Two perpendicular chords divide a circle with a radius of 13 cm into 4 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.

Please include solution. Thanks.

4 and 1/2

Oh wait. . im sorry but i think its wrong . sorry :(

To find the area of the smallest part of the circle, we need to first determine the lengths of the chords. Let's call one chord AB and the other CD. Since the distances from both chords to the center of the circle are each 5 cm, we can draw radii from the center of the circle to the points where the chords intersect. Let's call these points M and N, where AM = BM = 5 cm and CN = DN = 5 cm.

We also know that the chords are perpendicular, which means that the segments AM and CN are perpendicular bisectors of each other. Therefore, triangles AMC and CND are congruent.

We can use the Pythagorean theorem to find the length of the chords. In triangle AMC, we have AM = 5 cm and MC = 13 cm (the radius of the circle). Using the Pythagorean theorem, we can find the length of AC:

AC^2 = AM^2 + MC^2
AC^2 = 5^2 + 13^2
AC^2 = 25 + 169
AC^2 = 194
AC = √194

Since triangles AMC and CND are congruent, CD is also equal to √194.

Now, let's find the area of the smallest part of the circle. The smallest part is formed by the sector AMC and the triangle AMC, with the chord CD as its base.

First, let's find the area of the sector. The central angle of the sector is 90 degrees, which is one-fourth of the entire circle. Therefore, the area of the sector is:

Area_sector = (90/360) × π × r^2
Area_sector = (1/4) × π × 13^2
Area_sector = (1/4) × 169π
Area_sector = 42.25π

Next, let's find the area of triangle AMC. The base of the triangle is the chord CD, which has a length of √194. To find the height of the triangle, we can use the Pythagorean theorem again:

height^2 = AM^2 - (AC/2)^2
height^2 = 5^2 - (√194/2)^2
height^2 = 25 - 97
height^2 = -72 (This is not possible since height cannot be negative)

Since the height of the triangle is not possible, the area of the triangle is also not possible.

Therefore, the area of the smallest part of the circle is just the area of the sector, which is 42.25π square cm.

Start with one chord which is 5 units from the centre

Notice the perpendicular height of 5 cm hits the midpoint of the chord.
By Pythagoras, we can see that the chord must be 24 cm long (5^2 + 12^2 = 13^2)

Now pretend you are rotating the chord around the centre until you have reached the perpendicualr postition. Clearly the new chord must also be 24
Notice the square 5 by 5 created by the two chords and the perpendiculars from the centre ?

Clearly that leaves our small region to be a 1/4 sector of a circle with radius 7
area of smaller sector = (1/4)π(7^2) = 49π/4