Question

2-117. solve for the missing lengths in the sets of similar figures below
a. ( abcd sim jklm )
diagram: rectangle ( abcd ) with ( ab = 12 ) mm, ( ad = 6 ) mm; rectangle ( jklm ) with ( jm = 10 ) mm, ( jk = x )
hint (a):
diagram
b. ( \triangle nop sim \triangle xyz )
diagram: triangle ( xyz ) with ( xz = 39 ) mm, ( yz = w ); triangle ( nop ) with ( pn = 7 ) mm, ( on = 3 ) mm

Explanation:

Part (a)

Step1: Recall similarity ratio

Since \(ABCD \sim JKLM\), corresponding sides are proportional. The ratio of \(AD\) to \(JM\) is \(\frac{6}{10}=\frac{3}{5}\), so the ratio of \(AB\) to \(JK\) should be the same. Let \(JK = x\), then \(\frac{AB}{JK}=\frac{AD}{JM}\), so \(\frac{12}{x}=\frac{6}{10}\).

Step2: Solve for \(x\)

Cross - multiply: \(6x = 12\times10\). Then \(6x = 120\), divide both sides by 6: \(x=\frac{120}{6}=20\).

Part (b)

Step1: Determine the similarity ratio

Since \(\triangle NOP\sim\triangle XYZ\), the ratio of \(OP\) to \(XZ\) is \(\frac{7}{39}\)? Wait, no, let's check the corresponding sides. Wait, \(OP = 7\) mm, \(XZ = 39\) mm, and \(NP = 3\) mm. Let \(ZY = w\) (wait, no, \(XY\) or \(ZY\)? Wait, the triangles: \(\triangle NOP\) has sides \(NP = 3\), \(OP = 7\), and \(\triangle XYZ\) has \(XZ = 39\), and we need to find \(ZY\) (let's assume \(OP\) corresponds to \(XZ\) and \(NP\) corresponds to \(ZY\)? Wait, no, let's get the correspondence right. \(\triangle NOP\sim\triangle XYZ\), so \(\frac{NP}{ZY}=\frac{OP}{XZ}\). So \(NP = 3\) mm, \(OP = 7\) mm, \(XZ = 39\) mm. Let \(ZY = w\). Then \(\frac{3}{w}=\frac{7}{39}\)? Wait, no, maybe I got the correspondence wrong. Wait, maybe \(OP\) corresponds to \(XZ\) and \(NO\) (wait, no, the labels: \(\triangle NOP\) with vertices \(N\), \(O\), \(P\); \(\triangle XYZ\) with \(X\), \(Y\), \(Z\). So \(N\) corresponds to \(X\), \(O\) corresponds to \(Y\), \(P\) corresponds to \(Z\). So \(NP\) corresponds to \(XZ\), \(OP\) corresponds to \(YZ\), and \(NO\) corresponds to \(XY\). Wait, the given lengths: \(NP = 3\) mm, \(OP = 7\) mm, \(XZ = 39\) mm. We need to find \(YZ = w\). So the ratio of similarity is \(\frac{NP}{XZ}=\frac{3}{39}=\frac{1}{13}\)? Wait, no, that can't be. Wait, maybe \(OP\) corresponds to \(XZ\) and \(NP\) corresponds to \(XY\)? No, let's start over.

Since \(\triangle NOP\sim\triangle XYZ\), the ratio of corresponding sides is equal. Let's assume that \(OP\) corresponds to \(XZ\) and \(NP\) corresponds to \(XY\)? Wait, the problem shows \(\triangle NOP\) with \(P\) to \(N\) is 3 mm, \(P\) to \(O\) is 7 mm, and \(\triangle XYZ\) with \(X\) to \(Z\) is 39 mm, and we need to find \(Y\) to \(Z\) (length \(w\)). So if \(\triangle NOP\sim\triangle XYZ\), then \(\frac{OP}{XZ}=\frac{NP}{YZ}\). Wait, \(OP = 7\) mm, \(XZ = 39\) mm, \(NP = 3\) mm, \(YZ = w\). So \(\frac{7}{39}=\frac{3}{w}\)? No, that would be cross - multiplying \(7w = 117\), \(w=\frac{117}{7}\), which is not an integer. Wait, maybe the correspondence is \(\triangle NOP\sim\triangle YXZ\)? No, maybe I mixed up the sides. Wait, the first triangle \(\triangle NOP\): let's see the sides: \(NP = 3\), \(PO = 7\), and the second triangle \(\triangle XYZ\): \(XZ = 39\), \(ZY = w\). If the similarity ratio is \(\frac{NP}{XZ}=\frac{3}{39}=\frac{1}{13}\), then \(PO\) should correspond to \(ZY\), so \(ZY=PO\times13 = 7\times13 = 91\)? Wait, no, that would be if \(NP\) corresponds to \(XZ\) and \(PO\) corresponds to \(ZY\). Let's check the ratio: \(\frac{NP}{XZ}=\frac{3}{39}=\frac{1}{13}\), then \(\frac{PO}{ZY}=\frac{1}{13}\), so \(ZY = PO\times13=7\times13 = 91\). Wait, that makes sense. So the ratio of \(\triangle NOP\) to \(\triangle XYZ\) is \(\frac{1}{13}\), so each side of \(\triangle XYZ\) is 13 times the corresponding side of \(\triangle NOP\). So if \(NP = 3\), then \(XZ = 3\times13 = 39\) (which matches), and if \(PO = 7\), then \(ZY=7\times13 = 91\). So \(w = 91\) mm.

Answer:

a. The length of \(x\) is \(\boldsymbol{20}\) mm.
b. The length of \(w\) is \(\boldsymbol{91}\) mm.