Sides of a quadrilateral ABCD are 6cm,8cm,12cm,14cm respectively and angle B=90°. Find area of quadrilateral ABCD
Answers
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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