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

**Solution:**

We use the basic rules of construction to solve the question given.

Let's use the angle-sum property of a triangle to find the measure of ∠RPQ in ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°

Given that, m∠PQR = 105° and m∠QRP = 40°

∠PQR + ∠QRP + ∠RPQ = 180°.

105° + 40° + ∠RPQ = 180°

So, ∠RPQ = 35°

Now, let’s construct ΔPQR such that PQ = 5cm, ∠PQR = 105° and ∠RPQ = 35°, with the steps given below

**Steps of construction :**

- Draw a line segment PQ of length 5 cm.
- At P, draw a ray PX making 35° with PQ.
- At Q, draw a ray QY making 105° with PQ.
- Rays PX and QY will intersect at point R.
- Triangle PQR is now constructed.

**☛ Check: **NCERT Solutions for Class 7 Maths Chapter 10

**Video Solution:**

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

Class 7 Maths NCERT Solutions Chapter 10 Exercise 10.4 Question 2

**Summary:**

∆PQR is constructed with PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°.

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