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[tex]$ \triangle PQR $[/tex] is a triangle. A straight line parallel to side [tex]$QR$[/tex] of the triangle cuts [tex]$PQ$[/tex] at [tex]$M$[/tex] and [tex]$PR$[/tex] at [tex]$N$[/tex]. If [tex]$\frac{PN}{NR} = \frac{5}{8}$[/tex], find

(a) [tex]$\frac{PM}{MQ}$[/tex]

(b) [tex]$\frac{QM}{MP}$[/tex]

(c) [tex]$\frac{QM}{QP}$[/tex]

(d) [tex]$\frac{QR}{MN}$[/tex]

Sagot :

GinnyAnswer
Certainly! Let's analyze the problem step-by-step, addressing each part:

Given:
- \( \triangle PQR \)
- A line parallel to \( QR \) cuts \( PQ \) at \( M \) and \( PR \) at \( N \).
- \( \frac{PN}{NR} = \frac{5}{8} \)

### (a) \( \frac{PM}{MQ} \)

Since \( MN \) is parallel to \( QR \), triangles \( PMN \) and \( PQR \) are similar by the Basic Proportionality Theorem (Thales' Theorem).

Due to the similarity, the ratio of corresponding sides of similar triangles is equal.

Thus,
[tex]\[ \frac{PM}{MQ} = \frac{PN}{NR} = \frac{5}{8} \][/tex]

### (b) \( \frac{QM}{MP} \)

Since \( \frac{PM}{MQ} = \frac{5}{8} \), the inverse ratio will give us:
[tex]\[ \frac{QM}{MP} = \frac{MQ}{PM} = \frac{8}{5} \][/tex]

### (c) \( \frac{QM}{QP} \)

To find this ratio, we recognize that:
[tex]\[ QP = QM + MP \][/tex]

From the previous part:
[tex]\[ \frac{QM}{MP} = \frac{8}{5} \][/tex]

Let \( MP = x \). Then \( QM = \frac{8}{5}x \).

So,
[tex]\[ QP = QM + MP = \frac{8}{5}x + x = \frac{8}{5}x + \frac{5}{5}x = \frac{13}{5}x \][/tex]

Hence,
[tex]\[ \frac{QM}{QP} = \frac{\frac{8}{5}x}{\frac{13}{5}x} = \frac{8}{13} \][/tex]

### (d) \( \frac{QR}{MN} \)

Given \( \frac{PN}{NR} = \frac{5}{8} \),
[tex]\[ \frac{PN + NR}{NR} = \frac{5 + 8}{8} = \frac{13}{8} \][/tex]

Since the triangles are similar, corresponding segments of similar triangles are proportional. Therefore, \( \frac{QR}{MN} \) follows the same ratio:

[tex]\[ \frac{QR}{MN} = \frac{13}{5} \][/tex]

Summarizing:
(a) \( \frac{PM}{MQ} = \frac{5}{8} \approx 0.625 \)

(b) \( \frac{QM}{MP} = \frac{8}{5} = 1.6 \)

(c) \( \frac{QM}{QP} = \frac{8}{13} \approx 0.615 \)

(d) \( \frac{QR}{MN} = \frac{13}{5} = 2.6 \)

These results align with the given solutions and the applied mathematical principles.
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