A park, in the shape of quadrilateral ABCD ,has Angle C = 90 degree ,

1.	A park, in the shape of quadrilateral ABCD ,has Angle C = 90 degree , AB = 9 m , BC = 12 m, CD = 5m and AD = 8 m. how much area does it occupy.
Answer» 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

A park, in the shape of quadrilateral ABCD ,has Angle C = 90 degree , AB = 9 m , BC = 12 m, CD = 5m and AD = 8 m. how much area does it occupy.

Answer»

Answer:

Join BD in ΔBCD, BC and DC are GIVEN.

So, we can calculate BD by APPLYING Pythagoras theorem

⇒BD=

+CD

144+25

=13 m=BD

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

⇒Area of ΔBCD

×b×h=

×12×5

=30 m

⇒Area of ΔABD

s(s−a)(s−b)(s−c)

(Heron's formula)

⇒2S=9+8+13, S=

⇒S=15 m

⇒Area of ΔABD

15(15−9)(15−8)(15−13)

15×6×7×2

630×2

1260

=35.49m

⇒Area of Park = Quad ABCD

=30+35.49

=65.49 m

≈65.5 m