Question

construct pqr where PQ=6cm, PR=8cm, and P=120°.Also construct a bisector of P​

Answer 1

Answer:

nice question I have seen it in my earlier classes book thanks for old memories

Answer 2

Step-by-step explanation:

Steps of construction :

1. Draw a line segment PQ=6 cm

2. Draw an arc, using P as a centre and radius 8 cm

3. Draw another arc, using Q as a centre and radius 7 cm

4. Now, join PR and QR to get ΔPQR

5. Draw a ray PX by making an acute angle, angle QPX

6. Divide PX into 4 equal parts.

P¹,P²,P³,P⁴,P⁵such PP¹=P¹P²=P²P³=P³P⁴ =P⁴P⁵

7. Join P5Q

8. Draw a line P4Q′

which is parallel to QR

9. Similar to step 8, draw a line Q ′ R′

which is parallel to QR.

Therefore, ΔPQ R′is the required triangle.