Q3 of 24 Page 283

Rita draws a right angled triangle. Let us prove logically that the perpendicular bisectors of sides are concurrent and where the position of the circum centre (inside/outside/on the side) is

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PO is perpendicular to AB and OQ is perpendicular to BC


Since Δ ABC is right angled at B so POQB forms a rectangle


Hence we can say that


PO = BQ (Opposite sides of rectangle)


BP = QO(Opposite sides of a rectangle)


In Δ AOP and Δ QOC


∠APO = ∠OQC (OP and OQ are perpendiculars)


∠OAP = ∠COQ (Corresponding angles)


PO = BQ = CQ (Q is midpoint of BC)


So Δ AOP and Δ COQ are congruent by A.A.S. axiom of congruency


Hence AO = OC


So O is the midpoint of AC


Hence the perpendiculars are concurrent


The circum centre lies on the midpoint of the hypotenuse of the triangle.


More from this chapter

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1

I draw an acute angled triangle PQR and let us prove that the perpendicular bisectors PQ, QR and RP are concurrent. Hence let us write where the circumcentre lies (inside/outside/on the side) of the acute angle triangle.

 2

I draw an obtuse angled triangle and let us prove that the perpendicular bisectors of sides are concurrent. Let us write where the circumcentre lies (inside/outside/on the side) of the triangular region.

 1

Let us write what is the length of circum radius of a triangle lengths of sides of which are 6 cm, 8 cm and 10 cm.

 2

If the length of circum radius of a right-angled triangle is 10 cm, then let us write how much length of hypotenuse of triangle is.