Question

The sides of a quadrilateral taken in order are 5 m, 12 m, 14 m and 15 m respectively. The angle contained in the first two sides is a right angle. Find the area of the quadrilateral.
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Answer 1

Answer:

In quad. ABCD, AB=5 m, BC= 12 m CD= 14m, DA = 15m and ∠ABC=900 Join AC, 

Now in right ΔABC,

AC2=AB2+BC2=(5)2+(12)2

= 25+144=169=(13)2

∴AC=13m

Now area of right ΔABC

= 12×base×height

= 12×AB×BC

=12×5×12m2=30m2

and in ΔACD, sides are 13m, 14m, 15m

∴s=a+b+c2=13+14+152=422=21m

Now area of ΔACD

= √s(s−a)(s−b)(s−c)

= √21(21−13)(21−14)(21−15)

= √21×8×7×6.

= √3×7×2×2×2×7×2×3.

= 7×3×2×2=84m2

∴ Area of quad. ABCD = 30+84= 114m2