In quadrilateral ABCD, angleB = angleD= 90°, AB = 7 cm, AC = 25 cm, CD = 20 cm. Find the area of ABCD.
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Answer:
Correct option is
A
True
Area quadrilateral ABCD= area of triangle ABC+area of triangle ACD
ar△ABC=
s(s−a)(s−b)(s−c)
,s=
2
a+b+c
=
2
9+6+7
=11㎝
∴ar△ABC=
11(11−9)(11−6)(11−7)
=
11×2×5×4
=2
110
㎠
=10.48×2=20.96㎠
ar△ACD=
s(s−a)(s−b)(s−c)
,s=
2
a+b+c
=
2
15+9+12
=18㎝
∴ar△ACD=
18(18−15)(18−9)(18−12)
=
18×3×9×6
=
6×3×3×9×6
=3×3×6=54㎠
∴ar△ABCD=20.96+54=74.96㎠
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