Question

A park is in the shape of a quadrilateral ABCD has angle C is equal to 90 degree , BC is equal to 12 metre, AB is equal to 9cm, CD is equal to 5 metre ,AD equal to 8 metre how much area does it occupy?

Answer 1

Answer:Area of park is 65.49 sq-m

Solution:

See the shape of park in the attachment figure.

To find the area,

1) join BD

2) Now we know that ∆BCD is a right triangle

3) Find the length of DB with the help of Pythagoras Theorem

${(DB)}^{2} = {(DC)}^{2} + {(CB)}^{2} \\ \\ {(DB)}^{2} = {(5)}^{2} + {(12)}^{2} \\ \\DB = \sqrt{25 + 144} \\ \\ DB = \sqrt{169} \\ \\ DB = 13 \: m$
Now,Area of ∆ BCD=

$\frac{1}{2} \times BC \times DC \\ \\ = \frac{1}{2} \times 12 \times 5 \\ \\ = 30 \: {m}^{2}$
4) Area of ∆ ABD can be calculated with the help of Hero's Formula

$2s = 13 + 9 + 8 \\ \\ s = 15 \\ \\area \: \: ABD = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{15(15 - 13)(15 - 8)(15 - 9)} \\ \\ = > \sqrt{15 \times 2 \times 7 \times 6} \\ \\ = > \sqrt{1260} \\ \\ = > 35.49 \: {m}^{2}$

Area of Quadrilateral ABCD=

ar(∆ BCD)+ar(∆ABD)

=> 30+35.49 sq-m

=65.49 sq-m

Hope it helps you.

Note: I had assumed that you were mistakenly write AB= 9 cm

Answer 2

Answer

I hope it will help you.