Question

Q.8 Construct A PQR in which PQ = 7 cm, PR = 5 cm and Z P= 75°​

Answer 1

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1.Draw base BC=7cm

2.Let's draw ∠B=75 ∘

Let the ray be BX

3.Open the compass to length AB+AC=13cm.From point B as center, cut an arc on ray BX.Let the arc intersect BX at D

4.Join $$CD$

5.Now, we will draw perpendicular bisector of CD

6.Mark point A where perpendicular bisector intersects BD

7.Join AC

Result:△ABC is the required triangle

Answer 2

Step-by-step explanation:

∆PQR with sides PQ = 7 cm, QR = 8 cm, PR = 5 cm. Construction: (i) Draw the ∆PQR with the given measurements. (ii) Construct altitudes from any two vertices (R and P) to their opposite sides (PQ and QR) respectively. (iii) The point of intersection of the altitude H is the Orthocentre