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Question

Draw a right triangle ABC in which AC = AB = 4.5cm and = 90^ . Draw a triangle similar to with its sides equal to (5 4 )^th of the corresponding sides of .

Accepted Answer

Given that Construct a right triangle of sides AB = AC = 4.5cm, and $\angle\text{A}=90^\circ$ and then a triangle similar to it whose sides are $\Big(\frac{5}{4}\Big)^\text{th}$ of the corresponding sides of $\triangle\text{ABC}.$ We follow the following steps to construct the given

Step of construction:
Step I: First of all we draw a line segment $A B=4.5 cm$.
Step II: With A as centre and draw an angle $\angle A =90^{\circ}$.
Step III: With $A$ as centre and radius $A C=4.5 cm$.
Step IV: Join $B C$ to obtain $\triangle ABC$.
Step V: Below $A B$, makes an acute angle $\angle B A X=60^{\circ}$.
Step VI: Along $A X$, mark off five points $A_1, A_2, A_3, A_4$ and $A_5$ such that $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5$.
Step VII: Join $A _4 B$.
Step VIII: Since we have to construct a triangle each of whose sides is $\left(\frac{5}{4}\right)^{\text {th }}$ of the corresponding sides of $\triangle ABC$.
So, we draw a line $A_5 B^{\prime}$ on $A X$ from point $A_5$ which is $A_5 B^{\prime} \| A_4 B$, and meeting $A B$ at $B^{\prime}$.
Step IX: From $B^{\prime}$ point draw $B^{\prime} C^{\prime} \| B C$, and meeting $A C$ at $C^{\prime}$. Thus, $\triangle AB ^{\prime} C ^{\prime}$ is the required triangle, each of whose sides is $\left(\frac{5}{4}\right)^{\text {th }}$ of the corresponding sides of $\triangle ABC$.

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