mensuration area volumes Model Questions & Answers, Practice Test for ssc chsl tier 1 2023

Answer: (a)

If a parallelogram and a rectangle stand on the same base and on the same side of the base with the same height, then perimeter of parallelogram is greater than perimeter of rectangle.

∴ $I_1 > I_2$

Answer: (b)

Let the rise in water level = x m

Now, volume of pool = 40 × 90 × x = 3600 x $cm^3$

When 150 men take a dip, then displacement of water = $8m^3$

∴ ${3600 x}/{150} = 8⇒{900}/{150} x = 2⇒$ x = 33.33 cm

x = 33.33 cm

Answer: (b)

A quadrilateral ABCD, AP and BP are bisectors of ∠A and ∠B, respectively.

∴ ∠APB = 180° - $(1/2 ∠A + 1/2 ∠B)$

We know that sum of all angles of a quadrilateral = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

∴ $1/2 ∠A + 1/2 ∠B + 1/2 ∠C + 1/2 ∠D = {360°}/2$

⇒ $1/2∠C + 1/2 ∠D = 180° - (1/2 ∠A + 1/2B)$

⇒ $1/2$(∠C + ∠D) = ∠APB [from eq. (i)]

⇒ ∠C + ∠D = 2∠APB

Answer: (a)

Calculating area of triangle using heron's formula, we get

Semi perimeter,

s = ${a + b + c}/2 = {51 + 37 + 20}/2 = {108}/2$ = 54

Area of triangle = $√{s(s - a)(s - b)(s - c)}$

= $√{54(54 - 51)(54 - 37)(54 - 20)}$

= $√{54 × 3 × 17 × 34}$

= $√{3.3.3.2.3.17.17.2}$ = 3.3.2.17 = 306 sq. cm

Answer: (b)

Curved surface area of the sphere = 4π$r^2$

⇒ 616 = 4π $r^2$

⇒ $πr^2 = {616}/4 = 154 ⇒ r^2 = {154 × 7}/{22}$ = 49

∴ r = $√{49}$ = 7 cm

∴ Volume of the sphere = $4/3 πr^3$

= $4/3 × {22}/7 × 7 × 7 × 7 = {4312}/3 cm^3$ .