A park is in the shape of quadrilateral the sides of the park are 15m,20m,26m,17m and the angle between the first two sides is a right angle .Find the area of the park
Answers
Answer:
The area of the park =
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
Area of Triangle ABC= 1/2× BC× AB
=1/2×20×15=150 sqmt.
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={17+26+25}{2}=34
Therefore,
area of triangle ACD=sqrt{34(34-17)(34-26)(34-25)}
=240 sqmt.
Area of the park ABCD = Area of ΔABC + Area of ΔACD
= 150 + 240
= 390 meters square
Hence, The area of the park = 390 meters²