mensuration area volumes Model Questions & Answers, Practice Test for ssc mts paper 1 2023
ssc mts paper 1 2023 SYLLABUS WISE SUBJECTS MCQs
Number System
LCM & HCF
Ratio Proportion & Partnership
Percentages
Profit & Loss
Time & Work
Time & Distance
Simple Interest & Compound Interest
Mensuration: Area & Volumes
Algebraic Expressions
Trigonometric Ratios & Identity
Read the following information carefully and answer the given questions that follow.
A piece of land is in the form of a parallelogram and the perimeter of the land is 86m. The length of one side exceeds the other by 13 m and one of the diagonals is 41m.
What is the area of the parallelogram ?
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 $m^2$
A right circular cone is sliced into a smaller cone and a frustum of a cone by a plane perpendicular to its axis. The volume of the smaller cone and the frustum of the cone are in the ratio 64 : 61. Then their curved surface areas are in the ratio
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
Let ABCD be a parallelogram. Let X and Y be the mid– points of the sides BC and AD, respectively. Let M and N be the mid–points of the sides AB and CD, respectively.
Consider the following statements:
1. The straight line MX cannot be parallel to YN.
2. The straight lines AC, BD, XY and MN meet at a point.
Which of the above statements is/are correct ?
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.
In a circle of radius 2 units, a diameter AB intersects a chord of length 2 units perpendicularly at P. If AP > BP, then AP is equal to
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.
The difference between the area of a square and that of an equilateral triangle on the same base is 1/4 $cm^2$. What is the length of side of triangle?
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
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