Construct a $\triangle ABC$ in which $AB = 5\ cm, \angle B = 60^o$, altitude $CD = 3\ cm$. Construct a $\triangle AQR$ similar to $\triangle ABC$ such that side of $\triangle AQR$ is 1.5 times that of the corresponding sides of $\triangle ACB$.

Given:

A $\triangle ABC$ in which $AB = 5\ cm, \angle B = 60^o$, altitude $CD = 3\ cm$.

To do:

We have to construct a $\triangle ABC$ in which $AB = 5\ cm, \angle B = 60^o$, altitude $CD = 3\ cm$. Construct a $\triangle AQR$ similar to $\triangle ABC$ such that side of $\triangle AQR$ is 1.5 times that of the corresponding sides of $\triangle ACB$.

Solution:

Steps of construction:

(i) Draw a line segment $AB = 5\ cm$.

(ii) At $A$, draw a perpendicular and cut off $AE = 3\ cm$.

(iii) From $E$, draw $EF\ \parallel\ AB$.

(iv) From $B$, draw a ray making an angle of $60^o$ meeting $EF$ at $C$.

(v) Join $CA$.

$ABC$ is the required triangle.

(vi) From $A$, draw a ray $AX$ making an acute angle with $AB$ and cut off three equal parts making $AA_1= A_1A_2 = A_2A_3$.

(vii) Join $A_2$ and $B$.

(viii) From $A_3$, draw $A_3B’$ parallel to $A_2B$ and $B’C’$ parallel to $BC$.

$C’AB’$ is the required triangle.

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Updated on: 10-Oct-2022

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