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Question
In each of the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segment are marked in each figure. Determine x, y, z in each case.
Solution
Δ ABC is right angled triangle right angled at B
`AB^2+BC^2= AC ^2`
`x^2+z^2=(4+5)^2`
`x^2+z^2=9^2`
`x^2+z^2=81`...............(1)
Δ BAD is right triangle right angled at D
`BD^2+AD^2=AB^2`
`y^2+4^2=x^2`
`y^2+16=x^2`
`16=x^2-y^2`...............(2)
Δ BDC is right triangle right angled at D
`BD^2+DC^2=BC^2`
`y^2+25=z^2`
`25 = z^2-y^2`..................(3)
By canceling equation (1) and (3) by elimination method, we get
y canceling and by elimination method we get
`z^2=90/2`
`z^2=45`
`z=sqrt45`
`z=sqrt(3xx3xx5)`
`z=3sqrt5`
Now, substituting `z^2=45` in equation (iv) we get
`y^2+z^2=65`
`y^2+45=65`
`y^2=65-45`
`y^2=20`
`y = sqrt20`
`y=sqrt(2xx2xx5)`
`y=2sqrt5`
Now, substituting `y^2=20` in equation (ii) we get
`x^2-y^2=16`
`x^2-20=16`
`x^2=16+20`
`x^2=36`
`x=sqrt36`
`x=sqrt(6xx6)`
`x = 6`
Hence the values of x, y, z is `6,2sqrt5,3sqrt5`
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