Question

ABCD is a parallelogram with side a b = 10 CM. Its diagonal AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.

Answer 1

Given,
AB=10cm
AC=12cm
BD=16cm

To find: ar(ABCD)

Let O be the intersecting point of AC and BD

We know,
diagonals of a parallelogram bisect each other
=>OA=OC=1/2×AB=6cm
also, OB=OD=1/2×BD=8cm

we get,. in ∆AOB
OA=6cm
OB=8cm
AB=10cm(given)
By using Heron's formula
ar(∆AOB)=√s(s-a)(s-b)(s-c)
here,
s=(a+b+c)/2=(6+8+10)/
2=24/2=12cm
therefore,
ar(∆AOB)=√12(12-6)(12-8)(12-10)
=√12×6×4×2
=√6×2×6×2×2×2
=6×2×2=24cm^2
we know that the diagonals of a parallelogram divides it into four equal triangles
=>ar(∆AOB)=ar(∆BOC)=ar(∆COD)=ar(∆AOD)=
24cm^2
therefore,
ar(ABCD)=sum of the four ∆s
=24+24+24+24
=96cm^2

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Answer 2

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