Question

The sides of quadrilateral ABCD are 6 CM 8 cm 12 cm and 14 cm and the angle between the first two sides is right angle find its area

Answer 1

Let AB = 6cm , BC= 8 cm , CD = 12 cm , and AD = 14 cm

angle B be right angle

now join diagonal AC

therefore quadrilateral ABCD now devides into two triangles ABC and ACD

The sum of areas of these two triangles is area of quadrilateral

now area of triangle ABC = 1/2 × 8 ×6

= 24 sq cm

to find area of triangle ACD we use Heron's formula

the sides of triangle ACD are as

AD =14 cm. and CD = 12 cm

the third side AC will be 10 cm can be find using Pythagoras theorem in triangle ABC

So ,

s = 10 +12 + 14 /2

= 36 /2

18

by Heron's formula ,

area of triangle ACD = √ s(s-a)(s-b)(s-c)

= √18 *(18-10)(18-12)(18-14)

= √18 ×8 ×6 ×4

= √576 × 6

= 24√6

= 24 × 2.45

= 58.8 sq cm

therefore area of rectangle is , 24 + 58.8= 82.8 sq cm