In an isosceles triangle, the measure of each of the equal sides is 10 cm and the angle between them is 45°… · ABC26GN4478
Subject: General Aptitude · Chapter: Area · Exam: C.P.O. 2006
In an isosceles triangle, the measure of each of the equal sides is 10 cm and the angle between them is 45°. The area of the triangle is
(a)$25 \mathrm{cm}^{2}$
(b)$\frac{25}{2} \sqrt{2} \mathrm{cm}^{2}$
(c)$25 \sqrt{2} \mathrm{cm}^{2}$
(d)$25 \sqrt{3} \mathrm{cm}^{2}$
Answer
Explanation
Area of the triangle $=\frac{1}{2} a b \sin \theta=\left(\frac{1}{2} \times 10 \times 10 \times \sin 45^{\circ}\right) \mathrm{cm}^{2}$ $$=\left(\frac{1}{2} \times 10 \times 10 \times \frac{1}{\sqrt{2}}\right) \mathrm{cm}^{2}=\left(\frac{50}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\right) \mathrm{cm}^{2}=25 \sqrt{2} \mathrm{cm}^{2} .$$
· Browse this chapter in the app