Question

Construct a triangle ABC such that: BC = 6 cm, $\angle B={60}^{\circ }$ and $\angle C={45}^{\circ }$ In the same figure, find a point which is equidistant from the vertices of the triangle.Name this point. Draw the Circumcircle of the triangle.

• A. Circumcenter
• B. Incenter
• C. Midpoint
• D. Data insufficient

Answer 1

The correct option is A Circumcenter

Step (I) A line BC=6cm is drawn.

Step (II) $\angle ABC={60}^o\&\angle ACB={45}^o$ are drawn.

Step (III) OD, OE & OF are drawn as right bisectors of AB, BC & CA respectively.

The right bisector intersect at O.

So O is the circumcentre of $\Delta ABC$.

A circle is drawn with O as centre and either of OA, OB or OC as radius,

This is the circumcircle of$\Delta ABC$.

Justification-

Between the triangles OAF & OCF,

AF=FC (F is the mid point of AC),

OF is the common side,

$\angle OFA=\angle OFC\left(OF\&isrightbisectorofAC\right)\therefore \Delta OAF\equiv \Delta OCF.\therefore OA=OC..........\left(i\right)$

Again, between $\Delta AOD\&\Delta BOD$

$AD=BD\left(DisthemidpointofAB\right),ODisthecommonside.\angle ODA=\angle ODB={90}^o\therefore \Delta AOD\equiv \Delta BOD.\therefore OA=OB........\left(ii\right).$

Comparing (i) & (ii),

OA=OB=OC.

So we can draw a circle with o as centre and OA or OB or OC as radius.

So O is the circumcentre of $\Delta ABC$.