Problem: length of each equal side of an isosceles triangle is 10cm and the mcuded angle between those two sides is 45°. Find the area of the triangle.
View all: BANGLADESH KRISHI BANK LTD (OFFICER-CASH) WRITTEN QUESTION (MATH) SOLVE | 2018
Correct Answer: 22$\sqrt{2}$
Explanation:
Let, x be the perpendicular drawn from the vertex of triangle
We know, Sin 45 ° = $\frac{1}{\sqrt{2}}$
=> $\frac{1}{\sqrt{2}}$ = $\frac{x}{10}$
=> x = $\frac{10}{\sqrt{2}}$
Area of triangle = $\frac{1}{2}$ x Base x Height
= $\frac{1}{2}$ x 10 x $\frac{10}{\sqrt{2}}$ = 25$\sqrt{2}$
