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Question

Construct a $$\Delta PQR$$, in which $$PQ=6\,cm$$, $$QR=7\,cm$$ and $$PR=8cm$$. Then construct another triangle whose sides are $$\dfrac{4}{5}$$ times the corresponding sides of $$\Delta PQR$$


Solution

Steps of construction :

1. Draw a line segment $$PQ=6$$ cm

2. Draw an arc, using $$P$$ as a centre and radius $$8$$ cm

3. Draw another arc, using $$Q$$ as a centre and radius $$7$$ cm

4. Now, join $$PR$$ and $$QR$$ to get $$\Delta PQR$$

5. Draw a ray $$PX$$ by making an acute angle, angle $$QPX$$

6. Divide $$PX$$ into $$4$$ equal parts.
$$P_1,P_2,P_3,P_4,P_5$$ such $$PP_1=P_1P_2=P_2P_3=P_3P_4=P_4P_5$$

7. Join $$P5Q$$

8. Draw a line $$P4Q'$$ which is parallel to $$QR$$

9. Similar to step $$8$$, draw a line $$Q'R'$$ which is parallel to $$QR$$.
Therefore, $$\Delta PQ'R'$$ is the required triangle.

1537982_1713723_ans_b79c4ed3c49b4e04976b003b99e21eb8.png

Mathematics
RS Agarwal
Standard X

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