In the given figurw ,AC=12cm,BC=6cm,AD=8 cm . find the area of triangle abc and also the value of h
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ADC is a right angle triangle .
DC^2 = AC^2 - AD^2
= 144 - 64 = 80
DC = 4* sqrt(5) cm
Area of triangle ADC = 8* 4* sqrt(5)/2 cm^2
= 16 sqrt(5) cm^2.
DB = (4 sqrt(5) - 6) cm
Area of triangle ADB = 8*(4*sqrt5 -6)/2 = 16 sqrt5 - 24 cm^2
Area(triangle ABC)= 24 cm^2
( base = BC=6 cm. Height AD= 8cm)
h = 2*24 /12 = 4 cm.
DC^2 = AC^2 - AD^2
= 144 - 64 = 80
DC = 4* sqrt(5) cm
Area of triangle ADC = 8* 4* sqrt(5)/2 cm^2
= 16 sqrt(5) cm^2.
DB = (4 sqrt(5) - 6) cm
Area of triangle ADB = 8*(4*sqrt5 -6)/2 = 16 sqrt5 - 24 cm^2
Area(triangle ABC)= 24 cm^2
( base = BC=6 cm. Height AD= 8cm)
h = 2*24 /12 = 4 cm.
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4
Answer is 4 cm.......
Prashantkumar546:
with explanation
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