mensuration area volumes Model Questions & Answers, Practice Test for ibps po prelims 2023

Answer: (a)

In ΔABC

CD ⊥ AB

and AB × CD = CA × CB .... (i)

In ΔCDB

$BC^2 = BD^2 + CD^2$ ... (ii)

$CD^2 = BC^2 – BD^2$

From eq. (i) and (ii)

$1/{CD^2} = 1/{BC^2} - 1/{CA^2}$

Answer: (c)

Using cosine law

cos A = ${AD^2 + AB^2 - BD^2}/{2.AD.AB}$

⇒ cos 60° = ${AD^2 + (2AD)^2 - BD^2}/{2.AD.2AD}$

⇒ $1/2 = {AD^2 + 4AD^2 - BD^2}/{4AD^2}$

⇒ $2AD^2 = 5AD^2 – BD^2$

⇒ $BD^2 = 5AD^2 – 2AD^2$

⇒ BD = $3AD^2$

⇒ BD = $√3$ AD

So, option (c) is correct.

Answer: (d)

Canvas used = $CSA_{cylinder} + CSA_{cone}$

= 2πrh + πrl

= 1980 + ${22}/7 × 63 × 65 = 14850 m^2$

Answer: (d)

Join OB. OA = OB = OC

Then,

∠OAB = ∠OBA = 20°

∠OCB = ∠OBC = 30°

∠ABC = 50°

We know that ∠ABC = $1/2$ ∠AOC

∴ ∠AOC = 2 ∠ABC = 2 × 50° = 100°

Answer: (d)

Let radius and height of a cone be r and h

∵ $r/h = 3/4$

Volume of cone = $1/3 πr^2 h$

∴ 96π = $1/3 π × r^2 × {4r}/3$

⇒ $r^3 = {96 × 3 × 3}/4$ = 216 ⇒ r= 6 cm

and h = 8 cm

∴ Lateral surface area

= πr$√{r^2 + h^2}$

= π × 6$√{36 + 64}$

= 60π sq cm