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

Answer: (c)

External Radius(R) = $8/2$ = 4 cm and Internal Radius(r)

= $4/2$ = 2cm

∴ Volume of hollow sphere

= $4/3 π(R^3 - r^3)$

= $4/3 π (4^3 - 2^3)$

= ${4π}/3$ × 56

Let h = Height of the cone

According to question

Volume of cone = Volume of hollow sphere

∴ $1/3 πr_1^2 h = 4/3 π × 56$

⇒ $(4)^2$ h = 4 × 56

⇒ h = ${4 × 56}/{16}$

∴ h = 14 cm

Answer: (c)

Given radius = 9 cm

∴ diameter ABCD = 18 cm

Given that AB = BC = CD = 6cm

Area of the semicircle with diameter AD

= $π/2 (9)^2 = π {81}/2$

Area of the semicircle with diameter

AB = $π/2 (3)^2 = π × 9/2$

Area of the semicircle with diameter

BD ∼ = $π/2 (6)^2 = π/2 36$

Area of the shaded portion

= ${81π}/2 + {9π}/2 - {36π}/2 ∼ = {54π}/2 = 27π$

Answer: (a)

Given a circle with centre c.

OP and OQ are tangents to the circle from Q + O point O outside the circle.

Given OR × SQ = OS × RP

⇒ ${OR}/{RP} = {OS}/{SQ}$

⇒ RS || PQ (By Basic proportionality theorem)

(1) Also CP = CQ = radius of the circle.

A perpendicular drawn from P to Q,

Draw circle X and Y with centre O and radius OR and DS respectively.

Since RS || PQ

Here O is the center of circle X and Y both Radius OR and OS lies in the same circle.

⇒ OR = OS ⇒ Area of Circle X = Area of circle Y

⇒ X = Y

Statement (1) is true.

(2) Also we know that if two tangents are drawn to the circle then ∠POC = ∠QOC and ∠PCO = ∠QCO

Also we know that CP = CQ = radius

So ΔPtc and ΔQtc are similar by AA similarly.

i.e., ∠P = ∠Q = 45° ...(1)

and ∠t = ∠t = 90°

Also ∠PCO = ∠QOC (Alternate angles)

∠POC = ∠QCO (Alternate angles)

from (1) if ∠P = ∠Q = 45°

⇒ ∠QCO = ∠PLO = 45°

⇒ ∠POC + ∠QCO = 45° + 45° = 90°

Statement (2) is true.

∴ Option (a) is correct.

Answer: (b)

∴ Required area of the path,

EF = 30 + 4 = 34 m, GF = 16 + 4 = 20 m

∴ Area of path = Area of EFGH – Area of ABCD

= 34 × 20 – 30 × 16 = 680 – 480 = 200 $m^2$

Answer: (c)

Perimeter of ΔPQR = 1 + 2 + 3 = 6 units

Now, in ΔDEF, ∼

${DQ}/{DF} = 1/2 = {PQ}/{FE}$

So, 2 PQ = FE

Similarly,DF = 2 PR and DE = 2QR

∴ perimeter of ΔDEF = 2 × 6 = 12 units

Similarly, perimeter of ΔABC

= 2 × Perimeter of ΔDEF

= 2 × 12

= 24 units