WBBSE Class 9 Maths Mensuration Chapter 1 Area And Perimeter Of Triangle And Quadrilateral Useful Information And Formulae Multiple Choice Questions

WBBSE Class 9 Maths  Mensuration Chapter 1 Area And Perimeter Of Triangle And Quadrilateral Useful Information And Formulae Multiple Choice Questions

Example 1. The length of the diagonal of a square is 12√2 cm, and the area of the square is

1. 288 sq cm
2. 144 sq cm
3. 72 sq cm
4. 18 sq cm

Solution: \(\frac{12 \sqrt{2}}{\sqrt{2}}\) cm = 12 cm

Area = (12)² sq cm = 144 sq cm

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∴ The correct answer is 2. 144 sq cm

∴ The area of the square is 144 sq cm

Example 2. If the area of a square is A₁ sq units and the area of square drawn on the diagonal of that square is A₂ sq units then A₁:A₂ is

1. 1: 2
2. 2: 1
3. 1: 4
4. 4: 1

Solution: Let the length of the side of A₁ be a units

∴ length of the side of A₂ = √2a units

∴ Ratio a²: (√2a)² = 1:2

∴ The correct answer is 1. 1: 2

∴ A₁:A₂ is 1: 2

Example 3. If the rectangular place of which length and breadth are 6 mt & 4 mt is desired to prove it with 2 cm square tiles, then the no. of tiles will be

1. 120000
2. 240000
3. 60000
4. 180000

Solution: No. of tiles = \(\frac{600 \times 400}{2 \times 2}\) = 300 x 200 = 60000

∴ The correct answer is 3. 60000

Example 4. If a square and a rectangle having the same perimeter and areas are S and R, then

1. S = R
2. S > R
3. S < R

Solution: Let the length and breadth of the rectangle be x units and y units and length of the square be a units

∴ 2(x + y) = 4a

∴ x + y = 2a

Now, (x − y)² > 0

or,(x + y)² > 4xy

or, \(\left(\frac{x+y}{2}\right)^2\) > xy

or, a² > xy

∴ S > R

∴ The correct answer is 2. S > R

Example 5. If the length of the diagonal of a rectangle is 10 cm and the area is 62.5 sq cm, then the sum of their length and breadth is

1. 12 cm
2. 15 cm
3. 20 cm
4. 25 cm

Solution: If the length and breadth be x cm and y cm, then x² + y² = 100 and xy = 62.5

or, (x + y)² – 2xy = 100

or, (x + y)² = 100+ 2 x 62.5 = 100 + 125 = 225

∴ x + y = 15

∴ The correct answer is 2. 15 cm

∴  The sum of diagonal of a rectangle length and breadth is 15 cm

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Example 6. If each side of an equilateral triangle is 4 cm, the measure of height is

1. 4√3 cm
2. 16√3 cm
3. 8√3 cm
4. 2√3 cm

Solution: Height = \(\frac{\sqrt{3}}{2} \times 4^2 \mathrm{~cm}=2 \sqrt{2}\) cm

∴ The correct answer is 4. 2√3 cm

∴ Equilateral triangle height is 2√3 cm

Example 7. Isosceles right-angled triangle of which the length of each side of equal two sides of a units. The perimeter is

1. (1 + √2) a units
2. (2 + √2) a units
3. 3a units
4. (3 + 2√2 )a units

Solution: Hypotenuse AC = \(\sqrt{a^2+b^2} \text { units }=a \sqrt{2} \text { units }\)

 

 

 

 

 

 

 

perimeter = (a + a + a√2) units = (2 + √2) a units

∴ The correct answer is 2. (2 + √2) a units

∴ The perimeter of Isosceles right-angled triangle is  (2 + √2) a units

Example 8. If the area, perimeter, and height of an equilateral triangle are a, s, h then \(\frac{2a}{h}\) = 

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

Solution: \(\frac{2 a}{s h}=\frac{2 \times \frac{\sqrt{3}}{4} \times(\text { sidc })^2}{3 \times \text { side } \times \frac{\sqrt{3}}{2} \times \text { side }}=\frac{2 \sqrt{3}}{4} \times \frac{2}{3 \sqrt{3}}=\frac{1}{3}\)

∴ The correct answer is 3. \(\frac{1}{3}\)

Example 9. The length of each equal side of an isosceles triangle is 5 cm and length of base is 6 cm. The area is

1. 18 sq cm
2. 12 sq cm
3. 15 sq cm
4. 30 sq cm

Solution: area = \(\frac{1}{2} \times 6 \sqrt{5^2-\frac{6^2}{4}} sq cm\)

= 3\(\sqrt{25-9} \mathrm{sq} \mathrm{cm}=12 \mathrm{sq} \mathrm{cm}\)

∴ The correct answer is 2. 12 sq cm

The area is isosceles triangle is

Example 10. D is such a point on AC of triangle ABC so that AD: DC = 3:2 if the area of triangle ABC is 40 sq cm the area of ΔDBC is

1. 16 sq cm
2. 24 sq cm
3. 15 sq cm
4. 30 sq cm

Solution: \(\frac{A D}{D C}=\frac{3}{2}\)

\(\frac{\triangle \mathrm{ABC}}{\Delta \mathrm{BDC}}=\frac{\frac{1}{2} h \mathrm{AC}}{\frac{1}{2} h \mathrm{DC}}=\frac{5}{2}\)

 

 

 

 

 

 

 

 

or \(\frac{A D+D C}{D C}=\frac{3+2}{2}\)

or, \(\frac{40}{\triangle B D C}=\frac{5}{2}\)

or, \(\frac{\mathrm{AC}}{\mathrm{DC}}=\frac{5}{2}\)

or, \(\triangle \mathrm{BDC}=\frac{40 \times 2}{5}=16\)

∴ The correct answer is 1. 16 sq cm

∴ The area of ΔDBC is 16 sq cm

Example 11. The difference of length of each side of a triangle from its semi-perimeter are 8 cm, 7 cm and 5 cm respectively. The area is

1. 20√7 sq cm
2. 10√14 sq cm
3. 20√14 sq cm
4. 140 sq cm

Solution: s – a = 8

\(\begin{gathered}
s-b=7 \\
s-c=5 \\
\hline 3 s-s=20
\end{gathered}\)

 

\(\Delta=\sqrt{10(8)(7)(5)} \text { sq cm }\)

 

= \(\sqrt{400 \times 7} \mathrm{sq} \mathrm{cm}\)

= 20√7

∴ s = 10

∴ The correct answer is 1. 20√7 sq cm

∴ The area of a triangle is 20√7 sq cm

Example 12. The height of the parallelogram is 1/3 rd of its base. If the area is 192 sq cm, the height is

1. 4 cm
2. 8 cm
3. 16 cm
4. 24 cm

Solution: Base x height = 192

or, 3h × h = 192

or, h² = 64

h = 8

∴ The correct answer is 2. 8 cm

∴ The parallelogram height is 2. 8 cm

Example 13. If the length of one side of a rhombus is 6 cm, and one angle is 60° the area will be

1. 9√3 sq cm
2. 18√3 sq cm
3. 36√3 sq cm
4. 6√3 sq cm

Solution: ΔABC is an equilateral triangle

 

 

 

 

 

 

 

∴ Area = 2\(\frac{\sqrt{3}}{4}\) x 6² sq cm = 18√3 sq cm

∴ The correct answer is 2. 18√3 sq cm

∴ The area will be a rhombus 18√3 sq cm

Example 14. The length of one diagonal of rhombus is its 3 times, of another diagonal. If the area of field in the shape of rhombus is 96 sq cm. Then the length of long diagonal is

1. 8 cm
2. 12 cm
3. 16 cm
4. 24 cm

Solution: \(\frac{1}{2}\) x 3 x (shorter diagonal)² = 96

or, shorter diagonal = \(\sqrt{\frac{3296 \times 2}{3}}=\sqrt{64}\) = 8 cm

∴ length of the long diagonal = 24 cm

∴ The correct answer is 4. 24 cm

∴ The length of long diagonal of rhombus is 24 cm

Example 15. A rhombu and a square are on the same base. If the area of square is x² sq units and area of rhombus be y sq units then

1. y > x²
2. y < x²
3. y = x²

Solution: CP ⊥ EF

 

 

 

 

 

 

CF > CP

or, AB > CP (AB = CF)

∴ AB x AB > AB x CP

or, x² > y

∴ The correct answer is 2. y < x²

Example 16. Area of a field of a trapezium is 162 sq cm and height is 6 cm. If length of one parallel side is 23 cm, then the length of other parallel side is

1. 29 cm
2. 31 cm
3. 32 cm
4. 33 cm

Solution: \(\frac{1}{2}\)(23 + x)x 6 = 162

(Let length of other parallel side be x cm)

or, 23 + x = 54

∴ x = 31

∴ The correct answer is 2. 31 cm

∴ The length of other parallel side is 31 cm