Q. 34.0( 14 Votes )

Construct a ΔPQR, in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then, construct another triangle whose sides are 4/5 times the corresponding sides of ΔPQR.

Answer :

Steps of Construction:


1. First, we draw the triangle ABC of given dimensions.


2. Draw a line segment QR of length 7cm.


3. With Q as center draw an arc of radius 6cm(length of PQ).



4. With R as center draw an arc of radius 8cm(length of PR).



5. The point where the arcs intersect each other is P. Join PQ and PR.



6. Now we construct the triangle having dimensions of 4/5 of this triangle.


7. Draw a ray AZ at an acute angle to the line QR in the other direction of P.


8. Make 5 equal arcs along QZ, taking Q as starting point for the first arc, Q1 for second and so on till Q5. Join Q5R.



9. As we have to get the 4/5 dimensions of this triangle. We draw a line parallel to QR from Q4 till QR. This point is R’.



10. Now from R’, draw another line parallel to PR cutting PQ at P’.



11. P’QR’ is the required triangle.


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