Question

1. A park in the care of a quadrilateral ABCD. hasC=90. AB=0.30-
2. Find the area of a quadrilateral ABCD in which AB = 3 cm. BC=4c

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