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

Answer: (a)

1. On drawing the three straight lines through the three vertices of DABC, we get the following figure.

Here, AB||DF, BC || DE and AC || EF.

Clearly, A, B and C are the mid–points of DE, EF and DF respectively.

By mid–point theorem, BC = $1/2$ DE or DE = 2BC

Similarly, DF = 2AB and EF = 2 AC. Hence, Statement 1 is correct.

2. Also, area of ΔABC = $1/4$ area of ΔDEF or area of ΔDEF = 4 area of ΔABC. Hence, Statement 2 is also coreect.

Answer: (b)

Area of ΔABC, A = $1/2$ × b × AB

AB = ${2A}/b$ ... (i)

By Pythagoras theorem, $AC^2 = AB^2 + BC^2$

AC = $√{{4A^2}/b^2 + b^2}$

Again in ΔABC

A = $1/2$ × AC × BD

BD = ${2A}/{√{{4A^2}/{b^2} + b^2/1}} = {2A}/{√{{4A^2 + b^4}/b^2}}$

= ${2Ab}/{√{4A^2 + b^4}}$

Answer: (b)

$H^2 = P^2 + B^2$

∴ $AB^2 + BC^2 + CD^2 + AD^2 = AC^2 + BD^2$

⇒ $(10)^2 + (10)^2 + (10)^2 + (10)^2 = (16)^2 + (12)^2$

⇒ 400 = 400

Hence, both A and R are true and R is the correct explanation of A.

Answer: (c)

In ΔBCF,

Pythagoras theorem,

$(5)^2 = (3)^2 + (BF)^2$ ⇒ BF = 4cm

AB= 2 + 4 + 4 = 10 cm

Now in ΔACF, $AC^2 = CF^2 + FA^2 ⇒ AC^2 = 3^2 + 6^2$

AC= $√{45}$ cm

Similarly, BD = $√{45}$ cm

∴ Sum of diagonal = 2 × $√{45} = 2 × 3 √5$

= 6 $√5$ cm

Answer: (c)

According to question

Surface area of sphere = 3 (Volume of sphere)

⇒ 4π$r^2 = 3 × 4/3 πr^3$ ⇒ r = 1

∴ Diameter = 2r = 2 cm