can xome1 help me plzz im stuck
posted by hii on .
HELP PLZZ...A circle has radius 2 cm,(in that circle there is a traingle ABO)
O is the centre of the circle.
the chord AB has length 3 cm. .
BO is 2cm
AO is 2cm
(i) Show that ÚAOB 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.
draw 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.
This splits the triangle into two right triangles, each with
hypotenuse 2, and adjacent side = 1.5
cos ¦È = adjacent / hypotenuse = (1.5) / 2 = 0.75
¦È = arccos (0.75) ¡Ö 0.723 radians
Angle(AOB) = ¦Ð - 2¦È ¡Ö ¦Ð - 1.446 ¡Ö 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 ¦È = opposite / hypotenuse
opposite = (hypotenuse) * sin ¦È = 2 * sin (0.723 radians)
¡Ö 1.323 cm
So area of one of the small right triangles
¡Ö (1/2) (1.5) (1.323) ¡Ö 0.992 cm^2
Area of triangle AOB is two times this amount or ¡Ö 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%