ABCD is a rectangle and BCE is an isosceles triangle in which BC = CE. What is the area of the given figure?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\left(\frac{b}{2}+a\right) b\)
The figure is made of two parts: the rectangle ABCD and the right-angled triangle BCE. In the triangle, BC = b (the side of the rectangle opposite AD) and CE = BC = b, and the angle at C is a right angle because DCE is a straight line.
Rectangle: \(a\times b=ab\)
Triangle: \(\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\times b\times b=\frac{b^2}{2}\)
Total: \(ab+\frac{b^2}{2}=\left(a+\frac{b}{2}\right)b\)