in the adjoining figure ABC is an isosceles triangle with a b equal to AC I go to 7.5 cm and BC equal to 9 CM if the height ID from is a to b c is 6 cm find the area of triangle ABC the height from sea to a b
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by heron's formula
S=(a+ b+ c)/2
=(7.5+7.5+9)/2
=(15+9)/2
=(24)/2
=12
area of triangle=√S(S-a)(S-b)(S-c)
=√12(12-7.5)(12-7.5)(12-9)
=√12(5.5)(5.5)(3)
=√3×2×2×5.5×5.5×3
=3×2×5.5
=33 sq. cm
S=(a+ b+ c)/2
=(7.5+7.5+9)/2
=(15+9)/2
=(24)/2
=12
area of triangle=√S(S-a)(S-b)(S-c)
=√12(12-7.5)(12-7.5)(12-9)
=√12(5.5)(5.5)(3)
=√3×2×2×5.5×5.5×3
=3×2×5.5
=33 sq. cm
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