Question

solve it:in a parallelogram abcd. ab=16cm bc=12 cm and diagonal ac=20cm .find the area of the parallelogram.(with solution)

plzz as fast as u can

Answer 1

Let AC and BD meets at O. sides of OAB are 8,10 and 12 Area of triangle OAB, =) √15(15-8)(15-10)(15-12) =)15√7 =)(1/2)x12xOM, where OM is perpendicular to AB, OM=(5√7)/2 Distance between AB and CD=5√7 AREA of ABCD=2 x AREA of BCD =2(1/2)x12x5√7 =60√7 units

Answer 2

Question

1. In parallelogram ABCD; AB = 16 cm , BC = 12 cm and diagonal AC = 20 cm. Find the area of the parallelogram.

Solution

For triangle ABC,

AB = 16 cm, AC = 20 cm and BC = 12 cm

$\sf{s=\dfrac{a+b+c}{2}=\dfrac{16+20+12}{2}=24cm}$

$\sf{:⟹ Area \: of \: ∆ABC= \sqrt{s(s-a(s-b)(s-c)} }$

$\sf{= \sqrt{24(24-16)(24-20)(24-12)} }$

$\sf{= \sqrt{24×8×4×12} = \sqrt{9216} =96 {cm}^{2} }$

$\sf{✿Area \: of \: parallelogram \: ABCD=2×96 {cm}^{2} =192 {cm}^{2} }$