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

Answer: (d)

Perimeter of parallelogram land = 86 m and diagonal = 41m

Suppose one side of parallelogram be x m other side = (x + 13) m

∴ Perimeter = 2(x + x +13) = 86

⇒ 2x + 13 = ${86}/2$ = 43 ⇒ 2x = 43 – 13 = 30

∴ x = ${30}/2$ = 15

one side of parallelogram = 15 m

other side = 15 + 13 = 28 m

area of ΔABD = $√{s(s - a) (s - b) (s - c)}$

= $√{42(42 - 15)(42 - 28)(42 - 41)}$

$[∵ s = {15 + 41 + 28}/2 = {84}/2 = 42]$

= $√{42 × 27 × 14 × 1} = 126 m^2$

Required area of parallelogram = 2 × Area of ΔABD

= 2 × 126 = 252 $m^2$

Answer: (c)

Let volume of smaller cone be 64 unit and volume of frustum be 61 unit

Total volume of bigger cone = 64 + 61 = 125 units

Here smaller cone is cut from bigger cone then respective ratio of their radius, hight will be equal

∴ Respective ratio of area of bigger cone to that of smaller cone

= $(^3√125)^2 : (^3√64)^2$ = 25 : 16

Respective ratio of area of smaller cone to that of frustum

= 16 : 25 – 16 = 16 : 9

Answer: (c)

From Statement 1. Given, ABCD is a parallelogram. X and Y are mid–points of BC and AD, respectively. M and N are the mid–points of AB and CD, respectively.

From statement 2. Here join point A and C.

In ΔABC, M and X are mid–points of AB and BC.

∴ MX || AC and MX = $1/2$ AC ...(i)

In ΔADC, Y and N are mid–points of AD and CD.

∴ YN || AC and YN = $1/2$ AC ...(ii)

From equations (i) and (ii), we get MX || YN

From statement 2

So, Statement 1 is not correct.

Clearly, straight lines AC, BD, XY and MN meet at a point, So Statements 2 is correct.

Answer: (c)

Let us consider a circle of radius 2 units.

Diameter = AB = 2 × 2 = 4 units

QR be a chord of circle

then QR = 2 units

Let O be the centre of the circle

Given AP > BP

In right angled triangle POR

By applying Pythagoras theorem, we get

OP = $√{(RO)^2 - (PR)^2}$

OP = $√{(2)^2 - (1)^2} = √{4 - 1} = √3$

OP = $√3$

AP = AO + OP

= 2 + $√3$ (where OA is the radius of the circle)

∴ Option (c) is correct.

Answer: (a)

Let the side of an square be a cm.

By given condition,

Area of square – Area of an equilateral triangle = $1/4$

⇒ $a^2 - √3/4 a^2 = 1/4 ⇒ a^2(1 - √3/4) = 1/4$

⇒ $a^2 (4 - √3) = 1 ⇒ a^2 = 1/{4 - √3}$

∴ a = $(4 - √3)^{-1/2}$ cm