Question

two perpendicular chords divide a circle with a radiusof 13 cm. into four parts, if the perpendicular distances of both chords are 5 cm. each from the center of the circle , find the area of the smallest part

Answer 1

http://openstudy.com/users/sara12345#/updates/52835482e4b0218bc650b4e4

Answer 2

From the diagram, triangle AOG is right-angled at G, so AG = √(13^2 - 5^2) = 12 cm.
Therefore, x = AC = BC = 12 - 5 = 7 cm. Therefore, AB = 7√2 and AD = BD = 7√2/2.

Let θ = angle AOB. Then angle OAB = 90 - θ/2. In triangle AOB,
(7√2)/sin(θ) = 13/sin(90 - θ/2). Therefore, θ = 2*arcsin(7√2/26).

Area of sector AOB = (1/2)*13^2*2*arcsin(7√2/26) = 66.01259 cm^2.

By Pythagoras, OC = 5√2, so area of triangle OCB = (1/2)*5√2*13*(7√2/26) = 35/2.
If it's not clear, the multiplier, 7√2/26, is the value of sin(θ/2) (for angle DOB).
Area of triangle OCA = area of triangle OCB, so 2*(area of triangle AOC) = 35 cm^2.

Thus, grey area = Area of sector AOB - 2*(area of triangle AOC)
= 66.01259 - 35
= 31.01259 cm^2

General formula (R = radius of circle, d = distance of both chords from centre) :
Let k = √(R^2 - d^2) - d, and provided that d ≤ R/√2, then -
Area of smallest part is A = R^2*arcsin[k/(R√2)] - dk

In both k and A, replacing (d) with (-d) gives A = area of largest part.
The other two equal areas are then easily obtained by subtraction from πR^2

Answer 3

draw a figure first.

Answer 4

the wau that sir ??

Answer 5

did you get all those ???
i used the pythagoras theorem once