+566 READ FIRST *I am so sorry for posting multiple, but my computer is really glitchy and it wouldn't let me post, giving me a small box many times, so when I finally got it to work like once or twice, I though that I might as well put it all in one post to save struggles later on.*
1. The figure shows a square in the interior of a regular hexagon. The square and regular hexagon share a common side. What is the degree measure of ∠ABC?
2. As shown in the diagram, AB and CD are two chords of circle O. The extension of the two chords intersect outside the circle at P. If PC = CD = 5 and PA = 4, find AB.
3. Rectangle ABCD is inscribed in circle O. We know AB = 4 and BC = 6. What is the combined area of the gray regions?
+591 1.
we can first start off by figuring out how many degrees are in each interior angle of a hexagon: (6-2)*180$^\circ$=720$^\circ$ $\frac{720^\circ}{6}=120^\circ$ and we know that the triangle with B and C as the two base vertices is an isoscolese, because the square and the hexagon have equal side lengths, so angle B = angle C. so we have
$\frac{180-(120-90)}/2$= angle C = $75^\circ$
and because we want to find angle ABC, we have to subtract 75$^\circ$ from 120 to get angle ABC = 120-75 = $\boxed{45^\circ}$
2.
Because of power of point, we know that PC$\cdot$CD = PA$\cdot$AB = 25 = 4AB
so AB=$\boxed{\frac{25}4}$
3.
The diagonal of rectangle ABCD is the diameter of the circle, so
4$^2$+6$^2$= CD^2 = 52
CD= 2$\sqrt{13}$
so the area of the square is:
$(\frac{2\sqrt{13}}2)^2\cdot\pi=13\pi$
so
shaded region = $\boxed{13\pi-24}$
+128406 First one
Call the triangle BCD
CD = BD
Angle CDB = 120 - 90 = 30
So angle CBD = (180 - 30) / 2 = 150 /2 = 75
So angle ABC = 120 - 75 = 45°
+591 1.
we can first start off by figuring out how many degrees are in each interior angle of a hexagon: (6-2)*180$^\circ$=720$^\circ$ $\frac{720^\circ}{6}=120^\circ$ and we know that the triangle with B and C as the two base vertices is an isoscolese, because the square and the hexagon have equal side lengths, so angle B = angle C. so we have
$\frac{180-(120-90)}/2$= angle C = $75^\circ$
and because we want to find angle ABC, we have to subtract 75$^\circ$ from 120 to get angle ABC = 120-75 = $\boxed{45^\circ}$
2.
Because of power of point, we know that PC$\cdot$CD = PA$\cdot$AB = 25 = 4AB
so AB=$\boxed{\frac{25}4}$
3.
The diagonal of rectangle ABCD is the diameter of the circle, so
4$^2$+6$^2$= CD^2 = 52
CD= 2$\sqrt{13}$
so the area of the square is:
$(\frac{2\sqrt{13}}2)^2\cdot\pi=13\pi$
so
shaded region = $\boxed{13\pi-24}$
+2401 Hello iamhappy!
I'm gonna try the third one. :))
AC is the diameter of the circle, and with pythagreon theorum. 4^2+6^2 = 52, sqrt(52)
So, the raduis is sqrt(13) since it's half the diameter.
The area of the circle is 13pi (sqrt(13)^2*pi)
And the area of the white rectangle is 4*6 = 24
So the area of the grey is 13pi-24.
I hope this helped. :))
=^._.^=
