Question

find the area of a parallelogram ABCD in which AB equal to 14 cm BC is equals to 10 cm and ac equal to 16 cm given root 3 value equals to 1.73

Answer 1

Let ABCD be the given parallelogram.

Area of parallelogram ABCD

= area of ABC + area of ACD

= 2 area of ABC

Let, (AB) a = 14 cm, (BC) b = 10 cm, (AC) c = 16 cm

s=[a+b+c]/2  =[14+16+10]/2 =20

area of triangle ABC= √[s(s-a)(s-b)(s-c)]

area of triangle ABC= √[20(20-14)(20-10)(20-16)]

area of triangle ABC=  √[20*6*10*4]

area of triangle ABC= [5*4*2* √3]sq cm

area of triangle ABC= [40 ×√3]sq cm

Therefore, area of parallelogram ABCD

= 2 × area of triangle ABC = 2×40 √3

Therefore, area of parallelogram ABCD= 80 √3

=80×1.732  =138.56 sq cm