Question

please friends it's very important..​

Answer 1

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Given:-

• QR = 4.2 cm
• ∠Q = 120°
• PQ = 3.5 cm

To do:-

• To draw PM ⊥ QR

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Construction:-

• First we will draw a rough sketch of ΔPQR, then we will construct ΔPQR in the steps.

Steps of construction:-

• 1) Draw a line segment QR = 4.2 cm

• 2) Construct ∠RQX = 120°.

• 3) Along QX, set off QP = 3.5 cm.

• 4) join PR, Then ΔPQR is the required triangle

• 5) Produce RQ to S to the left.

• 6) With P as centre and with sufficient radius, draw an arc, cutting SQ at A and B.

• 7) With A as centre and radius more than half of AB, draw an arc. Now with B as centre and with the same radius draw another arc, cutting the previous arc at C.

• 8) Join PC, meeting RQ produced at M. Then, PM ⊥ QR.

$\red{\bigstar}$$\displaystyle\large\underline{\sf\red{Diagram}}$

• Refer to the attachment for the diagram of construction.

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