Question

a ∆pqr pqdraw=qr =7.8cm and pqr = 90° draw its line of symmetry​

Answer 1

Answer:

Draw a circle of radius 3.2 cm.

Answer:

Steps of construction:

(a) Open the compass for the required radius of 3.2 cm.

(b) Make a point with a sharp pencil where we want the centre of circle to be.

(c) Name it O.

(d) Place the pointer of compasses on O.

(e) Turn the compasses slowly to draw the circle.

Hence, it is the required circle.

Step-by-step explanation:

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Answer 2

Refer to attachment figure

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 PR.

Thus, △ PQR is the required triangle.