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# Could someone plz help me solve this.

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A triangle has side lengths measuring 30, 40 and 50 units. What is the length of its shortest altitude, in units?

Guest Nov 15, 2017
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#1
+91049
+2

A triangle has side lengths measuring 30, 40 and 50 units. What is the length of its shortest altitude, in units?

Here is the picture.

The altitudes are 30, 40, and h

h is the smallest because h is not the hypotenuse of either of the 2 created triangles.

$$x^2+h^2=30^2 \qquad \qquad h^2+(50-x)^2=40^2\\ x^2+h^2=900 \qquad \qquad h^2+2500-100x+x^2=1600\\ x^2+h^2=900 \qquad \qquad h^2+x^2=1600-2500+100x\\ x^2+h^2=900 \qquad \qquad x^2+h^2=100x-900\\ so\\ 100x-900=900\\100x=1800\\x=18\\~\\ h^2=900-x^2\\h^2=900-18^2\\h^2=576\\h=\sqrt{576}\\h=24\\$$

So the smallest altitude is 24

Melody  Nov 15, 2017
#2
+78753
+3

Thanks,  Melody....here's another approch

Area   = (1/2) 30 * 40  =  (1/2) 50  * h

600  = 25 h

24  = h  = the shortest altitude

CPhill  Nov 15, 2017

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