Question

Question 2
Construct ∆PQR if PQ = 5 cm, ∠PQR = 105° and ∠QRP = 40°.
(Hint : Recall angle sum property of a triangle)

Answer 1

$\triangle PQR$ is the required triangle.

Step-by-step explanation:

Given,

Construct $\triangle PQR$ where $PQ = 5\ cm,\angle PQR =105\ degree\ and\ \angle QRP=40\ degree$

We know,

$\angle P+\angle Q+\angle R=180$

⇒$\angle P=180-105-40=35\ degree$

Following steps:

• Draw a line segments $PQ=5\ cm$
• At $P$, draw an angle $35\ degree$ using protector like $\angle QPX=35\ degree$
• Also at $Q$, draw an angle $105\ degree$ using protector like $\angle PQY=105\ degree$
• Where line $PX\ and\ QY$ intersect at $R$

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

Answer 2

Construction of triangle PQR:

Given,

         ∠Q = 105°, ∠R = 40°

In ΔPQR,

         ∠P + ∠Q + ∠R = 180° [ angle sum property ]

         ∠P + 105° + 40° = 180°

         ∠P = 35°

Steps of construction:

• Draw a line segment PQ = 5 cm.
• Mark angle 105° at Q and draw a ray QX.
• Similarly draw a ray PY from P with an angle of 35°.
• Rays QX and PY intersect at a point.
• Mark this point as R.
• PQR is the required triangle.