Two sides of a triangle are 15 cm. The lenght of the altitude to the smaller side is 10 cm. Find the area of the triangle
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AB=AC = 15cm
angle ABC=angle ACB=x
AD=10cm
In∆ADC , angle ADC=90°
By Pythagorus Theorem,
AC²= AD²+DC²
15² = 10² +DC²
225-100=DC²
DC²=125
DC=√125
DC = 5√5
BC = BD + DC
= 5√5 + 5√5
= 10√5
Base = 10√5 and height = 10
Area of triangle = 1/2×base ×height
= 1/2×10√5×10
=10√5×5
= 50√5cm²
angle ABC=angle ACB=x
AD=10cm
In∆ADC , angle ADC=90°
By Pythagorus Theorem,
AC²= AD²+DC²
15² = 10² +DC²
225-100=DC²
DC²=125
DC=√125
DC = 5√5
BC = BD + DC
= 5√5 + 5√5
= 10√5
Base = 10√5 and height = 10
Area of triangle = 1/2×base ×height
= 1/2×10√5×10
=10√5×5
= 50√5cm²
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