Q13 of 141 Page 447

An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is


Given: Equilateral triangle of side 9 cm is inscribed in a circle.


Construction: Join OA, OB, OC and drop a perpendicular bisector from center O to BC.


Here,


Area (ΔABC) = 3× area (ΔOBC)


Area (ΔABC) = a2 = × 92 =


Now,


Area (ΔOBC) = × AC × OD = × 9 × OD


We know that,


Area (ΔABC) = 3× area (ΔOBC)


= × 9 × OD


OD =


Now, in ΔODC


By Pythagoras theorem


OC2 = OD2 + DC2


OC2 = 2 + 2


OC2 = + = = 27


OC =


∴ Radius = OC =

More from this chapter

All 141 →
11

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ∠AOC = 25° ∠ACD = 25°, then ∠AOD = ?

 12

In the given figure, AB is a chord of a circle with centre O and BOC is a diameter. If OD ⊥ AB such that OD = 6cm then AC = ?

 14

The angle in a semicircle measures

 15

Angles in the same segment of a circle area are