Math, asked by amansharma12184, 1 day ago

7. Construct a APQR in which ZP = 60° and 2Q = 45° and PQ = 4.8 cm. Write construction steps also.​

Answers

Answered by 31aliahmedzahidshaik
0

Answer:

Steps of Construction:

1. Draw AB of length 6.5 cm

2. Taking A as centre and radius 7.2 cm, draw an arc

3. Taking B as centre and radius 8.4 cm, draw another arc to cut the first arc at point C

4. Now, join AC and BC

Hence,

Triangle ABC is the required triangle

(ii) PQ = 4.8 cm, QR = 6.3 cm and PR = 5.5 cm

Answered by COMMANDED
0

Answer:

Draw the base QR = 6 cm.

At point Q draw a ray QX making an ∠ QXR = 60

o

.

Here, PR-PQ = 2cm

PR > PQ

The side containing the base angle Q is less than third side.

2) Cut the line segment QS equal to PR-PQ = 2 cm, from the ray QX extended on opposite side of base QR.

3) Join SR and draw its perpendicular bisector ray AB which intersect SR at M.

4) Let P be the intersection point of SX and perpendicular bisector AB. Then join PR.

Thus, △ PQR is the required triangle.

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