mensuration area volumes Model Questions & Answers, Practice Test for ssc chsl tier 1 2023

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 + $3/4 √5$

Answer: (b)

∵ triangle with side 12cm, 13cm and 5 cm is a right triangle

Area = $1/2 b × h = 1/2$ × 12 × 5 = 30

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

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.

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