Question

Constaut a triangle PQR given that
QR = 3cm, L PQR= 45 and QP- PR
=2 cm​

Answer 1

Answer:

ΔPQR, QR = 3 cm, ∠PQR = 45°

and QP – PR = 2 cm ( given).

To construct ΔPQR, use the following steps.

1.Draw the base QR of length 3 cm.

2.Make an angle XQR = 45° at point Q of base QR.

3.Cut the line segment QS =QP- PR = 2 cm from the ray QX.

4.Join SR and draw the perpendicular bisector of SR and name it AB.

5.Let bisector AB intersect QX at point P. Join PR Thus, ΔPQR is the required triangle.

Base QR and ∠PQR are drawn as given.

Since, the point P lies on the perpendicular bisector of SR.

PS = PR

Now, QS = PQ – PS

= PQ -PR

Hence this construction is justified.

Answer 2

Answer:

Steps of construction :

1. Draw a ray OX and cut off a line segment QR = 3cm.

2. At Q, construction ∠ YQR = 45° with the help of protractor .

3. On QY , cut off QS = 2cm.

4. Join RS.

5. Draw perpendicular bisector of RS to meet QY at P.

6. Join PR. Then PQR is the required triangle.