Question

construct a triangle PQR in which QR = 3 CM angle PQR = 45 degree and QP - PR = 2cm​

Answer 1

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

$\triangle PQR$ is the required triangle.

Step-by-step explanation:

Given,

Construct a $\triangle PQR$ where $QR=3cm$, $\angle PQR=45$° and $QP-PR=2cm$

Following steps:

• Draw a line $QR=3cm$ using scale.
• At point $Q$,draw an angle $45$° like $\angle XQR$ where $QR$ as a base.
• Cuts a point $A$ on $QX$ line like $QA=2cm$
• Join $AR$ and bisect the line $AR$

• Bisect line of $AR$,cuts the line $QX$ which is known as $P$
• Join $PR$

Now, $\triangle PQR$ is the required triangle.