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August 30, 2014

August 30, 2014

Posted by **hii** on Sunday, September 16, 2007 at 5:43pm.

O is the centre of the circle.

the chord AB has length 3 cm. .

BO is 2cm

AO is 2cm

(i) Show that AOB is 1.696 radians (to three decimal places).

(ii) Hence ﬁnd the area of the (smaller) sector AOB.

(iii) Find the area of the triangle AOB.

(iv) Hence ﬁnd the (smaller) area enclosed between the chord AB and the circle

(v) Express both the area of the sector and the area from part (a)(iv) as percentages of the area of the circle.

- can xome1 help me plzz im stuck -
**bobpursley**, Sunday, September 16, 2007 at 5:49pmdraw a perpendicular from O to AB. Let x be the intersection. You know xb, and bo lengths, so find the sine of obx, thence the measure of obx. Double it to get boa.

The rest come from that.

Iwill be happy to critique your thinking.

- wod this be right? -
**hi**, Monday, September 17, 2007 at 5:14pmThis splits the triangle into two right triangles, each with

hypotenuse 2, and adjacent side = 1.5

cos = adjacent / hypotenuse = (1.5) / 2 = 0.75

= arccos (0.75) 0.723 radians

Angle(AOB) = - 2 - 1.446 1.696 radians.

Area of one of the small right triangles is (1/2)bh

Need to find h (length of segment from 0 to midpt AB)

sin = opposite / hypotenuse

opposite = (hypotenuse) * sin = 2 * sin (0.723 radians)

1.323 cm

So area of one of the small right triangles

(1/2) (1.5) (1.323) 0.992 cm^2

Area of triangle AOB is two times this amount or 1.984 cm^2

for v) area oc circle , pie r ^2 = pie 2^2 = 12.56

so 100 / 12.56 x 1.984 = 15.796 which is 15%

- wod this be right? -

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