a field is in shape of a quadrilateral ABCD in which side a b is equals to 18 m a d is equals to 24 m b c is equals to 40 m b c is equals to 50 M and Angle A is equals to 90 degree find the area of the field
Answers
The quadrilateral ABCD is divided into two triangles. ABD and BCD.
ABD is a right angle triangle.
Area = 1/2 AB * AD as angle A is 90, AB is the altitude and AD is the base.
= 1/2 * 18 * 24 = 216 cm^2
BC^2 = AB^2 + AD^2 => BC = 30 cm
BCD triangle
s = semi perimeter = 60 cm
area = sqroot(60 * 30 * 20 * 10 ) = 600 cm^2
total area = 816 cm^2
Answer:
816 m²
Step-by-step explanation:
We have,
AB = 18 m, AD = 24 m, BC = 40 m, DC = 50 m, ∠A = 90°.
BD is joined.
In Δ BCD,
By applying Pythagoras theorem,
⇒ BD² = AB² + AD²
⇒ BD² = 18² + 24²
⇒ BD² = 900
⇒ BD = 30 m.
(i)
Area of ΔABD = (1/2) * base * height
= (1/2) * 24 * 18
= 216 m²
(ii)
Now,
Semi-perimeter of BCD = (50 + 40 + 30)/2
= 120/2
= 60 m
Using heron's formula:
Area of ΔBCD = √s(s - a)(s - b)(s - c)
= √60(60 - 30)(60 - 40)(60 - 50)
= √60 * 30 * 20 * 10
= √360000
= 600 m²
∴ Area of Quadrilateral ABCD = Area of ΔABD + Area of ΔBCD
= 216 + 600
= 816 m²
Therefore, Area of the field = 816 m².
Hope it helps!
