The perimeter of a right triangle is 24 cm. If its hypotenuse is 10 cm, find its area.
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Let the side be a, b, c and let c be the hypotenuse and a, b will be base or height
then a+b+c = 24
a+b+10 = 24
a+b = 24-10 = 14 ----------------- 1
a2+b2 = 102 ---------------- 2
From 1 we get a = 14-b
put the value of a in equation 2
(14-b)2+b2 = 100
196-28b+b 2+b 2 =100
2b2-28b+196-100 = 0
b2-14b+48 = 0
(b-6)(b-8) = 0 so b =6 or 8 then we will get a= 8 or 6 using equation 1
so the area of triangle is 1/2 bxh = 1/2 x 8x 6 = 24 cm2
then a+b+c = 24
a+b+10 = 24
a+b = 24-10 = 14 ----------------- 1
a2+b2 = 102 ---------------- 2
From 1 we get a = 14-b
put the value of a in equation 2
(14-b)2+b2 = 100
196-28b+b 2+b 2 =100
2b2-28b+196-100 = 0
b2-14b+48 = 0
(b-6)(b-8) = 0 so b =6 or 8 then we will get a= 8 or 6 using equation 1
so the area of triangle is 1/2 bxh = 1/2 x 8x 6 = 24 cm2
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