divide a line segment of length 14 cm internally in the ratio $2 : 5$. Also, justify your construction.


Given:

A line segment of length 14 cm.

 To do:

We have to divide a line segment of length 14 cm internally in the ratio $2 : 5$.

Solution:

Steps of construction:

(i) Draw a line segment $AB = 14\ cm$.

(ii) Draw a ray $AX$ making an acute angle with $AB$.

(iii) From $B$, draw another ray $BY$ parallel to $AX$.

(iv) Cut off two equal parts from $AX$ and five parts from $BY$.

(v) Join $A_2$ and $B_5$ which intersects $AB$ at $P$.

$P$ is the required point which divides $AB$ in the ratio of $2 : 5$ internally.

Justification: In $\triangle \mathrm{AA}_{2} \mathrm{P}$ and $\triangle \mathrm{BB}_{5} \mathrm{P}$,

$\angle A_{2} A P=\angle P B B_{5}$                  ($\angle A B Y=\angle B A X$)

$\angle \mathrm{APA}_{2}=\angle \mathrm{BPB}_{5}$       (Vertically opposite angles)

Therefore, by AA similarity,

$\triangle \mathrm{AA}_{2} \mathrm{P} \sim \Delta \mathrm{BB}_{5} \mathrm{P}$

This implies,

$\frac{A A_{2}}{B B_{5}}=\frac{A P}{B P}$

$\frac{A P}{B P}=\frac{2}{5}$ 

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Updated on: 10-Oct-2022

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