Question

ABCD is a trapezium in which AB=16 cm and DC=24 cm.If E and F are the midpoints of AD and BC .Prove that ar(ABFE)=9/11 ar(EFCD).

Answer 1

Answer:

The proved is given below.

Step-by-step explanation:

Given ABCD is a trapezium in which AB=16 cm and DC=24 cm.If E and F are the midpoints of AD and BC. We have to prove that

$ar(ABFE)=\frac{9}{11}ar(EFCD)$

Let the height of trapezium is 2x cm

As EF is the mid segment of trapezium ABCD therefore the two trapezium ABFE and EFCD are two trapezium with height x cm and length of EF is the average of AB and CD.

i.e $EF=\frac{AB+CD}{2}=\frac{16+24}{2}={40}{2}=20cm$

Now, $\text{Area of trapezium ABFE}=\frac{1}{2}\times (AB+EF)\times height$

                                           =$\frac{1}{2}\times (16+20)\times x=\frac{1}{2}\times 36\times x$

$\text{Area of trapezium EFCD}=\frac{1}{2}\times (EF+CD)\times height$

                                           =$\frac{1}{2}\times (20+24)\times x=\frac{1}{2}\times 44\times x$

Hence, $\frac{\text{Area of trapezium ABFE}}{\text{Area of trapezium EFCD}}=\frac{\frac{1}{2}\times 36\times x}{\frac{1}{2}\times 44\times x}=\frac{36}{44}=\frac{9}{11}$

⇒ $ar(ABFE)=\frac{9}{11}ar(EFCD)$