Draw a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm.

Question:

Draw a right triangle in which sides (other than the hypotenuse) are of lengths $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$. Then construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the first triangle.

Solution:

Given that

Construct a right triangle of sides let $A B=8 \mathrm{~cm}, A C=6 \mathrm{~cm}$, and $\angle A=90^{\circ}$ and then a triangle similar to it whose sides are $(3 / 4)^{\text {th }}$ of the corresponding sides of $\triangle A B C$.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment let $A B=8 \mathrm{~cm}$.

Step: II- With $A$ as centre and draw an angle $\angle A=90^{\circ}$.

Step: III-With $A$ as centre and radius $A C=6 \mathrm{~cm}$.

Step: IV-Join $B C$ to obtain right $\triangle A B C$.

 

Step: V-Below $A B$, makes an acute angle $\angle B A X=60^{\circ}$.

Step: VI- Along $A X$, mark off five points $A_{1}, A_{2}, A_{3}$, and $\mathrm{A}_{4}$ such that $A A_{1}=A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4}$

Step: VII- Join $A_{4} B$.

Step: VIII - Since we have to construct a triangle each of whose sides is $(3 / 4)^{\text {th }}$ of the corresponding sides of right $\triangle A B C$.

So, we draw a line $A_{3} B^{\prime}$ on $A X$ from point $A_{3}$ which is $A_{3} B^{\prime} \| A_{4} B$, and meeting $A B$ at $B$ '.

Step: IX- From $B^{\prime}$ point draw $B^{\prime} C^{\prime} \| B C$, and meeting $A C$ at $C^{\prime}$

 

Thus, $\triangle A B^{\prime} C^{\prime}$ is the required triangle, each of whose sides is $(3 / 4)^{\text {th }}$ of the corresponding sides of $\triangle A B C$.

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