Given triangle IMJ with altitude JL, JL = 32, and IL =24, find IJ, JM, LM, and IM.
https://api.agilixbuzz.com/Resz/~D_77DAAAAAg4s38iYZ7NxB.W6wAY0-_E7ysj8j7iXQi4C/63669058,F27,0,0,21/Assets/Screen%20Shot%202017-11-08%20at%2010.46.51%20AM.png

Answer :

check the picture below.

keeping in mind that, an altitude coming from a right-angle, like that in that triangle, will make two smaller "similar" triangles, thus you'd end up with 3 similar triangles, the Large one containing the two small ones, the Medium triangle and the Small one.

now, let's use those ratios to get first "x", and then "y" and "z",

[tex]\bf \cfrac{small}{medium}\qquad \cfrac{32}{x}=\cfrac{24}{32}\implies \cfrac{32}{x}=\cfrac{3}{4}\implies \cfrac{32\cdot 4}{3}=x\implies \boxed{\cfrac{128}{3}=x}\\\\ -------------------------------\\\\ \cfrac{small}{medium}\qquad \cfrac{24}{32}=\cfrac{z}{y}\implies \cfrac{3}{4}=\cfrac{z}{y}\implies y=\cfrac{4z}{3} \\\\\\ \cfrac{large}{small}\qquad \cfrac{24+x}{z}=\cfrac{y}{32}\implies \cfrac{(24+x)32}{y}=z\implies \cfrac{\left( 24+\frac{128}{3} \right)32}{\frac{4z}{3}}=z[/tex]

[tex]\bf \cfrac{\left( \frac{200}{3} \right)32}{\frac{4z}{3}}=z\implies \cfrac{\frac{6400}{3}}{\frac{4z}{3}}=z\implies \cfrac{6400}{\underline{3}}\cdot \cfrac{\underline{3}}{4z}=z\implies \cfrac{1600}{z}=z \\\\\\ 1600=z^2\implies \sqrt{1600}=z\implies \boxed{40=z} \\\\\\ \textit{now, we know that } y = \cfrac{4z}{3}\implies y=\cfrac{4(40)}{3}\implies \boxed{y=\cfrac{160}{3}}[/tex]
${teks-lihat-gambar} jdoe0001

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