Draw a ΔABC with side BC = 6 cm. AB = 5 cm and ∠ ABC = 60°

Question:

Draw a ΔABC with side BC = 6 cm. AB = 5 cm and ∠ ABC = 60°. Then, construct a triangle whose sides are (3/4)th of the corresponding sides of the ΔABC.

Solution:

Given that

Construct a $\triangle A B C$ of given data, $A B=5 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $\angle A B C=60^{\circ}$ 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.

Step: II- With as centre draw an angle.

Step: III- With as centre and radius, draw an arc.

Step: IV- Join AC to obtain.

Step: V -Below AB, makes an acute angle.

Step: VI -Along $A X$, mark off four points $A_{1}, A_{2} A_{3}$ and $A_{4}$ 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 $\triangle A B C$.

So, we take three parts out of four equal parts on $A X$ from point $A_{3}$ draw $A_{3} B^{\prime} \| A_{4} B$, and meeting $A B$ at $B$.

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

 

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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