Maharashtra State BoardSSC (English Medium) 9th Standard
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Construct δ Pqr Such that ∠ P = 70 ∘ , ∠ R = 50 ∘ , Qr = 7.3 Cm and Constructs Its Circumcircle. - Geometry

Diagram

Construct Δ PQR such that ∠P = 70° , ∠R = 50° , QR = 7.3 cm  and constructs its circumcircle.

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Solution

By angle sum property of triangles,

In Δ PQR,

∠P + ∠Q +∠R = 180°

⇒ 70° + ∠Q + 50° = 180°

⇒ ∠Q = 60°

Steps of construction:
1. Draw a line QR = 7.3 cm.
2. With Q as centre, draw an angle of 60° using the protractor. Similarly, draw an angle of 50° with R as centre. 
Where the rays RY and XQ meet, name the point as P. 
Thus, Δ PQR is obtained. 
3. Draw the perpendicular bisectors of the lines PR and QR. Let these perpendicular bisectors meet at point O. 
4. With O as centre and OP as radius, construct a circle touching all the vertices of the Δ PQR. 
This circle is thus the required circumcircle. 

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APPEARS IN

Balbharati Mathematics 2 Geometry 9th Standard Maharashtra State Board
Chapter 6 Circle
Practice Set 6.3 | Q 2 | Page 86
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