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

Answer: (d)

Total surface area of cone = $πR^2 + πLR$

= πR$[R + √{R^2 + H^2}] ∵ 1 = √{R^2 + H^2}$

= π × 6(6 + 10) = 96π $cm^2$

Answer: (d)

Since the conical cavity in the cylinder is hollowed out, therefore inner surface area of the cavity is curved surface area of the cone.

Therefore, CSA of the cone = πrl = π . 6. $√{8^2 + 6^2}$

= π.6. $√{64 + 36 = 100}$ = π . 6.10 = 60π

Answer: (c)

We draw a line segment parallel to BF

In ΔADG, ∼

BF or EF || DG and AE = ED (since, E is mid-point of ∼ AD) ∼

∴ AF = GC ... (i) ∼

Similarly, in ΔBCF, ∼

DG || BF and BD = DC ∼

FG = GC ... (ii) ∼

From Eqs. (i) and (ii), ∼

CF = $2/3$ AC

Answer: (c)

AB = 32, AD = 24 OD = ?

$DB^2 = 32^2 + 24^2$

= $8^2 [4^2 + 3^2]$

$DB^2 = 8[5^2] = (40)^2$

DB = 40

OD = 20

Answer: (b)

Area = $√3/4 × 20^2 + 1/2$ × 16 × 12

= 269.2