Question

Construct a triangle of sides 5 cm, 6cm and 7cm and then a triangle similar to it whose sides are $${\displaystyle \frac{7}{5}}$$ of the corresponding sides of the first triangle.

Answer 1

Hint: Here first we will construct the triangle with sides 5 cm, 6cm, and 7cm and then we will use the given fraction of the side to construct a similar triangle.

Complete step by step answer:
Let us first draw the triangle with sides 5 cm, 6cm and 7cm.
- First, we will draw the base AB of the triangle of length 5cm.
- Then A as the center we will mark an arc with the help of a compass at a distance of 6cm.
- Similarly, now with B as the center, we will mark an arc with the help of a compass at a distance of 7cm.
- Now finally we will mark the point so obtained as C and then join the points A and C and A and B.

Now we will construct a similar triangle of ABC whose sides are $${\displaystyle \frac{7}{5}}$$of the corresponding sides of$$\mathrm{\Delta }ABC$$.
We will first construct a ray AY such that it makes an acute angle with line AB on the opposite side of vertex C.

Now we will mark seven points (as 7 is greater than 5 in the fraction$${\displaystyle \frac{7}{5}}$$) at equal distances from each other on the ray AY i.e. $$A{A}_1=A_1{A}_2=A_2{A}_3=A_3{A}_4=A_4{A}_5=A_5{A}_6=A_6{A}_7$$

Now we will join $$A_5$$to $$B$$and then we will draw a line $$A_7$$ (as the side of the similar triangle is $${\displaystyle \frac{7}{5}}$$of the corresponding sides of$$\mathrm{\Delta }ABC$$) to the extended line AB such that it intersect AB at $$B^{{}^{\prime }}$$and is parallel to$$A_{5}B$$.

Now we will finally draw a line from $$B^{{}^{\prime }}$$to extended AC such that it intersects AC at $$C^{{}^{\prime }}$$and the line $$B^{{}^{\prime }}{C}^{{}^{\prime }}$$is parallel to the line BC.

Now the new triangle so formed i.e. $$\mathrm{\Delta }A{B}^{{}^{\prime }}{C}^{{}^{\prime }}$$ is the similar triangle to $$\mathrm{\Delta }ABC$$.

Note:
- Students should use compass and scale for perfect measurement and arcs an accurate distance to get the desired triangles.
- Also, we have drawn three arcs on the ray AY because we had to construct a similar triangle such that its sides are $${\displaystyle \frac{7}{5}}$$of the corresponding sides of the first triangle.