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, △ 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 ∥ Q₅R, meeting QR at S.
Step 9. From S, draw ST ∥ PR, meeting PQ at T.

Thus, △TQS is the required triangle, each of whose sides is $\frac{3}{5}$ the corresponding sides of ∆PQR.

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Whose sides are $\frac{3}{4}$ times the corresponding sides of $Δ$ PQR?

Q. Construct a triangle PQR with sides QR = 7 cm, PQ = 6 cm and

∠

PQR = 60º. Then construct another triangle whose sides are

\frac{3}{5}

of the corresponiding sides of ∆PQR. [CBSE 2014, 2015]

Q. Draw a triangle

P Q R

, right angled at

Q

such that

P Q = 3

cm,

Q R = 4

cm. Now construct

△ A Q B

similar to

△ P Q R

each of whose sides is

\frac{7}{5}

times the corresponding side of

△ P Q R

Constructa $Δ P Q R,$ in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then, construct another triangle whose sides are $\frac{4}{5}$ to,of the corresponding sides of $Δ P Q R$ .

Q. (a) Construct

Δ P Q R

, given

¯ ¯¯¯¯¯¯ ¯ P Q = 3.5 c m, ¯ ¯¯¯¯¯¯ ¯ P R = 4.5 c m

, and

¯ ¯¯¯¯¯¯¯ ¯ Q R = 5.5 c m

(b) Construct ΔPQR, when PQ = 4 cm, QR = 6 cm and ∠PQR = 60

^{\circ}

. [4 MARKS]

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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.