mensuration Model Questions & Answers, Practice Test for ssc mts tier 1 2024

Answer: (d)

Radius of conical flask = r

Height of conical flask = h

h r 2r x

∴ Volume of conical flask = $1/3 πr^2h$

Radius of cylindirical flask = 2r

Height of cylindirical flask = x

∴ Volume of cylindirical flask = π$(2r)^2$ x

= π$4r^2$x

According to question,

$1/3 πr^2h = 4πr^2x$

x = ${1.πr^2h}/{3 × 4πr^2} = h/{12}$

∴ Height of the cylinderical flask is . $h/{12}$

Answer: (d)

Diagonal of square = 12 $√2$cm ∼

Side of square = ∼ ${12 √2}/√2$ = 12cm ∼

Perimeter of square = 4 × 12 = 48 cm ∼

Perimeter of equilateral triangle = 48 cm ∼

Side of the equilateral triangle = ∼ ${48}/3$ = 16 cm

Area of triangle $√3/4 × (16)^2$

= $√3/4 × 256$

= 64$√3cm^2$ ∼

So, option (d) is correct. ∼

Answer: (c)

Given ∠OBD + ∠ODB = ∠CBD + ∠CDB

Let ∠OBD = ∠ODB = θ

and ∠DBC = $θ_1$ , ∠BDC = $θ_2$

θ + θ = $θ_1 + θ_2$ ... (i)

⇒ 2θ = $θ_1 + θ_2$

∴ ∠BOD = 180° – 2θ

⇒ ∠BCD = ${360° - (180° - 2 θ)}/2$

(by properties of circle)

⇒ 180° – $(θ_1 + θ_2)$ = 90° + θ

⇒ 180° – 2θ = 90° + θ

⇒ 90° = 3θ

⇒ θ = 30°

∴ ∠BOD = 120°

∠BAD = 60°

Answer: (a)

Let two circles with center O and O' intersect each other at point A and B, such that AB is a common chord.

Then AB = 10 $√3$ cm (Given)

Since both passes through the center of each other as shown in figure

So, $O_1$O is the radius of both circle

Let $O_1O = r = AO_1$ = AO

AX = AB/2 = 5 $√3$ cm

(Since OX is perpendicular to chord AB, so bisects it)

AO$O_1$ forms an equilateral triangle with side = radius = r

sin 60 = $√3/2 = {AX}/{AO} = {5√3}/r$

⇒ So r = 10 cm

⇒ diameter = 20 cm

Answer: (b)

Let height of triangle be = h

as ΔABDE = 5 × h

ar ΔBDE = $1/2$ × 5 × h

ar ΔBCD = $1/2$ × 7 × h

Ratio = 5h : ${5h}/2 : {7h}/2$

= 10 : 5 : 7