SSC CGL 2022 PYQ — Question 52

The correct answer is option 1 i.e. 20

Concept Used : 

Area of a triangle = 1/2 × base × height

Circumradius, R = ABC/4Δ

where A, B, C are the sides of the triangle and Δ is area of triangle.

Given : 

Equal sides of isosceles triangle = 16$\sqrt{5}$  cm

Length of unequal side = 32 cm

Solution : 

Let MD be the perpendicular on side ND of the triangle,

Hence ND = DP = 32/2 = 16 cm

Applying Pythagoras theorem in ∆MND, 

(MD)² = (MN)² - (ND)²

(MD)² = (16$\sqrt{5}$ )² - (16)²

(MD)² = 32 cm

Now area of ∆MNP = 1/2 × NP × MD

                                        = 1/2 × 32 × 32

                                        = 512 cm²

Now the circumradius R of ∆MNP = ABC/4∆

                                                                     = (16$\sqrt{5}$ × 16$\sqrt{5}$ × 32)/(4 × 521)

                                                                     = 20 cm

Hence the radius of the circle circumscribing ∆MNP is 20 cm