Area of quadilateral
Answers
Note:
1) Pythagoras theorem:-
According to this theorem, the square of the hypotenuse is equal to the sum of squares of the remaining two sides.
2) In a right angled triangle, the side opposite to the right angle is its hypotenuse.
3) Area of triangle = (1/2)•base•height
4) Area of ∆ABC denoted by ;
Ar(∆ABC).
Solution:-
Clearly,
∆BCD is a right angled triangle,
right angled at B.
Thus,
DC is hypotenuse , BC is base and BD is perpendicular .
Now,
Applying Pythagoras theorem in ∆BCD,
We have;
=> DC^2 = BC^2 + BD^2
=> 17^2 = 8^2 + BD^2
=> BD^2 = 17^2 - 8^2
=> BD^2 = 289 - 64
=> BD^2 = 225
=> BD = √225
=> BD = 15
Thus, length of BD is 15 cm.
Also,
∆ABD is a right angled triangle,
right angled at A
Thus,
BD is hypotenuse , AB is base and AD is perpendicular .
Now,
Applying Pythagoras theorem in ∆ABD,
We have;
=> BD^2 = AB^2 + AD^2
=> AB^2 = BD^2 - AD^2
=> AB^2 = 15^2 - 9^2
=> AB^2 = 225 - 81
=> AB^2 = 144
=> AB = √144
=> AB = 12
Thus, the length of AB is 12 cm.
Now,
Ar(∆BCD) = (1/2)•BC•BD
= (1/2)•8•15
= 4•15
= 60 cm^2
Ar(∆ABD) = (1/2)•AB•AD
= (1/2)•12•15
= 6•15
= 90 cm^2
Also,
Ar(quad.ABCD) = Ar(∆BCD) + Ar(∆ABD)
= (60 + 90) sq. cm.
= 150 cm^2
Hence,
The required area of the given quadrilateral is 150 cm^2