Find the area of a right angled triangle, whose hypotenuse is 25 and perimeter is 60
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Let, base be b and perpendicular be p,
Then,b+p+25=60
:- b+p=35 - equation 1
Also, b^2+p^2=25^2
b^2+p^2=625 - equation 2
Squaring equation 1
b^2+p^2+2bp = 1225
:-b^2+p^2=1225-2bp -equation3
comparing equation 2&3
625=1225-2bp
2bp=600
bp=300 - equation 4
Since, area of right triangle:-
1/2 × bp,
From equation 4
:- 1/2 × 300
:- 150
Hence area = 150
Then,b+p+25=60
:- b+p=35 - equation 1
Also, b^2+p^2=25^2
b^2+p^2=625 - equation 2
Squaring equation 1
b^2+p^2+2bp = 1225
:-b^2+p^2=1225-2bp -equation3
comparing equation 2&3
625=1225-2bp
2bp=600
bp=300 - equation 4
Since, area of right triangle:-
1/2 × bp,
From equation 4
:- 1/2 × 300
:- 150
Hence area = 150
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