Question

In the given figure, F and E are points on the side AD of $\Delta ABD$. Through F a line is drawn parallel to AB to meet BD at the point C. Area of quadrilateral BCEF is equal to ________ .

• A. area of $\Delta ACE$
• B. area of $\Delta BFC$
• C. area of $\Delta ABC$
• D. area of $\Delta ABD$

Answer 1

The correct option is A

area of $\Delta ACE$

(i) $AB\parallel FC$ .... (given)

(ii) area $\Delta FCB$ = area $\Delta FCA$ ..... (triangles on the same base and between the same parallels are equal in area)

(iii) adding area FCE to both sides we get,
area $\Delta FCB$ + area $\Delta FCE$ = area $\Delta FCA$ + area $\Delta FCE$

From figure we can notice BCEF can be divided into triangles FCB and FCE. Similarly, triangle ACE can be split into triangle FCA and FCE.

$\therefore$ Area of quadrilateral BCEF = area of $\Delta ACE$