Question

ABCD is a parallelogram with sides AB=12cm, BC=10cm and diagonal AC= 16cm. Find the area of the parallelogram. Also find the distance between it's shorter sides.
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Answer 1

hope u find it clear by my solution

Answer 2

Area of $ABCD$ parallelogram is $119.84cm^{2}$ and Distance between it's shorter sides is $9.98cm$

Step-by-step explanation:

Given,

$ABCD$ is a parallelogram where $AB=12cm,BC=10cm$ and diagonal $AC=16cm$

In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. That is, each diagonal cuts the other into two equal parts.

∴ Diagonal $AC$ divide parallelogram into two equal triangle.

From figure,

Using Heron's Formula,

Area of $\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}$   where $s=\frac{a+b+c}{2}$

∴ $s=\frac{10+12+16}{2}=19$

⇒Area of  $\triangle ABC=\sqrt{19(19-10)(19-12)(19-16)}$

                             $=\sqrt{19\times 9\times 7\times 3}$

                            $=59.92$$cm^{2}$

∴ Area of $ABCD$ parallelogram$=(2\times59.92)=119.84cm^{2}$

Also known area of parallelogram,

  ∴ $(AB\times h)=119.84$

 ⇒ $h=\frac{119.84}{12}$

 ⇒ $h=9.98cm$

∴ Distance between it's shorter sides is $9.98cm$