Question

Find the area of the quadrilateral ABCD whose sides are 5m, 12m, 15m, and 14m and angle between two first sides is 90°.

Answer 1

Hope dis might b da correct ans.....

Answer 2

Answer:

114 m^2

Step-by-step explanation:

In triangle ABC

=1 /2 × 5× 12

= 30m^2

but

AC = √ AB ^2 +BC ^2

= √5 ^2 + 12 ^2

=√ 144 + 25

=√ 169

= 13 m

Now ,

In triangle ACD ,

s = a +b+c /2

= 14 + 13 + 15 / 2

= 42 / 2

= 21 m

Now,

Area of triangle = √ s (s-a) (s-b) (s-c)

=√ 21 ( 21 -13)(21-15)(21-14)

=√21 ×8× 6×7

=√ 7×3×2×2×2×2 ×3×7

=√ 7^2 × 3^ 2 × 2^2 × 2^2

= 7× 3 ×2×2

= 84 m^2

Now Read Area = 84 m^2 + 30 m^2

=114 m^2

Here is your answer ..