Question

The length of four sides of a quadrilateral are 5cm, 12cm, 14cm,and 15 cm. The angle between first two adjacent sides is 90°. calculate the area of the quadrilateral​

Answer 1

Answer:

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The sides of a quadrilateral ABCD are 6cm,8cm,12cm and 14cm (taken in order).

i.e. ∠ABC=90o

Now, let us join the points A & C.

So, we get two triangles namely △ABC (a right angled triangle) and △ACD.

Applying Pythagoras theorem in △ABC, we get

AC=AB2+BC2

=62+82

=36+64

=100

=10cm

So, the area of △ABC=21×base×height

=21×AB×BC

=21×6×8=24

Now, in △ACD, we have

AC=10cm,CD=12cm,AD=14cm.

According to Heron's formula the area of triangle (A)=[s(s−a)(s−b)(s−c)]

where, 2s=(a+b+c).

Here, a=10cm,b=12cm,c=14cm

s=2(10+12+14)=236=18

Area of △ACD=[18×(18−10)(18−12)(18−14)]

=(18×8×6×4)

=(2×3×3×2×2×2×2×3×2×2)

=[(2×2×2×2×2×2×3×3)×2×3]

=2×2×2×3×6

=246

So, total area of quadrilateral ABCD =△ABC+△ACD

=24+246

=24(6

+1)