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Construction of a Perpendicular Bisector

Draw Δ PQR su...

Question

$D r a w ∆ P Q R s u c h t h a t Q R = 5.5 c m, ∠ P = 50 ° a n d P Q + P R = 9.5 c m .$

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Solution

Steps of construction:
1. Draw QD = 9.5 cm.
2. Draw ray DX, such that $∠ QDX = 25 ° (i . e . \frac{1}{2} ∠ P)$ .
3. With Q as centre and radius 5.5 cm, draw an arc, intersecting ray QX at point R.
4. Join R and Q.
5. Draw the perpendicular bisector of seg RD, intersecting QD at point P.
6. Hence, $△$ PQR is the required triangle.

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