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

Answer: (d)

Given, ∠AOB = 100°

∴ Reflex ∠AOB = 360° – ∠AOB

= 360° – 100° = 260°

∴ ∠ACB = ${\text"Reflex ∠AOB"}/2 = {260°}/2$

= 130°

Answer: (d)

Let the side of the equilateral triangle be a cm.

Area of the triangle ABC = ${√3 a^2}/{4}$

Height of the triangle = ${√3 a}/{2}$ cm

As G is circumcentre

Area of the square EFGC = $(a/{√3})^2 = {a^2}/{3}$ sq. cm.

Ratio area of ΔABC to EFGC

${√3 a^2}/{4} : {a^2}/{3}⇒3 √3 : 4$.

Answer: (d)

Here ΔABC ∼ ΔDEF :

Then ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

⇒ ${\text"are of ΔABC"}/{\text"area of ΔDEF"} = ({AB}/{DE})^2$

= $({√2 + 1}/√3)^2$

= ${2 + 1 + 2√2}/3 = {3 + 2 √2}/3$

= 3 + 2$√2$ : 3

Answer: (b)

Given that, AB = 5 cm, BC = 4 cm, CA = 4.2 cm and DE = 10 cm, EF = 8 cm and FD = 8.4 cm

Now, ${AB}/{DE} = 5/{10} = 1/2$

${BC}/{EF} = 4/8 = 1/2$

and ${CA}/{FD} = {4.2}/{8.4} = 1/2$

∴ ${AB}/{DE} = {BC}/{EF} = {CA}/{FD}$

∴ ΔABC ∼ ΔDEF

We know that,

${AB}/{DE} = {AL}/{DM}$

⇒ ${AL}/{DM} = 1/2$

Alternate Method

Every sides of second is double of first one.

Hence, ${AL}/{DM} = 1/2$.

Answer: (c)

Let the position of person = A

We have to find out the distance between A and E.

AB = 7 km then CD = 7 km

BC = 3 m then AD = 3 km

DE = CE – CD = 13 – 7 = 6 km

Draw AD ⊥ CE

In right angled triangle AED, we get

$(AE)^2 = AD^2 = DE^2$

= $(3)^2 + (6)^2$

AE = $√{9 + 36}$

= $√{45} = √{5 × 9}$

AE = 3 $√5$km

∴ Option (c) is correct.