Question

in the diagram below, $\triangle ijk \sim \triangle lmk$. find $r$.

diagram with right triangles: $\triangle ijk$ with legs $ij = 6$ m, $jk$ (vertical), right angle at $j$; $ik = 10$ m. $\triangle lmk$ with leg $lm = 12$ m, $mn$ (vertical, right angle at $m$), $lk = r$.

write your answer as a whole number or a decimal. do not round.
$r = \square$ meters

Explanation:

Step1: Use the property of similar triangles

Since \(\triangle LJK\sim\triangle LMK\), the ratios of corresponding sides are equal. That is \(\frac{LJ}{LM}=\frac{LK}{LJ}\).

Step2: Calculate the length of \(LJ\)

First, find \(LJ\). We know that \(LJ = 10 + r\).

Step3: Substitute into the proportion

Substitute into \(\frac{LJ}{LM}=\frac{LK}{LJ}\), we get \(\frac{10 + r}{12}=\frac{12}{10 + r}\). Cross - multiply gives \((10 + r)^2=12\times12\). Expand the left - hand side: \(100+20r+r^{2}=144\). Rearrange to a quadratic equation: \(r^{2}+20r + 100-144=0\), so \(r^{2}+20r - 44 = 0\).
Another way: Using the geometric mean theorem for right - angled triangles (since \(\triangle LJK\sim\triangle LMK\) and \(\angle N = 90^{\circ}\), \(\angle J=90^{\circ}\)). The length of the leg of a right - angled triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg.
We know that \(LJ^{2}=LK\times LT\) (where \(LT = 10 + r\), \(LK = 12\)). Wait, no, more accurately, for similar right - angled triangles \(\triangle LJK\) and \(\triangle LMK\) (right - angled at \(J\) and \(N\) respectively), we have \(\frac{LJ}{LM}=\frac{LK}{LJ}\).
Cross - multiplying: \(LJ^{2}=LM\times LK\). Let \(LJ=x\), then \(x^{2}=12\times(10 + x)\). But we can also use the fact that if \(\triangle IJK\sim\triangle LMK\), and by Pythagoras in \(\triangle IJK\), \(IJ = 8\), \(IK = 10\), then \(JK=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6\).
Since \(\triangle LJK\sim\triangle LMK\), \(\frac{LM}{LJ}=\frac{LJ}{LK}\). \(LM = 12\), \(LK=r\), \(LJ = 10 + r\). Wait, no, correct proportion:
For similar triangles \(\triangle LJK\) and \(\triangle LMK\) (right - angled), \(\frac{LM}{LJ}=\frac{LJ}{LK}\).
We know that in \(\triangle IJK\), \(IJ = 8\) (by Pythagoras \(IJ=\sqrt{10^{2}-6^{2}} = 8\)).
Since \(\triangle LJK\sim\triangle LMK\) (AA similarity: \(\angle L=\angle L\), \(\angle J=\angle N = 90^{\circ}\)), we have \(\frac{LM}{LJ}=\frac{LJ}{LK}\).
\(LM = 12\), \(LK=r\), \(LJ\) can be found from \(\triangle IJK\) (where \(IJ = 8\), \(IK = 10\), \(JK = 6\)) and using the property of similar triangles.
The correct proportion is \(\frac{LM}{LJ}=\frac{LJ}{LK}\). Let \(LJ=x\), then \(x^{2}=12\times(10 + x)\) is wrong. Wait, no, using the property of similar right - angled triangles:
If \(\triangle LJK\sim\triangle LMK\) (right - angled), then \(\frac{LM}{LJ}=\frac{LJ}{LK}\).
We know that \(LM = 12\), \(LK=r\), and \(LJ\) (the hypotenuse of \(\triangle IJK\) and part of the hypotenuse of \(\triangle LMK\)).
Another approach:
Since \(\triangle IJK\sim\triangle LMK\) (AA: \(\angle IKJ=\angle LKM\) (common angle), \(\angle IJK=\angle LMK = 90^{\circ}\)).
\(\frac{IJ}{LM}=\frac{IK}{LK}\).
We know \(IJ = 8\) (by Pythagoras in \(\triangle IJK\): \(IJ=\sqrt{IK^{2}-JK^{2}}=\sqrt{10^{2}-6^{2}}=\sqrt{100 - 36}=8\)), \(LM = 12\), \(IK = 10\), \(LK=r\).
Substitute into \(\frac{IJ}{LM}=\frac{IK}{LK}\), we get \(\frac{8}{12}=\frac{10}{r}\).

Step4: Solve the proportion

Cross - multiply: \(8r=12\times10\). Then \(8r = 120\). Divide both sides by 8: \(r=\frac{120}{8}=15\).

Answer:

\(r = 15\) meters