Question

. A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20

m, 26 m and 17 m and the angle between the first two sides is a right angle.

Find the area of the park
​

Answer 1

Answer:

area of the park would be 360

Step-by-step explanation:

The area of the park = 390 meters²

Step-by-step explanation:

Area of the garden ABCD = Area of ΔABC + Area of ΔACD

To find Area of ΔABC : First find length of AC

Since, the angle between first two sides of ABCD is right angle

⇒ m∠ABC = 90°

By using Pythagoras theorem in ΔABC :

AC² = AB² + BC²

AC² = 15² + 20²

⇒ AC = 25 m

\begin{gathered}\text{Area of }\Delta ABC = \frac{1}{2}\times Base\times Height\\\\Area=\frac{1}{2}\times 15\times 20\\\\\bf\implies\textbf{Area of }\Delta ABC = 150\textbf{ meters square}\end{gathered}

Area of ΔABC=

2

1

×Base×Height

Area=

2

1

×15×20

⟹Area of ΔABC=150 meters square

To find Area of ΔACD : we use Heron's formula :

Area=\sqrt{s(s-a)(s-b)(s-c)}Area=

s(s−a)(s−b)(s−c)

where s is the semi-perimeter of the triangle, and a, b, c are length of three sides of the triangle.

\begin{gathered}s=\frac{17+26+25}{2}=34\thinspace m\\\\\text{Area of }\Delta ACD =\sqrt{34(34-17)(34-26)(34-25)}\\\\\bf\implies\textbf{Area of }\Delta ABC = 204\textbf{ meters square}\end{gathered}

s=

2

17+26+25

=34m

Area of ΔACD=

34(34−17)(34−26)(34−25)

⟹Area of ΔABC=204 meters square

Area of the park ABCD = Area of ΔABC + Area of ΔACD

= 150 + 240

= 390 meters square

Hence, The area of the park = 390 meters²