In an isosceles triangle ABC, AB = AC, and AD is…

In an isosceles triangle ABC, AB = AC, and AD is perpendicular to BC. If AD = 5 cm and the perimeter of △ABC is 50 cm, then twice the length of BC (in cm) is:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. 48.

Let BC = 2x

⇒ BD = DC = x

⇒ AD = 5 cm

In △ABD

⇒ AB² = AD² + BD²

⇒ AB = $\sqrt{5^2+x^2}$ = $\sqrt{25+x^2}$

⇒ AB + BC + CA = 50

⇒ 2AB + 2x = 50 [AB = AC]

⇒ 2 $\sqrt{25+x^2}$ + 2x = 50

⇒ $\sqrt{25+x^2}$ + x = 25

⇒ $\sqrt{25+x^2}$ = 25 - x

Squaring both sides

⇒ 25 + x² = (25 - x)²

⇒ 25 + x² = 625 + x² - 50x

⇒ 50x = 600

⇒ x = 12

BC = 2x = 24 cm

Twice of BC = 24 × 2 = 48 cm

SSC Selection Post 2022 PYQ — Question 70