mensuration area volumes Model Questions & Answers, Practice Test for ssc chsl tier 1 2023
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Mensuration: Area & Volumes
A field is divided into four regions as shown in the given figure. What is the area of the field in square metres ?
![mensuration area and volume aptitude mcq 24 99](https://careericons.com/adminicon/bunch/images/mensuration-area-and-volume-aptitude-mcq-24-99.png)
Answer: (a)
S = ${2 + 3 + 4}/2 = 9/2$ as r = 2 × 1 = 2 $m^2$
Area = ${√{9/2 (9/2 - 2) (9/2 - 3) (9/2 - 4)}}/{√{9/2[5/2](3/2)(1/2)}}$
= $3/4 √{15}m^2$
Area = $1/2 × 2 × 3 = 3 m^2$ Area = $1/2$ [1 + 3] × 2 = 4
Total area = 2 + 3 + 4 + $3/4 √5$
9 + $3/4 √5$
What is the area of a triangle with sides of length 12 cm, 13 .cm and 5 cm ?
Answer: (b)
∵ triangle with side 12cm, 13cm and 5 cm is a right triangle
Area = $1/2 b × h = 1/2$ × 12 × 5 = 30
In the figure given above, PT is a tangent to a circle of radius 6 cm. If P is at a distance of 10 cm from the centre O and PB = 5 cm, then what is the length of the chord BC?
Answer: (b)
Given, PO = 10 cm, radius OT = 6 cm
and PB = 5 cm
In ΔOTP,
$(OP)^2 = (PT)^2 + (OT)^2$
⇒ $(10)^2 = (PT)^2 + 6^2$
⇒ PT = 8 cm
From properties of circle,
$(PT)^2$ = PB × PC
⇒ $8^2$ = 5 × (BC + PB)
⇒ 64 = 5 (BC + 5) ⇒ 5BC = 39
∴ BC = 7.8 cm
Consider the following statements
I. Let ABCD be a parallelogram which is not a rectangle. then, $2(AB^2 + BC^2) ≠ AC^2 + BD^2$
II. If ABCD is a rhombus with AB = 4cm, then $AC^2 + BD^2 = n^3$ for some positive integer n. Which of the above statements is/are correct?
Answer: (c)
I. ABCD is a parallelogram, then
$AC^2 + BD^2 = 2(AB^2 + BC^2)$
So it is not true.
II. ABCD is a rhombus and diagonals AC and BD ∼ bisect each other.
∴ AO = OC
and OB = OD
In ΔAOB, $AB^2 = AO^2 + OB^2$
$(4)^2 = ({AC}/2)^2 + ({BD}/2)^2$
∴ $AC^2 + BD^2$ = 64
= $(4)^3 i.e., n^3$
So only II is true.
A rectangular plot 15 m ×10 m, has a path of grass outside it. If the area of grassy pathway is 54 $m^2$ , find the width of the path.
Answer: (a)
Let the width of the path = W m
then, length of plot with path = (15 + 2W) m
and breadth of plot with path = (10 + 2 W) m
Therefore, Area of rectangular plot (wihout path)
= 15 × 10 = 150 $m^2$
and Area of rectangular plot (with path)
= 150 + 54 = 204 $m^2$
Hence, (15 + 2W) × (10 + 2W) = 204
⇒$4W^2$ + 50 W – 54 = 0
⇒$2W^2$ + 50W –54 = 0
⇒(W – 1) (2W + 27)
Thus W = 1 or –27
∴ with of the path = 1 m
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