If E and F are any two points lying on the sides DC and AD, respectively of a parallelogram ABCD, then

B F + C F = A E + B E

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A E = B F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A r e a (A E B) = A r e a (B F C)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A r e a (A D E) = A r e a (B E C)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $A r e a (A E B) = A r e a (B F C)$
Given: ABCD is a parallelogram

Construction: Join BF,CF,AE,BE.

Proof: $A r e a (△ A E B) = \frac{1}{2} \times A r e a (p a r a l l e l o g r a m A B C D)$ (a triangle and the parallelogram on the same base and between same parallels have same area)....(1)

Similarly,

$A r e a (△ B F C) = \frac{1}{2} \times A r e a (p a r a l l e l o g r a m A B C D)$ (a triangle and the parallelogram on the same base and between same parallels have same area).....(2)

From equation 1 and 2 we get,
$A r e a (△ A E B) = A r e a (△ B F C)$

Suggest Corrections

Similar questions

Q. In trapezium

A B C D

A B | | D C

E

and

F

are points on non-parallel sides

A D

and

B C

respectively such that

E F ∥ A B

. Prove that

\frac{A E}{E D} = \frac{B F}{F C}

Q. ABCD is trapezium with AB||DC. E and F are points on non-parralel sides AD and BC respectively such that EF||AB. Show that AE/ED=BF/FC.

ABCD is a trapezium with $A B ∥ D C$ . E and F are points on non-parallel sides AD and BC respectively such that $E F ∥ A B$ . Show that $\frac{A E}{E D} = \frac{B F}{F C}$ .

Q. ABC is an equilateral triangle. Points D, E, F are taken in sides AB, BC, CA respectively, so that AD=CF. Then AE, BF, CD enclosed a triangle which is :

Q. If P and Q are any two points lying respectively on the sides DC and AD of a parallelogram ABCD then show that ar(∆APB) = ar(∆BQC).

Join BYJU'S Learning Program