The sides of a triangle are of lengths 8 cm, 15cm and 17cm. Find the area of the triangle. Find the length of altitude drawn on the side with length 17cm
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Answered by
4
s=(8+15+17)/2=40/2=20
area=√(20-8)(20-15)(20-17)
=√(12×5×3)
=180
1/2×altitude×17=area
1/2×altitude×17=180
altitude=360/7cm
area=√(20-8)(20-15)(20-17)
=√(12×5×3)
=180
1/2×altitude×17=area
1/2×altitude×17=180
altitude=360/7cm
Answered by
7
LET ABC IS A TRIANGLE.
AB = 8CM , AC = 15CM , BC = 17CM
ALL THREE SIDES OF TRIANGLE ARE DIFFERENT IT MEANS TRIANGLE IS ISOSCELES TRIANGLE..
THEREFORE,
S = 1/2(AB+BC+AC)
S = 1/2(8+15+17)
S = 1/2×40
S = 20 CM
THEREFORE,
S - AB = 20-8 = 12CM
S-AC = 20-15 = 5CM
S-BC = 20-17 = 3CM
THEREFORE,
AREA OF TRIANGLE = UNDER ROOT S(S-AB) (S-AC) ( S-BC)
PUT VALUES AND SOLVE U WILL GET THE AREA OF TRIANGLE..
AB = 8CM , AC = 15CM , BC = 17CM
ALL THREE SIDES OF TRIANGLE ARE DIFFERENT IT MEANS TRIANGLE IS ISOSCELES TRIANGLE..
THEREFORE,
S = 1/2(AB+BC+AC)
S = 1/2(8+15+17)
S = 1/2×40
S = 20 CM
THEREFORE,
S - AB = 20-8 = 12CM
S-AC = 20-15 = 5CM
S-BC = 20-17 = 3CM
THEREFORE,
AREA OF TRIANGLE = UNDER ROOT S(S-AB) (S-AC) ( S-BC)
PUT VALUES AND SOLVE U WILL GET THE AREA OF TRIANGLE..
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