mensuration area volumes Model Questions & Answers, Practice Test for ssc mts paper 1 2023

Answer: (d)

$√{4 - x^2} + √{9 - x^2} = 3 ⇒ √{9 - x^2} = 3 - √{4 - x^2}$

on squaring both sides, we have

9 – $x^2 = 9 + 4 – x^2 – 6 √{4–x^2}$

⇒ $√{4 - x^2} = 2/3$

⇒ 4 – $x^2 = 4/9 ⇒ x^2 = {32}/9 ⇒ x = {4√2}/3$

⇒ Area = $1/2 (4 + 7) {4 √2}/3 = {22 √2}/3 cm^2$

Answer: (a)

Side of the room = 10m

Area of the room = $(10)^2 = 100m^2$

Side of table = 2m

Area of table = $√3/4 × (2)^2 = √3 = 1.73 m^2$

Area of 4 Book-shelves = 4 × (4 × 1) = 16$m^2$

Area of remaining room = 100 – 1.73 – 16 = 82.27 $m^2$

Required cost = ${82.27}/2$ × 100 = 4113

Answer: (d)

The centre of three lower balls form an equilateral triangle. Let C be the centroid of this triangle. C is at the distance of ${2r}/√3$ form each vertex of lower ball.

Let h = height of centre of 4th ball from C.

then, h = $√{(2r)^2 - ({2r}/√3)^2} = 2r √{2/3}$

∴ height of centre of 4th ball from plane

= h + r = 2r $√{2/3} + r = r({2√2 + √3}/√3)$ units

Answer: (d)

in ΔBCD

${\text"sin30°"}/{BD} = {\text"sin B"}/{CD}$

⇒ $1/{2BD} = {\text"sin B"}/{CD}$

⇒ $1/{2BD} = {12}/{13CD} ∵ sin B = {12}/{13}$

CD = ${24}/{13}$BD ...(i)

in ΔADC

${\text"sin 60°"}/{AD} = {\text"sin A"}/{CD} ∵ sin A = 5/{13}$

CD = $5/{13} × 2/√3$ AD

⇒ CD = ${10}/{13 √3}$ (13 - BD) ...(ii)

equating both eq (i) and (ii)

${10}/{13 √3} (13 - BD) = {24}/{13}BD$

BD = ${65}/{5 + 12 √3}$

CD = ${24}/{13} {65}/{5 + 12√3} = {120}/{5 + 12 √3}$

Answer: (d)

We know altitude of equilateral ΔABC is $√3/2$ a .

∴ Length of OC = $√3/2 a × 2/3 = a/√3$ = radius

Also, DF = b ⇒ DE = $b/2$

In ΔODE, cos 60° = ${DE}/{OD} = {b/2}/{a/√3}$

⇒ $1/2 = {√3b}/{2a} ⇒ a = √3b$

∴ $a^2 = 3b^2$