Circles: parts, circumference and area

(a) A radius joins the centre of the circle to any point on the edge.

(b) A straight line joining two points on the circumference (not through the centre) is a chord.

The labelled line starts at the centre and stops at the circle.

A line like that is a radius. A diameter goes right across the circle through the centre, so the 'Diameter' label is wrong.

(a) Go across 3, then up 2: $A = (3, 2)$.

(b) Go 4 left from $O$, then 3 up, and mark a cross labelled B.

(c) Set compasses to 4 cm (4 squares). Put the point on $(1, -1)$ and draw the circle; it passes through $(5, -1)$, $(-3, -1)$, $(1, 3)$ and $(1, -5)$.

(a) A diameter is a straight line from one side of the circle to the other that passes through the centre.

(b) Draw any chord (a line joining two points on the circle), then shade the smaller region between the chord and the curved edge. That region is a segment.

(a) A line from the centre to the edge is a radius.

(b) A line that touches the circle at just one point is a tangent.

$2\pi r = 10$, so $r = \frac{5}{\pi}$

Area $= \pi r^2 = \pi \times \frac{25}{\pi^2} = \frac{25}{\pi}$ m²

The diagonal of the rectangle is a diameter.

$d = \sqrt{8^2 + 10^2} = \sqrt{164} = 12.806...$ cm

Circumference $= \pi d = \pi \times 12.806... = 40.23...$

$= 40.2$ cm (3 s.f.)

$\pi d = 44$, so $d = \frac{44}{\pi} = 14.0056...$ cm

Area $= \frac{1}{2} d^2 \sin 60° = \frac{1}{2} \times 14.0056^2 \times \sin 60°$

$= 84.9$ cm² (3 s.f.)

Let the diameters be $a$ (hypotenuse), $b$ and $c$. Pythagoras: $a^2 = b^2 + c^2$

A semicircle with diameter $d$ has area $\frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8}$

Multiply Pythagoras by $\frac{\pi}{8}$: $\frac{\pi a^2}{8} = \frac{\pi b^2}{8} + \frac{\pi c^2}{8}$

So area A = area B + area C.

Smaller hexagon: made of 6 equilateral triangles of side $r$, so perimeter $= 6r$.

Larger hexagon: the circle's radius is the height of each equilateral triangle, so half a side $= r\tan 30° = \frac{r}{\sqrt3}$, side $= \frac{2r}{\sqrt3}$, perimeter $= \frac{12r}{\sqrt3} = 4\sqrt3\,r$.

The circle lies between them: $6r < 2\pi r < 4\sqrt3\,r$

Divide by $2r$: $3 < \pi < 2\sqrt3$