Question

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Given right triangle ABC with altitude BD drawn from vertex B to hypotenuse AC, find the lengths of AD, BD, and CD.
Keep as improper fraction

Answer 1

Answer:

AD = 3

BD = 10

CD = 17

Step-by-step explanation:

Answer 2

The requires side lengths are:

$AD = \frac{49}{25}$

$BD = \frac{168}{25}$

$CD = \frac{576}{25}$

Considering triangle BDC, we have:

$BC^2 = BD^2 +CD^2$

This gives

$24^2 = BD^2 +(25 - AD)^2$

$576 = BD^2 +(25 - AD)^2$ --------- (1)

Considering triangle ABD, we have:

$7^2 = BD^2 + AD^2$

$49 = BD^2 + AD^2$ ---- (2)

Subtract (2) from (1)

$576 - 49 = BD^2 -BD^2 +(25 - AD)^2 - AD^2$

$527 = (25 - AD)^2 - AD^2$

Expand

$527 = 625 - 50AD + AD^2 - AD^2$

$527 = 625 - 50AD$

Collect like terms

$50AD = 625 - 527$

$50AD = 98$

Divide both sides by 50

$AD = \frac{98}{50}$

Reduce fraction

$AD = \frac{49}{25}$

Recall that:

$49 = BD^2 + AD^2$

So, we have:

$49 = BD^2 + (49/25)^2$

$49 = BD^2 + 2401/625$

Collect like terms

$BD^2 = 49 - 2401/625$

Take LCM

$BD^2 = \frac{625 *49 - 2401}{625 }$

$BD^2 = \frac{28224}{625}$

Take the square roots of both sides

$BD = \frac{168}{25}$

Also, we have:

$CD = 25 - AD$

This gives

$CD = 25 - \frac{49}{25}$

Take LCM

$CD = \frac{625 - 49}{25}$

$CD = \frac{576}{25}$

Hence, the requires side lengths are:

$AD = \frac{49}{25}$

$BD = \frac{168}{25}$

$CD = \frac{576}{25}$

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