mensuration Model Questions & Answers, Practice Test for ibps clerk prelims 2024

Answer: (a)

Statements I and II are both true, because the locus of points which are equidistant from two parallel lines is a line parallel to both of them and draw mid way between them.

Also, it is true that the perpendicular distances of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.

Answer: (d)

Since, OR is a bisector of ∠PRQ.

∴ ∠PRO = ∠ORQ = 45°

Also, OP = OR

∴ ∠OPR = 45°

In ΔORS,

OR = OS ⇒ ∠ORS = ∠OSR = ${180° - 44°}/2$ = 68°

∴ ∠MRS = 68° – 45° = 23°

⇒ ∠PRS = 90° + 23° = 113°

By properties of cyclic quadrilateral.

∠PRS + ∠PQS = 180°

⇒ ∠PQS = 180°– 113° = 67°

In ΔPTQ,

∠QPT + ∠PQT + ∠PTQ = 180°

⇒ ∠PTQ = 180° – 45° – 67° = 68°

Answer: (c)

Given BC = 38 cm

QB = 27 cm

DC = 25 cm

AD ⊥ DC

We know that tangents are always equal, when they drawn to the circle from a point outside the circle.

∴ BQ = BR = 27 cm

RC = BC – BR = 38 – 27 = 11 cm

RC = PC = 11 cm

DC = 25 cm

DP = DC – PC = 25 – 11 = 14 cm

DP = OT = OP

∴ Radius of the circle = 14 cm

Answer: (b)

Let ABC is triangle, which have maximum area, while

AC is 2r

But OB = OC = r

By Pythagoras theorem

$OB^2 + OC^2 = BC^2$

BC = $√2$r = AC

Area of triangle = $1/2 × √2r × √2r = r^2$

Answer: (b)

Circumference of circle = 2π × 42

= $2 × {22}/7 × 42$ = 264 cm

Perimeter of square = 4x ⇒ 264 = 4x

x = 66 cm