Mensuration Model Questions Set 2 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

Construction : In quadrilateral ABCD, form A to C.

Now, in ΔABC

∵ AB = BC ...(Given)

∴ ∠BAC = ∠BCA

(angles opposite to equal side)

In ΔADC,

∵ CD > AD

∴ ∠DAC> ∠DCA

(since in a triangle, angle opposite to greater side is bigger than the angle opposite to smaller side)

On adding eqs. (i) and (ii), we get

∠BAD > ∠BCD

Answer: (d)

Let the length, breadth and height of the cuboid be x, 2x and 3x, respectively.

Therefore, volume = x × 2x × 3x = $6x^3$

New length, breadth and height = 2x, 6x and 9x, respectively.

New volume = $108x^3$

Thus, increase in volume = $(108 – 6)x^3 = 102x^3$

${\text"Increase in volume"}/{\text"Original volume"} = {102x^3}/{6x^3}$ = 17

Answer: (d)

Let O is centre of big circle, and O' is centre of smaller circle. Both are touch internally each other.

OA = 6 cm O'A = 4cm

Here PR is longest chord of big circle

P.S = ${PR}/2$

OS = AS – OA

= 8 – 6 = 2 cm.

In ΔPSO

$(PS)^2 + (OS)^2 = (OP)^2$

⇒ $(PS)^2 + (2)^2 = (6)^2$

PS = $√{36 - 4} = √{32} = 4 √2$

Now,

PS = ${PR}/2$

PR = 2 × PS

= 2 × $4 √2 = 8 √2$

Answer: (c)

Area of parallelogram = 6 × Area of ΔNPR

∴ NR × PL= 6 × $1/2$ × NR × PR

⇒ PL = 3PR (here, PL = PR + RL)

⇒ PR + RL= 3PR

⇒ RL= 2PL = 2 × 6 = 12 cm

Answer: (c)

Given, AC = BC

In ΔABC, ∠ABC = ∠CAB

∠ABC = ∠CAB = 38° (∴ AC = BC)

∠ACB = 180° – (∠ABC + ∠CAB)

= 180° – (38° + 38°) = 180° – 76° = 104°

In ΔACD, ∠ACD = 180° – 104° = 76°

and ∠ACD = ∠CAD = 76°

(∵ CD = AD)

∴ ∠ADC = 180° – (∠ACD + ∠CAD)

= 180° – (76° + 76°) = 28°