Construct a triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$ and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of first triangle.
Steps of Construction :
Step 1. Draw a line segment BC = 4 cm.
Step 2. With B as centre, draw an angle of 90o.
Step 3. With B as centre and radius equal to 3 cm, cut an arc at the right angle and name it A.
Step 4. Join AB and AC.
Thus, △ ABC is obtained .
Step 5. Extend $B C$ to $D$, such that $B D=\frac{7}{5} B C=\frac{7}{5}(4) \mathrm{cm}=5.6 \mathrm{~cm}$.
Step 6. Draw DE ∥ CA, cutting AB produced to E.
Thus, $\triangle E B D$ is the required triangle, each of whose sides is $\frac{7}{5}$ the corresponding sides of $\triangle A B C$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.