Question

ABCD is a parallelogram with sides a b equal to 12 CM BC equal to 10 cm and diagonal AC equal to 16 cm find the area of the parallelogram also find the distance between its shorter sides​

Answer 1

Question:

ABCD is a parallelogram with sides AB = 12cm BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram also find the distance between its shorter sides.

Given:

• ABCD is a parallelogram.
• AB = 12 cm
• BC = 10 cm
• AC = 16 cm

To Find:

1. Area of the parallelogram.
2. Distance between it's shorter sides.

Answer:

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

Solution:

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, 【For figure see the attachment】

Using Heron's Formula,

Area of $\triangle abc = \sqrt{s(s - a)(s - b)(s - c)}$

Where,

$s = \frac{a + b + c}{2}$

$s = \frac{12 + 10 + 16}{2}$

$s = \frac{ \cancel{38}}{ \cancel2}$

$s = 19$

So,

$\implies \sqrt{19(19 - 12)(19 - 10) (19 - 16)}$

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

$\implies {59.92 \: cm}^{2}$

∴ Area of ABCD parallelogram

$\implies 2 \times 59.92$

$\huge \boxed {\implies 119.84 \: {cm}^{2}}$

Also known area of parallelogram,

∴ $(AB \times h) = 119.84</p><p>[tex] \implies h = \frac{119.84}{12}$

【AB = 12 cm】{Given}

$\implies h = 9.98 \: cm$

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