Question 2 - Constructing a triangle when 2 angles and side between is known (ASA) - Practical Geometry

Last updated at Dec. 16, 2024 by

Ex 10.4, 2 - Construct PQR if PQ = 5 cm, PQR = 105 and QRP = 40

Ex 10.4, 2 - Chapter 10 Class 7 Practical Geometry - Part 2
Ex 10.4, 2 - Chapter 10 Class 7 Practical Geometry - Part 3
Ex 10.4, 2 - Chapter 10 Class 7 Practical Geometry - Part 4
Ex 10.4, 2 - Chapter 10 Class 7 Practical Geometry - Part 5 Ex 10.4, 2 - Chapter 10 Class 7 Practical Geometry - Part 6

Go Ad-free

Transcript

Question 2 Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°. (Hint: Recall angle-sum property of a triangle) First we draw a rough sketch To draw Δ PQR, we need to find ∠ P Finding ∠ P In Δ PQR, Sum of angles = 180° ∠ P + ∠ Q + ∠ R = 180° ∠ P + 105° + 40° = 180° ∠ P + 145° = 180° ∠ P = 180° – 145° ∠ P = 35° Now, let’s construct it Steps of construction 1. Draw a line segment PQ of length 5 cm 2. Now, we draw 35° from point P (using Protractor) 3. Now, we draw 105° from point Q (Using protractor) 4. Let point R be the point where the two rays intersect and label the sides Thus, Δ PQR is the required triangle

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo