Mensuration Model Questions Set 2 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Since we are given that a and c are co-prime i.e. HCF of a and c is 1, therefore we can say that a definitely divides d exactly.

So, a is a factor of d.

Answer: (b)

We know that when two triangles are similar then ratio of their areas is equal to square of corresponding sides.

${\text"area of ΔABC"}/{\text"area of ΔXYZ"} = {AB^2}/{XY^2} ⇒ {32}/{60.5} = {AB^2}/{(7.7)^2}$

⇒ ${32 × 59.29}/{60.5} = AB^2 ⇒ 31.36 = AB^2$

∴ AB = $√{31.36}$ = 5.6 cm

Answer: (b)

Since, DADB is a right angled triangle at D.

∴ ∠DAB = 180° – (90° + 75°)

⇒ ∠DAB = 15°

Also, ABCD is cyclic quadrilateral.

∴ ∠CAB + ∠BDC = 180°

⇒ ∠BDC = 180° – (35° + 15°) = 130°

Answer: (c)

Side of square = $2/√π$

radius of circular disc = $2/√π × 1/2 = 1/√π$

area of disc = $π r^2 = π (1/√π)^2$ = 1 unit 2

Answer: (d)

The sum of opposite angles in cyclic quadrilateral is always 180°.

∴ ∠ACQ + ∠APQ = 180°

75° + ∠APQ = 180°

∴ ∠APQ = 180° – 75° = 105°

∠ACQ + ∠QCR = 180° (∵ Straight line)

75° + ∠QCR = 180°

∠QCR = 180° – 75° = 105°

∠CQR = 180° – 105° – 30° = 45°

Since, ∠APQ + ∠BPQ = 180° (Straight line)

∴ 105° + ∠BPQ = 180°

∠BPQ = 75°

In ΔBPQ ∠B + ∠P + ∠Q = 180°

∠B + 75° + 45° = 180°

⇒ ∠B = 60° ∴ ∠CBA = 60°

Answer: (b)

Let the height of canal = h.

Then, area of canal = $1/2$ × h(9 + 7)

or = 1280 = $1/2$ h(16)

∴ h = ${1280 × 2}/{16}$ = 160m