A circle is inscribed in a triangle ABC , such that it touches the side AB, BC,AND CA at point D, E and F respectively. If length of one side AB,BC and CA AREA 12cm, 8cm and 10am respectively, find the length of AD,BE And CF .
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We know that tangents drawn to an external point are equal.
So,AD=AF=x
BD=BE=y
CE=CF=z
Now, AB=12cm; BC=8cm; AC=10cm
We get, x+y=12; y+z=8 and z+x=10
(x+y) + (y+z) + (z+x)= 12+ 8+ 10
2(x+y+z)=30
x+y+z=15
We already know that ,
x+y=12 and x+y+z=15. SO we get that(by substituting) 12+z=15 =z=3
Similarly we can find the rest
The end ans: x=7; y=5; z=3
So,AD=AF=x
BD=BE=y
CE=CF=z
Now, AB=12cm; BC=8cm; AC=10cm
We get, x+y=12; y+z=8 and z+x=10
(x+y) + (y+z) + (z+x)= 12+ 8+ 10
2(x+y+z)=30
x+y+z=15
We already know that ,
x+y=12 and x+y+z=15. SO we get that(by substituting) 12+z=15 =z=3
Similarly we can find the rest
The end ans: x=7; y=5; z=3
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