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In the diagram below, $K, M$ and $N$ respectively are points on sides $P Q$, PR and $Q$ R of $\Delta$ PQR. $K P=1,5 ;$ PM $=2 ;$ KM $=2,5$; $\mathrm{MN}=1 ; \mathrm{MR}=1,25$ and $\mathrm{NR}=0,75$

Prove that $\Delta \mathrm{KPM}|\mathrm{I}| \Delta \mathrm{RNM}$

in Mathematics by Diamond (75,918 points) | 26 views

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Answer:

\begin{array}{l}
\frac{\mathrm{KP}}{\mathrm{RN}}=\frac{1,5}{0,75}=2 ; \frac{\mathrm{PM}}{\mathrm{NM}}=\frac{2}{1}=2 ; \frac{\mathrm{KM}}{\mathrm{RM}}=\frac{2,5}{1.25}=2 \\
\therefore \frac{\mathrm{KP}}{\mathrm{RN}}=\frac{\mathrm{PM}}{\mathrm{NM}}=\frac{\mathrm{KM}}{\mathrm{RM}} \\
\therefore \Delta \mathrm{KPM}|\mathrm{I}| \Delta \mathrm{RNM} \quad[\text { Sides of } \Delta \text { in prop }]
\end{array}

 

OR

\begin{array}{l}
\frac{\mathrm{RN}}{\mathrm{KP}}=\frac{0,75}{1,5}=\frac{1}{2} ; \frac{\mathrm{NM}}{\mathrm{PM}}=\frac{1}{2} ; \frac{\mathrm{RM}}{\mathrm{KM}}=\frac{1,25}{2,5}=\frac{1}{2}\\
\therefore \frac{\mathrm{RN}}{\mathrm{KP}}=\frac{\mathrm{NM}}{\mathrm{PM}}=\frac{\mathrm{RM}}{\mathrm{KM}}\\
\therefore \Delta \mathrm{KPM}|\mathrm{L}| \Delta \mathrm{RNM} \quad[\text { Sides of } \Delta \text { in prop }]
\end{array}

 

 

Explanation:

In this question you are given the lengths of the sides of triangles KPM and RNM respectively. You need to reflect on typical examples done in class, and realize that they can prove that $\Delta K P M / / / \Delta R N M$ by showing that corresponding sides of these two triangles are in the same proportion. This demonstration involves a well-known procedure of calculating the ratio of the corresponding sides by taking the longest side length of $\Delta K P M$ and dividing by the longest side length of $\triangle R N M$, then dividing shortest side length of $\Delta K P M$ by the shortest side length of $\Delta R N M$, followed the division of the remaining side length $\Delta K P M$ by the remaining side length of $\Delta R N M$, and finally concluding that the all three ratios of corresponding sides are equal ( if they indeed are so). As this constitutes a well-known procedure embracing a simple application of a similar triangle theorem and calculations which involve a few steps, which are familiar and generally similar to those encountered in class.

by Diamond (75,918 points)

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