What is the difference in area (in cm²) of △ ABC having sides of 10 cm, 20 cm, and 20 cm, and a right-angled triangle △ PQR with a hypotenuse of 13 cm and one of the perpendiculars of 12 cm? Note: √(2)=1.41 , √(3)=1.73 , √(7)=2.65 , √(1)3=3.61 , √(1)5=3.87 , √(2)1=4.58

What is the difference in area (in cm²) of △ ABC having…

What is the difference in area (in cm²) of $\triangle ABC$ having sides of 10 cm, 20 cm, and 20 cm, and a right-angled triangle $\triangle PQR$ with a hypotenuse of 13 cm and one of the perpendiculars of 12 cm?

Note: $\sqrt 2=1.41$ , $\sqrt 3=1.73$ , $\sqrt 7=2.65$ , $\sqrt 13=3.61$ , $\sqrt 15=3.87$ , $\sqrt 21=4.58$

Show answer & explanation

Correct answer: C

Explanation

The correct answer is option 3 i.e. 66.75.

In ΔABC,

Semi-perimeter (s) = (10 + 20 + 20)/2 = 25 cm

Area using Heron's formula = √[s(s - a)(s - b)(s - c)]

= √[25(25 - 10)(25 - 20)(25 - 20)]

= √[25 × 15 × 5 × 5]

= √9375

= 25√15

= 96.75 cm²

In ΔPQR,

Let the base = x cm

Using the Pythagorean theorem: 12² + x² = 13²

⇒ 144 + b² = 169

⇒ b² = 169 - 144 = 25

⇒ b = √25 = 5 cm

Area of the right-angled triangle = (1/2) × Base × Height = (1/2) × 12 × 5 = 30 cm²

Difference in area = 96.75 - 30 = 66.75 cm²

Therefore, the difference in area between the two triangles is 66.75 cm².

SSC CGL 2024 PYQ — Question 63