Question

A park in the shape of a quadrilateral ABCD has angle C as 90 degree AB=10m, BC=8m, CD=6m and AD=6m. Prove that area of the quadrilateral is equal to 3(8+91)m^2

Answer 1

Answer:

Join BD in ΔBCD, BC and DC are given.

So, we can calculate BD by applying Pythagoras theorem

⇒BD=

BC

2

+CD

2

=

12

2

+5

2

=

144+25

=13 m=BD

⇒Area of □ABCD= Area of ΔABD+ Area of ΔBCD

⇒Area of ΔBCD

=

2

1

×b×h=

2

1

×12×5

=30 m

2

⇒Area of ΔABD

=

s(s−a)(s−b)(s−c)

(Heron's formula)

⇒2S=9+8+13, S=

2

30

⇒S=15 m

⇒Area of ΔABD

=

15(15−9)(15−8)(15−13)

=

15×6×7×2

=

630×2

=6

1260

=35.49m

2

⇒Area of Park = Quad ABCD

=30+35.49

=65.49 m

2

≈65.5 m

2

solution