sqm

576 + 246 = 816

24b = 816-576

24b = 240 = \(\frac{240}{24}\)

b ⇒ b=10

∴ The breadth of the room is 10 meters.

Class 7 Arithmetic Formulas

Question 10. what biggest size of square stones may be used to pave the floor of a room 20m long and 15m wide? Also, find the number of stones needed.
Solution:

Given the length of the room = 20m

width of room = 15m

prime factorization.

20= 2×2×5

15=3×5

The GCD of 20, 15 is 5.

So, the largest square Stone that can be used has a side length of 5 meters.

Area of room = 20m x 15m = 300 Sqm.

Area of one square stone = 5m×5m= 25 sq.m

Number of Stones = \(=\frac{\text { Total Area }}{\text { Area of one Stone }}\)

= \(\frac{300}{25}\)

= 12

The biggest size of square stones that may be used to pave the Floor is 5 meters.

The number of such stones needed to cover the floor is 12.

Arithmetic Chapter 6 Time And Distance

Question 1. Choose the correct answer:-

1. The time taken to travel 48kmat 36 km/h is.

1. 1hr somin.
2. 1hr 20min
3. 1hr 40 min
4. None of these.

Solution:

Distance = 48Km

speed = 36 km/h

Time =?

Speed = \(\frac{\text { Distance }}{\text { Time }}\)

Time = \(\frac{\text { Distance }}{\text { speed }}\)

⇒ \(\frac{48}{36}\)

= 1.33

Time = 1-33hr ⇒ 1hr.30 min

∴ The option (1) Thy 30 min is the correct answer.

Class 7 Maths Arithmetic Chapter 6 Solutions

Read and Learn More Class 7 Maths Solutions

2. The distance covered in 4.5hɣmpat 8.4km/hr is.

1. 37.8km
2. 36 km
3. 36.8 km
4. 37Km.

Time (T) = 4.5hrs

Distance =?

Speed(s) = 8.4 km/hr.

Speed\(=\frac{\text { Distance covered }}{\text { Time taken. }}\)

Distance covered = Speed x Time taken.

= 8.4×4.5

Distance covered = 37.8 Km

∴ The option (1) 37.8 km is the correct answer.

3. If the bus covers a distance of 50 km in 3hrs 45 mins. then its speed is.

1. 45km/hr.
2. 42.5km/hr
3. 40km/hr
4. None of these.

Distance = 150 KM.

Time = 3hrs 45mins ⇒ 1hr = 60 min

= 45 ⇒ \(\frac{45}{60} \times 1\) = 0.75

Speed =?

Speed = \(\frac{\text { Distance covered }}{\text { time taken }}\)

⇒ \(\frac{150}{3.75} \mathrm{~km} / \mathrm{hr}\)

= 40Km/hr

∴ The option (3) 40km/hr is the correct answer.

Question 2. Write true or False:

1. Speed remaining fixed the distance traveled and time taken are inversely proportional.

→ False

Speed = \(\frac{\text { Distance }}{\text { time }}\)

Speed ∞ Distance

⇒ \(\text { speed } \infty \frac{1}{\text { time. }}\)

2. Speed = Distance travelled × Time taken.

→ False

Speed = \(\frac{\text { Distance traveled. }}{\text { Time taken }}\)

3. when two objects move in the same direction their relative speed will be different of their actual speed.

→ True

Class 7 Arithmetic Chapter 6 Questions

Question 3. Fill in the blanks.

1. If two objects move in opposite directions, then relative speed is equal to the sum of their speed

2. \(4 \frac{1}{2}\) km/hr = 1.25 m/sec

⇒ \(4 \frac{1}{2} \Rightarrow \frac{9}{2}\) = 4.5km/hy

1Km = 1000meters

1hr = 60min = 3600sec

1 minute = 60 sec

60minute = ? 3600sec

4KM → 4000 meters

⇒ \(\frac{1}{2}\) km → 500 meters.

Total distance = 4500 meters

1hr = 3600 sec

m/sec= \(\frac{4500}{3600}\)

= 1.25

3. The time taken to travel 60km at 15 km/hr is 4 hr.

Distance = 60Km

Speed = 15km/hr

Time =?

Speed \(=\frac{\text { Distance covered }}{\text { Time taken }}\)

Time = \(\frac{\text { Distance covered }}{\text { speed }}\)

⇒ \(\frac{60}{15}\)

Time = 4hr

The time taken to travel 60 km at 15km/hr is 4hr

Question 4. A train of length 150m, moving with a speed of 75km/hr passes a tree calculate how long will it take to do so.
Solution:

Given:

Length of train = 150m

Speed of train = 75km/hr

⇒ \(75 \times \frac{5}{18}\) = 20.83 m/sec.

Time \(=\frac{\text { Distance covered }}{\text { Time taken }}\)

⇒ \(\frac{150}{20.83}\)

Time = 7.2 seconds.

So it will take approximately 7.2 seconds for the train to pass the tree.

Question 5. A train with a length of 100m takes 20 sec to pass a light post calculate the speed of the train.
Solution:

Length of the train = 100m

Time (t) = 20sec.

speed= ?

Speed \(=\frac{\text { Distance covered }}{\text { time taken }}\)

⇒ \(\frac{100 \mathrm{~m}}{20 \mathrm{sec}}\)

Speed = 5 m/sec

⇒ \(5 \times \frac{18}{5}\)

Speed = 18km/hr

∴ The speed of the train is 18km/hr

Class 7 Arithmetic Textbook Solutions

Question 6. A 120m long train with a speed of 45km/hr passes a platform in 30 sec. calculate the length of the Pat platform.
Solution:

Length of train = 120m

speed of train = 45 km/hr ⇒ \(45 \times \frac{5}{18}\) = 12.5m/s

Time required to pass the platform = 30 sec.

Let ‘x’ be the length of the platform.

The total length to pass the platform is (x+120)m

Speed \(=\frac{\text { Distance covered. }}{\text { Time taken to cover Distance }}\)

12.5 = \(\frac{x+120}{30}\)

375 = X+120

x = 375-120

x = 255m

∴ The length of the platform is 255m

Question 7. A train takes 6sec to pass a lamp post and 30 sec to pass a 280m long platform. Find the length of the train and also its speed.
Solution:

Let L be the length of the train

S be the speed of the train

For passing a lamp post

L = S × 6

For passing the platform

L+280 = S×30

6S+280 = S×30

280 = 30S-6S

280 = 24S

S = \(\frac{280}{24}\)

S = 11.67 m/s

S = \(11.67 \times \frac{18}{5}\)

S = 42.01Km/h

L = 6×S ⇒ 6 × 11.67

L≈70m

∴ The Speed of the train is 11.67mls of 42 km/h

The length of the train is 70m

Class 7 Maths Arithmetic Problems

Question 8. Two trains 120m and 105m long respectively come running at the rate of 60 km/hr and 45km/hr. respectively. How long will they take to cross each other if the two trains are running in the same direction?
Solution:

After the two trains meet, they would pass each other ie, the two trains would simultaneously pass a distance equal to the sum of their own length

The distance two trains will cover (120+105) mo8 225m

The relative speed. in the same direction

If two objects are moving in the same direction then the

relative speed = difference of their speeds.

= 60-45

relative speed = 15 km/hr

The time to cross each other.

T \(=\frac{\text { Total length }}{\text { Relative speed }}\)

⇒ \(\frac{225}{15}\)

= 15 sec

The time to cross each other is 15 sec

Question 9. A train passes two platforms of length. 275m and 140m In 28sec and 16 sec respectively calculate the length and Speed of the train
Solution:

Let the length of the train be xm

when the train passes a platform of length 275m Then the train has to cover a distance of length (275+2) m

ly

To pass the second platform of length 140m

Then the train has to cover a distance of length (140+x) m

Mathematically represent

Speed remains constant, and time and distance are in direct proportion.

28:16:: (x+275): (X +140).

⇒ \(\frac{28}{16}=\frac{x+275}{x+140}\)

⇒ 28(x+140) = 16(2+275)

⇒  28(x+140) = 16(x+275)

⇒ 28x+ 28×140 = 16x+16×275

⇒ 28x+3920 = 16x +4400

⇒ 28x-16X = 4400-3920

⇒ 12x = 480

x = \(\frac{480}{12}\)

x = 40 meters.

∴ The length of the train is 40 meters

In 28 sec the train travels (40+275) =315m

In 1 Sec the train travels \(\frac{315}{28} \mathrm{~m}\)

In 1hr or 3600 sec the train travels = \(\frac{315 \times 3600}{28}\)

⇒ \(\frac{1134,000}{28}\)

⇒ 40,500meters

⇒ 40.5km/hr.

∴ The length of the train is 40 meters.

The Speed of the train is 40.5 km/hr.

Class 7 Arithmetic Formulas

Question 10. Two trains of equal length running in opposite directions, pass a man standing by a side of railway line in 18 secs and 12 secs respectively. At what time will the two trains cross each other?
Solution:

Let’s denote the length of each train by L meters.

Let the speed of the first train be V₁ meters per second

The speed of the second train be V₂ meters per second

First train

L = V₁ x 18 ⇒ V₁ = \(\frac{L}{18}\)

second train

L = V₂ x 12 ⇒ V₂ = \(\frac{L}{12}\)

Relative speed is the sum of their speeds.

The relative speed V_r is given by

V_r = V₁+V₂

V_r = \(\frac{L}{18}+\frac{4}{12}\)

⇒ \(\frac{2 L+3 L}{36}\)

⇒ \(\frac{5 L}{36}\)

The total length to be covered when the two trains cross each other is the sum of their lengths

Total length = L + L = 2L

The time taken to cross each other is the total length divided by the relative speed:

T \(=\frac{\text { Total length }}{\text { Relative speed }}\)

⇒ \(\frac{2 L}{\frac{5 L}{36}}\)

⇒ \(2 L \times \frac{36}{5 L}\)

⇒ \(\frac{24}{\frac{5 L}{36}}\)

⇒ \(2 L \times \frac{36}{5L}\)

⇒ \(\frac{36 \times 2}{5}\)

= 14.4 Sec

∴ The two trains cross each other at 14.4 Sec