mensuration Model Questions & Answers, Practice Test for ssc chsl tier 1 2024

Answer: (c)

Perimeter of a rectangle = 82 m

∴ 2(Length + Breadth) = 82 m

⇒ Length + Breadth = 41 m

⇒ l + b = 41 m ...(i)

Also, its area = 400 $m^2$

⇒ l × b = 400 ...(ii)

Now, $(l – b)^2 = (l + b)^2$ – 4lb

= $(41)^2$ – 4 (400)

= 1681 – 1600 = 81

∴ l – b = 9 ...(iii)

From Eqs. (i) and (iii)

2l = 50 ⇒ l = 25 m and b = 16 m

∴ Required breadth (b) = 16 m

Answer: (c)

Let the radii of two circles are $r_1$ and $r_2$ , respectively.

Given, ${\text"Circumference of Ist circle"}/{\text"Circumference of IInd circle"} = 2/3$

⇒ ${2 π r_1}/{2 π r_2} = 2/3 ⇒ r_1/r_2 = 2/3$

⇒ $(r_1/r_2)^2 = 4/9$ ...(i)

∴ ${\text"Area of Ist circle"}/{\text"Area of IInd circle"} = {π r_1^2}/{π r_2^2} = (r_1/r_2)^2 = 4/9$

Answer: (c)

a + b + c + d = 360°

(1 + 2 + a) + (3 + 4 + b) + (5 + 6 + c) + (7 + 8 + d) = 180° × 4

∴ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 360°

Answer: (b)

In ΔADE and ΔABC

∠A = ∠A

DE || BC

ΔADE ∼ ΔABC

${AD}/{AB} = {AE}/{AC}$

⇒ ${AD}/{BD} = {AE}/{EC}$

$2/1 = 3/{EC} ⇒ EC = 3/2$ = 1.5 cm

Answer: (c)

Here, ΔABC forms an equilateral triangle.

where, AGFH form a rhombus and ΔHDE is also an equilateral triangle.

∴ Area of rhombus

= (Area of ΔAGF + Area of ΔGFH)

= $√3/4 (x/3)^2 + √3/4 (x/3)^2 = 2 × √3/4 (x/3)^2$

Now, area of ΔHDE = $√3/4 (x/3)^2$

and area of ΔABC = $√3/4 x^2$

By given condition,

${\text"Area of rhombus AGHF+ Area of ΔHDE"}/{\text"Area of ΔABC"}$

= ${2 × √3/4 (x/3)^2 + √3/4 (x/3)^2}/{√3/4 x^2}$

= ${3 × √3/4 (x/3)^2}/{√3/4 x^2} = 3/9 = 1/3$