# Mensuration Model Questions Set 1 Practice Questions Answers Test With Solutions & More Shortcuts

#### MENSURATION PRACTICE TEST [2 - EXERCISES]

Question : 1

What is the area of an equilateral triangle having altitude equal to 2 \$√3\$ cm?

a) 3 \$√3\$ sq cm

b) \$√3\$ sq cm

c) 2 \$√3\$ sq cm

d) 4 \$√3\$ sq cm

Area of equilateral triangle = \$√3/4 a^2\$

⇒ \$1/2 × a × h = √3/4 a^2\$

⇒ h = \$√3/2 a^2\$

⇒ 2\$√3 = √3/2 a^2\$

⇒ \$a^2\$ = 2 × 2 = 4

a = 2 cm

Area of equilateral triangle

= \$√3/4 a^2\$

= \$√3/4 (a)^2 = √3/4 (2)^2\$

= \$√3 cm^2\$

Question : 2

The base of an isosceles triangle is 300 unit and each for its equal sides is 170 units. Then the area of the triangle is

a) 12000 square units

b) 9600 square units

c) 10000 square units

d) None of the above

Let ABC be an isosceles triangle.

Area = \$1/2\$ × AD × BC

= \$1/2 × √{(170)^2 - (150)^2} × 300\$

= \$1/2 × √{28900 - 22500} × 300\$

(∵ ΔADC is a right angled triangle then by pythagoras theorem, we find AD)

= 150 × \$√{6400}\$ = 150 × 80 = 12000 units.

∴ Option (a) is correct.

Question : 3

The hypotenuse of a right-angled triangle is 10 cm and its area is 24 cm2 . If the shorter side is halved and the longer side is doubled, the new hypotenuse becomes

a) \$√265\$ cm

b) \$√245\$ cm

c) \$√255\$ cm

d) \$√275\$ cm

Let shorter and longer side of right angle triangle are x and y cm respectively.

Then, \$x^2 + y^2 = (10)^2 ⇒ x^2 + y^2\$ = 100 ...(i)

and Area = \$1/2 xy = 24 ⇒ x = {48}/y\$

Plug. in ∼ x = \${48}/y\$ into equation (i), we get

\$({48}/y)^2 + y^2 = 100\$

\$(48)^2 + y^4 = 100 y^2\$

\$y^4 – 100y^2 + (48)^2\$ = 0

On solving, we get y = 6 or 8

∴ x = 6, y = 8

when, x becomes half and y becomes double then,

x' = 3, y' = 16

Hypotenuse 2 2 = \$√{(x)^2 + (y')^2}\$

= \$√{3^2 + (16)^2} = √{265}\$ cm

Question : 4

Statement I Let the side DE of a ΔDEF be divided at S. so that \${DS}/{DE} = 1/√2\$. If a line through S parallel to EF meets DF at T, then the area of ΔDEF is twice the area of the ΔDST.
Statement II The areas of the similar triangles are proportional to the squares on the corresponding sides. Which one of the following is correct in respect of the above statements?

a) Statement I is true but Statement II is false

b) Both Statements I and II are true and Statement II is the correct explanation of Statement I

c) Both Statements I and II are true but Statement II is not the correct explanation of Statement I

d) Statement II is true but statement I is false

Given that, \${DS}/{DE} = 1/√2\$

DE = \$√2\$ × DS

When two triangles ΔDTS and ΔDEF are similar then their ratio of area is equal to square of corresponding sides.

⇒ \${ΔDST}/{ΔDEF} (1/√2)^2 = 1/2 ⇒ ΔDEF = 2ΔDST\$

So, both Statements I and II are true and Statement II is the correct explanation of Statement I.

Question : 5

A right circular cylinder just encloses a sphere. If p is the surface area of the sphere and q is the curved surface area of the cylinder, then which one of the following is correct?

a) 2p = q

b) P = q

c) p = 2q

d) 2p = 3

p = 4π\$r^2\$

q = 2πr.h = 2πr. 2r

= 4π \$r^2\$

Hence, P = q.

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