Question

ABCD is a parallelogram in which 'O' is a itersection point if it's diagonals if ar(∆AOD)=4cm² and the ar(||gm ABCD)=24cm² find ar(∆AOB)​

Answer 1

Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O.

∴ O is the mid - point of AC as well as BD.

Now, in △ADB , AO is its median

∴ ar(△ADB) = 2 ar(△AOD) [∵ median divides a triangle into two triangles of equal areas]

So, (△ADB) = 2 × 4 = 8 cm2

Now, △ADB and ||gm ABCD lie on the same base AB and lie between same parallel AB and CD .

∴ ar(ABCD) = 2 ar(△ADB)

= 2 × 8

= 16 cm2

Answer 2

The area of triangle AOB is 8 cm sq.

• Given:

• ABCD is a parallelogram
• O is the intersection point of two diagonals
• The area of triangle AOD = 4 cm sq.
• The area of the parallelogram is 24 cm sq.

We know that

One diagonal divides the area of the parallelogram into two equal parts.

So, ar(ΔABD) = ar(ΔBCD) = 1/2 * ar(||gm ABCD) = 12cm sq.

Now,

We get more triangles,

ΔADO and ΔAOB

They form a big triangle = ΔABD

(From given, ar(ΔAOD) = 4 cm sq.)

So,

⇒ 4 + ar(AOB) = 12 cm sq

⇒ ar(AOB) = 12 - 4

[By transporting 4 to RHS]

• Thus, are(AOB) = 8 cm sq.