Question

Construct ∆PQR in which PQ = 5.7cm, OR = 7.5cm, PQR = 1000

. Draw its incircle and

circumcircle.​

Answer 1

$\Large\mathsf{Answer}$

[ fig. is in the attachment]

Given: 

In ∆PQR, QR= 6 cm, PR-PQ=2 cm & ∠Q= 60°

Steps of construction:

1. Draw the base QR=6 cm.

At Point Q draw a ray QX making

an  ∠XQR=60°

Here , PR -PQ= 2cm

PR>PQ

The side containing the base

angle Q is less than third side.

 

2. Cut the line segment QS equal

to PR-PQ=2 cm , from the ray QX extended on 

opposite side of base QR.

3. Join SR and draw its

perpendicular bisector ray AB which intersect SR at M.

4. Let P be the intersection point

of SX and perpendicular bisector AB. Then join PR.

Thus ∆PQR is the required Triangle.

========================================================

Answer 2

Steps of construction:

1. Draw the base QR=6 cm.

At Point Q draw a ray QX making

an ∠XQR=60°

Here , PR -PQ= 2cm

PR>PQ

The side containing the base

angle Q is less than third side.

2. Cut the line segment QS equal

to PR-PQ=2 cm , from the ray QX extended on

opposite side of base QR.

3. Join SR and draw its

perpendicular bisector ray AB which intersect SR at M.

4. Let P be the intersection point

of SX and perpendicular bisector AB. Then join PR.

Thus ∆PQR is the required Triangle.

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Fig. is in attachment

$\huge \pink {sushant2141}$