Question

construct a angle pqr in which QR is equal to 6 CM angle q is equal to 60 degree and PQ + PR is equal to 10 cm​

Answer 1

Answer:

1) Draw the base QR = 6 cm.

At point Q draw a ray QX making an ∠ QXR = 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.

♥️$\color{green}{THANKYOU! }$♥️

Answer 2

$\huge\colorbox{yellow}{Answer ✔}$

Angle PQR is the required triangle.

$\huge\colorbox{yellow}{Explaination}$

1.Draw the base QR = 6 cm.At point Q draw a ray QX making an ∠ QXR = 60°.Here, PR-PQ = 2cmPR > 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.

$\huge\colorbox{yellow}{Hope it helps you ☺️✔}$