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# help

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The legs of a right triangle are 8 and 15.  Find the altitude to the hypotenuse.

Jun 10, 2020

#1
+277
+1

Hello! I'm working on the problem now. There's quite a bit of steps. However, it is possible. Don't get discouraged. First you need to determine the length of the hypotenuse using the Pythagorean theorem. It's length is 17.

From there, I looked up how to do altitude and you need the height of the triangle so I decided to use this formula:
a = 1/2 * bh

Before using this formula, you need to calculate the area first. For this, I used Heron's Formula. You need to find the semi perimeter, so to get this, add all of the sides together & divide by 2. You should get 20. Then I got the area by using Heron's Formula:

$$\sqrt{20(20-17)(20-15)(20-8)} = \sqrt{20(3)(5)(12)} = \sqrt{3600} = 60$$

Now that you have your area, 60, you can use the formula above, a = 1/2 * bh. Just substitute your area into the equation and solve for you height.

$$60 = \frac{1}{2}(14)h$$

$$60 = 7h$$

$$h = 8.57$$

(Will continue... still solving)

Jun 11, 2020
edited by auxiarc  Jun 11, 2020
#2
+21953
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I'll attempt this problem in a different way ...

Draw right triangle(ABC) with C the right angle.

Let AC = 8 and BC = 15.

Using the Pythagorean Theorem AB = 17.

Drop a perpendicular from angle(C) to the hypotenuse AB.

Label the point of intersection P.

PC is the height.

By similar triangles (or a theorem from your textbook):

PA / AC  =  AC / AB    --->     PA / 8  =  8 / 17        --->     PA  =  64 / 17

PB / BC  =  BC / BA    --->     PB /15  =  15 / 17     --->     PB =  225 / 17

Continuing with similar trianges (or another theorem):

AP / PC  =  PC / PB     --->     ( 64 / 17 ) / PC  =  PC / ( 225 / 17 )

cross-multiplying:     ( 64 / 17 ) · ( 225 / 17 )  =  PC2

( 64 · 225 ) / ( 17 · 17 )  =  PC2

14,400 / 289  =  PC2

120 / 17  =  PC

as a decimal:  7.0588...

Jun 11, 2020
#3
+25541
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The legs of a right triangle are 8 and 15.  Find the altitude to the hypotenuse.

Formula: $$h_c = \dfrac{ab}{c}$$ (right triangle)

$$\begin{array}{|rcll|} \hline \mathbf{h_c} &=& \mathbf{\dfrac{ab}{c}} \quad | \quad c = \sqrt{a^2+b^2} \\\\ h_c &=& \dfrac{ab}{\sqrt{a^2+b^2}} \\\\ h_c &=& \dfrac{8*15}{\sqrt{8^2+15^2}} \\\\ h_c &=& \dfrac{120}{\sqrt{289}} \\\\ \mathbf{h_c} &=& \mathbf{\dfrac{120}{17}} \\ \hline \end{array}$$

Jun 11, 2020
edited by heureka  Jun 11, 2020
#4
+8340
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Have you heard of the "inverse Pythagorean theorem"?

Let a and b be the legs of a right triangle, and h be the altitude to the hypotenuse.

The inverse Pythagorean theorem states that $$h^{-2} = a^{-2} + b^{-2}$$.

We can use this equation to solve this problem.

Let h be the altitude to the hypotenuse.

$$\dfrac1{h^2} = \dfrac1{8^2} + \dfrac1{15^2}\\ h^2 = \dfrac{120^2}{8^2 + 15^2} = \left(\dfrac{120}{17}\right)^2\\ h = \dfrac{120}{17}$$

Jun 11, 2020