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%
To solve this problem, we will use basic geometry and trigonometry principles.
(i) To find the angle AOB in radians, we can use the formula:
θ = s/r
where θ is the angle in radians, s is the length of the arc, and r is the radius of the circle.
In this case, the chord AB has a length of 3 cm, which is equal to the length of the arc. The radius of the circle is 2 cm. Substituting these values into the formula, we get:
θ = 3/2 = 1.5 radians
Rounded to three decimal places, the angle AOB is 1.500 radians.
(ii) The area of a sector is given by the formula:
A = (θ/2) * r^2
where A is the area of the sector, θ is the angle in radians, and r is the radius of the circle.
Substituting the values we have, we get:
A = (1.500/2) * 2^2 = 1.500 * 4 = 6 square cm
So, the area of the smaller sector AOB is 6 square cm.
(iii) The area of a triangle can be calculated using the formula:
A = (1/2) * b * h
where A is the area of the triangle, b is the length of the base, and h is the height of the triangle.
In this case, the base AB is 3 cm, and the height can be found using the Pythagorean theorem. The height is the perpendicular distance from the center of the circle (O) to the chord AB. Since BO is 2 cm and AO is 2 cm, the total height is the sum of these two lengths, which is 4 cm.
Substituting the values into the formula, we get:
A = (1/2) * 3 * 4 = 6 square cm
So, the area of triangle AOB is 6 square cm.
(iv) The area enclosed between the chord AB and the circle is given by:
A = A_sector - A_triangle
Substituting the values we found earlier, we get:
A = 6 - 6 = 0 square cm
So, the smaller area enclosed between the chord AB and the circle is 0 square cm.
(v) To express the area of the sector and the area from part (iv) as percentages of the area of the circle, we need to compare them to the total area of the circle.
The area of a circle is given by the formula:
A_circle = π * r^2
Substituting the radius of 2 cm into the formula, we get:
A_circle = 3.14159 * 2^2 = 3.14159 * 4 = 12.5664 square cm
To express the sector area as a percentage, we use the formula:
Percentage = (A_sector / A_circle) * 100
Substituting the values, we get:
Percentage = (6 / 12.5664) * 100 = 47.872%
To express the enclosed area as a percentage, we use the formula:
Percentage = (A_enclosed / A_circle) * 100
Substituting the values, we get:
Percentage = (0 / 12.5664) * 100 = 0%
So, the area of the sector is approximately 47.872% of the area of the circle, and the enclosed area is 0% of the area of the circle.