Question

A park is in the shape of a quadrilateral. The sides of the park are 15m,20m,26m and 17m and the angle between the first two sides os a right angle. Find the area of the park.

Answer 1

Answer:

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

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

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

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

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

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

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

                                           = 150 + 240

                                           = 390 meters square

Hence, The area of the park = 390 meters²

Answer 2

Answer:

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

\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}

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

Area=\sqrt{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.

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}

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

= 150 + 240

= 390 meters square

Hence, The area of the park = 390 meters²