perimeter of triangle is 60 centimetre is hypotenuse is 25 cm find the area of the angle
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perimeter of triangle=60
hypo.+sum of two side=60
hypo.=25
sum of two side =35
let the one side be a
and other side =35-a
in a right angle triangle by phytha . gor. thm.
a^2+(35-a)^2=25^2
a^2+1225+a^2-70a=625
2a^2-70a+600=0
a^2-35a+300=0
a^2-20a-15a+300=0
a(a-20)-15(a-20)=0
(a-20)(a-15)=0
a=20,15
Area of triangle= 1/2 B×H
=1/2×20×15
=150cm^2
hypo.+sum of two side=60
hypo.=25
sum of two side =35
let the one side be a
and other side =35-a
in a right angle triangle by phytha . gor. thm.
a^2+(35-a)^2=25^2
a^2+1225+a^2-70a=625
2a^2-70a+600=0
a^2-35a+300=0
a^2-20a-15a+300=0
a(a-20)-15(a-20)=0
(a-20)(a-15)=0
a=20,15
Area of triangle= 1/2 B×H
=1/2×20×15
=150cm^2
Answered by
1
Hypotenuse^2 = Base^2 + Altitude ^2
C^2 = A^2 + B^2 .....(1)
C = 25 cm
Perimeter = A + B + C = 60
A + B + 25 = 60
A + B = 35 ....(2)
(1) ==> C^2 = A^2 + B^2
We know that ( a + b )^2 = a^2 + b^2 + 2ab
Then, a^2 + b^2 = ( a +b )^2 - 2ab
So applying this theory in (1)
C^2 = ( A + B )^2 - 2AB
Substitute the value of C and the value of A + B from (2)
25^2 = 35^2 - 2AB
2AB = 1225 - 625
2AB = 600
AB = 300
Area = 1/2 × AB
so Area = 1/2 × 300 = 150 cm^2
C^2 = A^2 + B^2 .....(1)
C = 25 cm
Perimeter = A + B + C = 60
A + B + 25 = 60
A + B = 35 ....(2)
(1) ==> C^2 = A^2 + B^2
We know that ( a + b )^2 = a^2 + b^2 + 2ab
Then, a^2 + b^2 = ( a +b )^2 - 2ab
So applying this theory in (1)
C^2 = ( A + B )^2 - 2AB
Substitute the value of C and the value of A + B from (2)
25^2 = 35^2 - 2AB
2AB = 1225 - 625
2AB = 600
AB = 300
Area = 1/2 × AB
so Area = 1/2 × 300 = 150 cm^2
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