Question

ABCD is a parallelogram in which the diagonals AC and BD intersect at O if area of ABCD 68 CM square find the area of triangle AOB​

Answer 1

$\huge\bold\color{red}{:- Answer}$

Area of Parellelogram :: 68cm²

First , we had to Find the Height and base of the parellogram ::

$Area\:of\:parellelogram=>H×B$

$68cm²=>AB×BC$

Let,both Height and base be (x) cm:-

$68cm²=x²$

$√68=x$

$8.24=x$

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Hence,(AB)Base=8.24cm

(BC)Height=8.24cm

Now,we had to find the Area of ∆ AOB

Given=>

Height=$\frac{1}{2}of\:8.24$

$\:\:\:\:\:\:\:\:\frac{1}{2}×8.24$

$\:\:\:\:\:\:\:\:\:\:\:\:4.12$

Base=$8.24$

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Area of ∆ AOB =$\frac{1}{2}×H×B$

Area of ∆ AOB =$\frac{1}{2}×4.12×8.24$

Area of ∆ AOB = $=>16.97cm²$

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