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

Answer: (a)

Statement 1

AD divides ΔABC in equal area of two parts.

Then O is point on anywhere on AD

So area of triangle

ΔABO = ΔAOC,

So statement 1 is true.

Statement 2

G is the point of concurrence of the medians then

area of ΔABG = area of ΔBCG = Area of ΔACG

Both are true.

Answer: (a)

Suppose a horse is tied at vertex A. Then, area available grazing field is ADE.

Now, area of curve

ADE = ${π r^2 θ}/{360°} = {22 × (4.2)^2 × 60°}/{7 × 360°} = 9.24 m^2$

and area of equilateral Δ ABC = $√3/4 (side)^2$

= $√3/4 × (6)^2$ = 15.57

∴ Required percentage

= ${9.24}/{15.57}$ × 100 = 59.34% = 59% (approx)

Answer: (d)

From tangent properties

CF = CG = 16

BE = BG = 25 H

∴ BC = 16 + 25 = 41 cm

From ΔBCH,

CH = diameter = $√{(41)^2 – (9)^2}$

= $√{1681 – 81}$

= $√{1600}$ = 40cm

Answer: (b)

Given that,

Height of bucket = 25 cm

Radii of top (R) = 20 cm

radii of bottom (r) = 10 cm

∴ Capacity of bucket

= $π/3 h(R^2 + r^2 + Rr)$

= $π/3 × 25 (400 + 100 + 200) cm^3$

= $π/3 × 25 × 700 cm^3$

= $π/3 × {175 × 100}/{1000} = {17.5 π}/3$L

Answer: (c)

${\text"Shaded Area"}/{\text"Total Area"} = {1/2 (π × 9^2 - π × 6^2)}/{π × 9^2} = 5/{18}$

∴ ${\text"Shaded area"}/{\text"Unshaded area"} = 5/{18 - 5} = 5/{13}$