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A park, in the shape of a quadrilateral ABCD, has ?C = 90? , AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy?
Answer :

In the quadrilateral ABCD,
we have, right angled triangle ?BCD,

image

We have,
BD2=BC2+CD2
(By Pythagoras theorem)

= 122+52=144+25=169
? BD2=132
? BD = 13m

Also, In ?ABD, we have,

AB = 9m, BD = 13m, DA = 8m

Now, we know that,
s=a+b+c2
? s=9+13+82=302=15m

Now, Area of triangle
= 15(15?9)(15?13)(15?8)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 15×6×2×7
= 3×5×3×2×2×7
= 3×25×7m2
= 635m2 = 6 × 5.9m2 = 35.4m2(approx) ...(i)

Since, ?BCD is an right angled triangle,

Area of ?BCD = 12 × BC × CD
(Since, Area of triangle = 12 × Base × Height)

= 12×12×5=30m2 ...(ii)

Area of quadrilateral ABCD
= Area of ?ABD + Area of ?BCD

Hence from (i) and (ii),

Area of quadrilateral ABCD = 35.4m2+30m2=65.4m2