Question
Find the area of a quadrilateral whose sides are 9 cm, 40 cm, 28cm and 15 cm and the angle between the first two sides is a right angle
Hint:
Using pythagoras theorem, find the length of side AC.
area of quadrilateral ABCD = area of Δ ABC + area of Δ ADC
Find area of area of Δ ADC by heron's formula for finding the area of triangle
The correct answer is: 306 m2
Ans :- 306 (in sq.centimeters)
Explanation :-
Step1 :- Find the length of side Ac using pythagoras theorem,
In Δ ABC , AC2 = AB2 + BC2
We know that AB = 9 and BC = 40
Substituting the values we get AC2 = 92 + 402 = 81+1600 = 1681 =412
We get AC = 41 cm
Step2:- Find the area of Δ ABC,
Here,Δ ABC is a right angle triangle .
Area of right angle triangle = ½ (product of lengths of perpendicular sides)
Area of Δ ABC = ½ (40 × 9)
Area of Δ ABC = 180 cm2
Step 3:- Find the area of Δ ADC from heron's formula
Here, side length of triangle are 15,28 and 41 cm
Heron's formula for area of triangle with side length
a,b,c are Area
Where s
Let a = 28 ;b = 15 ;c = 41 We get s42
Area of Δ ADC =
=
=
We get area of Δ ADC = 126 (cm2)
Step 4:- Find the area of the Patch
area of patch ABCD = area of Δ ABC + area of Δ ADC
= 180 cm2 + 126 cm2
= 306 cm2
Therefore area of patch ABCD = 306 cm2.
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