the sides of a quadrilateral ABCD taken in order are 6 CM 8 cm 12 cm and 14 cm respectively and the angle between the first two sides is a right angle find its area
Answers
Answer:
area of the rectangle = 82.8 sq m
Step-by-step explanation:
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 divides 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