Construct a ∆PQR in which QR = 6 cm, PQ = 5 cm and ∠PQR = 60°. Now , construct a triangle similar to ∆PQR such that each of its sides is $\frac{3}{5}$ the corresponding sides of ∆PQR.

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Solution

Steps of Construction :

Step 1. Draw a line segment QR = 6 cm.
Step 2. With Q as centre, draw an angle of 60^o .
Step 3. With Q as centre and radius equal to 5 cm, draw another arc, cutting the previous arc at P.
Step 4. Join PQ and PR.
Thus, â–³ PQR is obtained .
Step 5. Below QR, draw an acute $∠ R Q X$ .
Step 6. Along QX, mark off five points Q₁, Q₂, Q₃, Q₄, Q₅, such that Q₁Q₂ = Q₂Q₃ = Q₃Q₄₌Q₄Q₅ .
Step 7. Join Q₅R.
Step 8. From Q₃, draw Q₃S âˆ¥ Q₅R, meeting QR at S.
Step 9. From S, draw ST âˆ¥ PR, meeting PQ at T.

Thus, â–³TQS is the required triangle, each of whose sides is $\frac{3}{5}$ the corresponding sides of âˆ†PQR.

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Construct a ∆PQR in which QR = 6 cm, PQ = 5 cm and ∠PQR = 60°. Now , construct a triangle similar to ∆PQR such that each of its sides is 35 the corresponding sides of ∆PQR.

Construct a ∆PQR in which QR = 6 cm, PQ = 5 cm and ∠PQR = 60°. Now , construct a triangle similar to ∆PQR such that each of its sides is $\frac{3}{5}$ the corresponding sides of ∆PQR.