NCERT Solutions · Class 9 Mathematics Measuring Space: Perimeter and Area

A sector is just a "pizza slice" of the circle — its area is the same fraction of the whole circle's area as its angle is of the whole $\displaystyle 360$° angle at the centre.Step $\displaystyle 1$: Write down what a sector's area formula is.For a sector of a circle, \[\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 \] where: This is really just: (fraction of the circle you have) × (area of the full circle). If the sector were the whole circle, \(\displaystyle \theta \) would be $\displaystyle 360$° and the fraction would be $\displaystyle 1$ — check that the formula gives back \(\displaystyle \pi r^2 \) in that case, which it does.Step $\displaystyle 2$: Write down what you're given. Step $\displaystyle 3$: Substitute into the formula.\[\text{Area} = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 \]Step $\displaystyle 4$: Simplify the angle fraction first.\[\frac{60}{360} = \frac{1}{6} \]Aside — this is the step people rush and get wrong. \(\displaystyle \theta \) must be in degrees here, and it's the fraction of a full turn ($\displaystyle 360$°), not of $\displaystyle 180$°. Since $\displaystyle 60$° is one-sixth of $\displaystyle 360$°, the sector really is one-sixth of the full circle.So now: \[\text{Area} = \frac{1}{6} \times \frac{22}{7} \times 7^2 \]Step $\displaystyle 5$: Deal with the \(\displaystyle 7^2 \) and the \(\displaystyle \div 7 \) together — don't turn \(\displaystyle \frac{22}{7} \) into a decimal yet.\[7^2 = 49, \qquad \frac{22}{7} \times 49 = 22 \times 7 = 154 \]Aside — this is the second place people trip up: they compute \(\displaystyle \frac{22}{7} \) as \(\displaystyle 3.14... \) too early and then the numbers stop cancelling nicely. Keep it as the fraction \(\displaystyle \frac{22}{7} \) until the $\displaystyle 7$ in the denominator can cancel with a $\displaystyle 7$ from \(\displaystyle r^2 \) — that's exactly what just happened.So: \[\text{Area} = \frac{1}{6} \times 154 = \frac{154}{6} \text{ cm}^2 \]Step $\displaystyle 6$: Simplify the fraction.\[\frac{154}{6} = \frac{77}{3} \text{ cm}^2 \]As a decimal (rounding only now, to $\displaystyle 2$ decimal places): \[\frac{77}{3} = 25.67 \text{ cm}^2 \text{ (approximately)} \]Answer: The area of the sector is \(\displaystyle \frac{77}{3} \) cm², which is approximately \(\displaystyle 25.67 \) cm².

Find the radius from the circumference first, then take one-quarter of the circle's area.A "quadrant" of a circle is a quarter of it — like one slice when you cut a circular pizza into $\displaystyle 4$ equal pieces with two cuts through the centre. So the plan is: get the radius from the circumference, find the area of the whole circle, then divide by 4.Step $\displaystyle 1$: Find the radius \(\displaystyle r\) from the circumference.The formula for circumference is \[C = 2\pi r \] where \(\displaystyle C\) is the circumference and \(\displaystyle r\) is the radius. We are told \(\displaystyle C = 44\) cm, and the instructions say to use \(\displaystyle \pi = \frac{22}{7}\).\[44 = 2 \times \frac{22}{7} \times r \]\[44 = \frac{44}{7} \times r \]Multiply both sides by \(\displaystyle \frac{7}{44}\) to isolate \(\displaystyle r\):\[r = 44 \times \frac{7}{44} = 7 \text{ cm} \]So the radius is \(\displaystyle r = 7\) cm. (This is a common trap spot — make sure you divide out the \(\displaystyle 2\pi\) correctly and don't accidentally use the area formula here. Circumference uses \(\displaystyle 2\pi r\); area uses \(\displaystyle \pi r^2\). Mixing them up is the single most common mistake in this type of question.)Step $\displaystyle 2$: Find the area of the full circle.The formula for the area of a circle is \[A = \pi r^2 \] where \(\displaystyle A\) is the area and \(\displaystyle r\) is the radius (the same \(\displaystyle r\) we just found).\[A = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 \]Since \(\displaystyle 49 = 7 \times 7\), one of those $\displaystyle 7$'s cancels with the $\displaystyle 7$ in the denominator:\[A = 22 \times 7 = 154 \text{ cm}^2 \]Step $\displaystyle 3$: Take one quarter of that area, since a quadrant is \(\displaystyle \frac{1}{4}\) of the circle.\[\text{Area of quadrant} = \frac{1}{4} \times A = \frac{1}{4} \times 154 \]\[\text{Area of quadrant} = \frac{154}{4} = 38.5 \text{ cm}^2 \](Watch out here: it's easy to stop at Step $\displaystyle 2$ and give $\displaystyle 154$ cm² as your final answer — but the question asks for the quadrant, not the whole circle. Always re-read what shape the question actually wants area of before you write your final line.)Answer: The area of the quadrant is \(\displaystyle 38.5 \text{ cm}^2\).

The minute hand sweeps out a sector of a circle — figure out what fraction of the full circle $\displaystyle 10$ minutes is, then take that fraction of the circle's area.Step $\displaystyle 1$: What shape does the minute hand trace?As the minute hand moves, its tip traces a circle. The length of the hand is the radius of that circle.So radius \(\displaystyle r = 7 \) cm.Step $\displaystyle 2$: How much of the circle does it cover in $\displaystyle 10$ minutes?The minute hand takes $\displaystyle 60$ minutes to go all the way around once — that's the full circle, \(\displaystyle 360^\circ \).In $\displaystyle 10$ minutes, the fraction of the full round it covers is: \[\frac{10}{60} = \frac{1}{6} \]Careful: this fraction is minutes-out-of-$\displaystyle 60$, not minutes-out-of-360. You don't need to convert to degrees at all here — the fraction \(\displaystyle \frac{1}{6} \) already tells you what part of the circle's area you want.Step $\displaystyle 3$: Formula for the area of a circle\[\text{Area of circle} = \pi r^2 \]where \(\displaystyle r \) is the radius.Using \(\displaystyle \pi = \dfrac{22}{7} \) as instructed: \[\text{Area of full circle} = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 \]Since \(\displaystyle 49 \div 7 = 7 \): \[= 22 \times 7 = 154 \text{ cm}^2 \]Step $\displaystyle 4$: Take the \(\displaystyle \frac{1}{6} \) fraction — this is the area sweptThe region the minute hand sweeps in $\displaystyle 10$ minutes is a sector (a pizza-slice piece) of this circle, covering \(\displaystyle \frac{1}{6} \) of it: \[\text{Area swept} = \frac{1}{6} \times 154 \text{ cm}^2 \]Careful: keep this as a fraction rather than dividing early and rounding — \(\displaystyle 154 \div 6 \) does not come out even, and rounding partway through would throw off the final answer.\[= \frac{154}{6} = \frac{77}{3} \text{ cm}^2 \]Step $\displaystyle 5$: Convert to a decimal\[\frac{77}{3} = 25.6\overline{6} \approx 25.67 \text{ cm}^2 \](rounded to two decimal places — the exact value is the fraction \(\displaystyle \frac{77}{3} \) cm², which is the same as \(\displaystyle 25\frac{2}{3} \) cm²).Answer: The area swept by the minute hand in $\displaystyle 10$ minutes is \(\displaystyle \dfrac{77}{3} \) cm\(\displaystyle ^2\), i.e. \(\displaystyle 25\dfrac{2}{3} \) cm\(\displaystyle ^2 \approx 25.67 \) cm\(\displaystyle ^2\).

A sector is a "pizza slice" of a circle: the slice takes exactly the same fraction of the circle's area as its angle takes of the full \(\displaystyle 360^\circ \) turn.What the question gives you. The circle has radius \(\displaystyle r = 10 \text{ cm} \). A chord has been drawn, and the two radii going to its endpoints make an angle of \(\displaystyle 90^\circ \) at the centre.The chord's only job here is to fix that \(\displaystyle 90^\circ \) angle. A sector is bounded by two radii and an arc, so the chord itself is not part of either sector's boundary and never enters the working below. (A chord does the real work when a question asks for a segment — the piece the chord cuts off. This question asks for sectors, so look only at the angle.)Which value of \(\displaystyle \pi \). This book normally uses \(\displaystyle \pi \approx \dfrac{22}{7} \), but this question tells you to use \(\displaystyle \pi \approx 3.14 \). So \(\displaystyle 3.14 \) is used everywhere below, and every decimal on this page comes from it.The two formulas.Area of a whole circle:\[A = \pi r^2 \]where \(\displaystyle r \) is the radius of the circle — the distance from the centre out to the edge.Area of a sector of that circle:\[\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 \]where \(\displaystyle \theta \) is the angle the sector makes at the centre, and \(\displaystyle \pi r^2 \) is the whole circle's area from the first formula. The fraction \(\displaystyle \dfrac{\theta}{360^\circ} \) is the share of the circle you are taking.Here \(\displaystyle r = 10 \text{ cm} \). The question says radius $\displaystyle 10$ cm, so \(\displaystyle 10 \text{ cm} \) is the number that goes into \(\displaystyle r^2 \).(i) The minor sector, \(\displaystyle \theta = 90^\circ \)\[\text{Area} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \pi \times (10 \text{ cm})^2 \]The fraction is \(\displaystyle \dfrac{90}{360} = \dfrac{1}{4} \), because a right angle is one quarter of a full turn around the centre.Squaring the radius:\[(10 \text{ cm})^2 = 100 \text{ cm}^2 \]So\[\text{Area} = \frac{1}{4} \times \pi \times 100 \text{ cm}^2 = 25\pi \text{ cm}^2 \]Now put in \(\displaystyle \pi \approx 3.14 \):\[\text{Area} = 25 \times 3.14 \text{ cm}^2 = 78.5 \text{ cm}^2 \](ii) The major sector, \(\displaystyle \theta = 270^\circ \)The major sector is everything left over — you get to it by turning through the other \(\displaystyle 360^\circ - 90^\circ = 270^\circ \).\[\text{Area} = \frac{270^\circ}{360^\circ} \times \pi r^2 = \frac{3}{4} \times \pi \times (10 \text{ cm})^2 \]The fraction is \(\displaystyle \dfrac{270}{360} = \dfrac{3}{4} \) — divide the top and the bottom by 90.\[\text{Area} = \frac{3}{4} \times \pi \times 100 \text{ cm}^2 = 75\pi \text{ cm}^2 \]Putting in \(\displaystyle \pi \approx 3.14 \):\[\text{Area} = 75 \times 3.14 \text{ cm}^2 = 235.5 \text{ cm}^2 \]A check on the arithmetic. The two sectors together make up the whole circle, since \(\displaystyle 90^\circ + 270^\circ = 360^\circ \). So the two areas should add up to the area of the whole circle:\[78.5 \text{ cm}^2 + 235.5 \text{ cm}^2 = 314 \text{ cm}^2 \]\[A = \pi r^2 = 3.14 \times 100 \text{ cm}^2 = 314 \text{ cm}^2 \]The two agree, so the fractions \(\displaystyle \frac{1}{4} \) and \(\displaystyle \frac{3}{4} \) were used correctly and the multiplication by \(\displaystyle 3.14 \) was done correctly in both parts.Be clear about what that check can and cannot catch. Because \(\displaystyle \dfrac{90}{360} + \dfrac{270}{360} = 1 \) exactly, the two sector areas add up to \(\displaystyle \pi r^2 \) whatever numbers you feed in for \(\displaystyle r \) and \(\displaystyle \pi \). So this check tests the arithmetic, not the starting numbers. For example, a student who wrongly squared the diameter \(\displaystyle 20 \text{ cm} \) would get \(\displaystyle 314 \text{ cm}^2 \) and \(\displaystyle 942 \text{ cm}^2 \), which sum to \(\displaystyle 1256 \text{ cm}^2 \), and their whole-circle area would be \(\displaystyle 3.14 \times 400 \text{ cm}^2 = 1256 \text{ cm}^2 \) — the check passes, even though both answers are four times too big.So check the radius separately, before you start. Read the question again and ask which length you have been handed. Here it says "a circle of radius $\displaystyle 10$ cm", so \(\displaystyle r = 10 \text{ cm} \) and \(\displaystyle r^2 = 100 \text{ cm}^2 \). If a question hands you the diameter instead, halve it first to get \(\displaystyle r \), and if it hands you the radius, do not halve it again.Answer: area of the minor sector \(\displaystyle = 78.5 \text{ cm}^2 \); area of the major sector \(\displaystyle = 235.5 \text{ cm}^2 \).

Split the pie-slice (sector) into a triangle plus a segment — the segment is what is left over once you cut the triangle away from the sector.A chord cuts the circle into two pieces: the smaller piece (bounded by the chord and the short arc) is the minor segment, and the leftover bigger piece (bounded by the chord and the long arc) is the major segment. To find a segment's area you always go through the sector first:\[\text{minor segment} = \text{sector} - \text{triangle} \]Given: radius \(\displaystyle r = 15 \) cm, angle at the centre \(\displaystyle \theta = 60^\circ \).The question tells you exactly which approximations to use here: \(\displaystyle \pi \approx 3.14 \) and \(\displaystyle \sqrt{3} \approx 1.73 \). (The usual default for this chapter is \(\displaystyle \pi \approx \frac{22}{7} \), but always use the value a question hands you directly — it overrides the default.)Step $\displaystyle 1$ — Area of the whole circle. Formula: \(\displaystyle A_{\text{circle}} = \pi r^2 \), where \(\displaystyle r \) is the radius. \[A_{\text{circle}} = 3.14 \times 15^2 = 3.14 \times 225 = 706.5 \text{ cm}^2 \]Step $\displaystyle 2$ — Area of the sector cut off by the $\displaystyle 60$° angle. Formula: \(\displaystyle A_{\text{sector}} = \dfrac{\theta}{360^\circ} \times \pi r^2 \), where \(\displaystyle \theta \) is the angle the two radii make at the centre. \[A_{\text{sector}} = \frac{60}{360} \times 706.5 = \frac{1}{6} \times 706.5 = 117.75 \text{ cm}^2 \] (Aside — this is where marks get lost: \(\displaystyle \theta \) must be in degrees here and divided by \(\displaystyle 360^\circ \), the full angle of the circle. Don't divide by \(\displaystyle 180^\circ \) or forget the division altogether.)Step $\displaystyle 3$ — The triangle inside the sector is equilateral, not just isosceles. The sector is bounded by two radii (\(\displaystyle OA \) and \(\displaystyle OB \), each \(\displaystyle = 15 \) cm) and the chord \(\displaystyle AB \). Triangle \(\displaystyle OAB \) has two sides equal to \(\displaystyle r \), so it is isosceles, and its included angle is \(\displaystyle 60^\circ \). The other two angles must be equal too, and all three angles add to \(\displaystyle 180^\circ \): \[\angle A = \angle B = \frac{180^\circ - 60^\circ}{2} = 60^\circ \] All three angles are \(\displaystyle 60^\circ \) — so triangle \(\displaystyle OAB \) is equilateral, meaning the chord \(\displaystyle AB \) also equals \(\displaystyle 15 \) cm. (Aside — this is the step people skip. Without noticing the triangle is equilateral, you'd try to use \(\displaystyle \frac{1}{2}r^2\sin\theta \), which needs a value of \(\displaystyle \sin 60^\circ \) the question hasn't given you. Because it gave you \(\displaystyle \sqrt{3} \approx 1.73 \) instead, that is your signal to use the equilateral-triangle area formula.)Step $\displaystyle 4$ — Area of the equilateral triangle. Formula: \(\displaystyle A_{\triangle} = \dfrac{\sqrt{3}}{4} a^2 \), where \(\displaystyle a \) is the side of the equilateral triangle (here \(\displaystyle a = r = 15 \) cm). \[A_{\triangle} = \frac{1.73}{4} \times 15^2 = \frac{1.73}{4} \times 225 = \frac{389.25}{4} = 97.3125 \text{ cm}^2 \]Step $\displaystyle 5$ — Area of the minor segment. \[A_{\text{minor segment}} = A_{\text{sector}} - A_{\triangle} = 117.75 - 97.3125 = 20.4375 \text{ cm}^2 \] Rounded to two decimal places: \(\displaystyle 20.44 \text{ cm}^2 \).Step $\displaystyle 6$ — Area of the major segment. The major segment is everything in the circle outside the minor segment — so subtract the minor segment from the whole circle, not from the sector. \[A_{\text{major segment}} = A_{\text{circle}} - A_{\text{minor segment}} = 706.5 - 20.4375 = 686.0625 \text{ cm}^2 \] Rounded to two decimal places: \(\displaystyle 686.06 \text{ cm}^2 \).

Region	Exact value	Rounded
Minor segment	\(\displaystyle 20.4375 \text{ cm}^2 \)	\(\displaystyle 20.44 \text{ cm}^2 \)
Major segment	\(\displaystyle 686.0625 \text{ cm}^2 \)	\(\displaystyle 686.06 \text{ cm}^2 \)

Check: the two segments should add back up to the whole circle — \(\displaystyle 20.4375 + 686.0625 = 706.5 \text{ cm}^2 \), which matches \(\displaystyle A_{\text{circle}} \) from Step 1. That confirms the arithmetic.Answer: Area of the minor segment \(\displaystyle \approx 20.44 \text{ cm}^2 \) (exactly \(\displaystyle 20.4375 \text{ cm}^2 \)); area of the major segment \(\displaystyle \approx 686.06 \text{ cm}^2 \) (exactly \(\displaystyle 686.0625 \text{ cm}^2 \)).

The area swept by one wiper blade is a sector of a circle — find the area of one sector, then double it for two wipers.When a wiper blade of length \(\displaystyle r \) sweeps through an angle \(\displaystyle \theta \) (in degrees), it traces a sector of a circle whose radius equals the blade's length. The area of a sector is:\[\text{Area of sector} = \frac{\theta}{360°} \times \pi r^2 \]where \(\displaystyle \theta \) is the angle swept and \(\displaystyle r \) is the radius (here, the length of the blade).Step $\displaystyle 1$: Write down what's given.\[r = 28 \text{ cm}, \qquad \theta = 120°, \qquad \pi = \frac{22}{7} \]Careful: the blade length is the radius here — a wiper blade pivots at one end and sweeps like the spoke of a circle, so the tip traces a circle of radius $\displaystyle 28$ cm.Step $\displaystyle 2$: Find the area swept by one blade.\[\text{Area}_1 = \frac{120°}{360°} \times \frac{22}{7} \times (28)^2 \]First simplify the fraction of the circle:\[\frac{120}{360} = \frac{1}{3} \]Now square the radius:\[28^2 = 784 \]Put it together:\[\text{Area}_1 = \frac{1}{3} \times \frac{22}{7} \times 784 \text{ cm}^2 \]Cancel $\displaystyle 7$ into $\displaystyle 784$ first (this keeps the numbers small and avoids a messy decimal mid-calculation):\[784 \div 7 = 112 \]\[\text{Area}_1 = \frac{1}{3} \times 22 \times 112 = \frac{2464}{3} \text{ cm}^2 \]So one blade cleans \(\displaystyle \dfrac{2464}{3} \text{ cm}^2 = 821\dfrac{1}{3} \text{ cm}^2 \approx 821.33 \text{ cm}^2 \).Step $\displaystyle 3$: Double it for two blades.The problem says the two wipers do not overlap — that's the key phrase. It means the two swept regions are completely separate, so you just add the two areas (you are not subtracting any shared overlap):\[\text{Total area} = 2 \times \text{Area}_1 = 2 \times \frac{2464}{3} = \frac{4928}{3} \text{ cm}^2 \]Step $\displaystyle 4$: Convert to a decimal.\[\frac{4928}{3} = 1642.666\ldots \approx 1642.67 \text{ cm}^2 \](Rounded to two decimal places, only at this final step — not earlier.)A common slip here: forgetting that "two wipers" means the sector area must be doubled, and stopping at the single-blade answer of \(\displaystyle 821\dfrac{1}{3} \text{ cm}^2 \). Always re-read what the question asks for — here it's the total area cleaned by both blades together.Answer: The total area cleaned at each sweep is \(\displaystyle \dfrac{4928}{3} \text{ cm}^2 \approx 1642.67 \text{ cm}^2 \).

Split the region into two shapes you already know how to measure: the pizza-slice sector \(\displaystyle OAB\) minus the flat triangle \(\displaystyle OAB\) sitting inside it.Let \(\displaystyle O\) be the centre of the circle and let chord \(\displaystyle AB\) subtend \(\displaystyle \angle AOB = 60°\) at \(\displaystyle O\). Since \(\displaystyle OA\) and \(\displaystyle OB\) are both radii, \(\displaystyle OA = OB = r\).The sector \(\displaystyle OAB\) (the "pizza slice" bounded by the two radii and the arc) is made up of exactly two pieces: the triangle \(\displaystyle OAB\), and the minor segment sitting between the chord and the arc. So:\[\text{Area of minor segment} = \text{Area of sector } OAB - \text{Area of } \triangle OAB \]Step $\displaystyle 1$ — find the shape of the triangle. In \(\displaystyle \triangle OAB\), \(\displaystyle OA = OB = r\), so the triangle is isosceles and the two base angles are equal. Call each base angle \(\displaystyle x\). The angles of any triangle add to \(\displaystyle 180°\): \[60° + x + x = 180° \implies 2x = 120° \implies x = 60° \] All three angles are \(\displaystyle 60°\), so \(\displaystyle \triangle OAB\) is equilateral, which means all three sides are equal too: \[AB = OA = OB = r \] Watch out: it's tempting to treat \(\displaystyle AB\) as some unknown length you'd need extra work to find. You don't — a \(\displaystyle 60°\) central angle always makes the chord exactly equal to the radius.Step $\displaystyle 2$ — area of the sector. Formula for the area of a sector, where \(\displaystyle \theta\) is the angle at the centre in degrees: \[\text{Area of sector} = \frac{\theta}{360°}\times \pi r^{2} \] Here \(\displaystyle \theta = 60°\), so: \[\text{Area of sector } OAB = \frac{60°}{360°}\times \pi r^{2} = \frac{1}{6}\pi r^{2} = \frac{\pi r^{2}}{6} \]Step $\displaystyle 3$ — area of the triangle, using Heron's formula. All three sides of \(\displaystyle \triangle OAB\) equal \(\displaystyle r\): \(\displaystyle a = b = c = r\). Heron's formula needs the semi-perimeter \[s = \frac{a+b+c}{2} = \frac{r+r+r}{2} = \frac{3r}{2} \] Then \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] Each of \(\displaystyle s-a,\ s-b,\ s-c\) equals \(\displaystyle \dfrac{3r}{2}-r = \dfrac{r}{2}\), so \[\text{Area of }\triangle OAB = \sqrt{\frac{3r}{2}\cdot\frac{r}{2}\cdot\frac{r}{2}\cdot\frac{r}{2}} = \sqrt{\frac{3r^{4}}{16}} = \frac{\sqrt{3}}{4}r^{2} \] Notice there is no \(\displaystyle \pi\) anywhere in this line. The triangle is three straight edges — it has nothing to do with the circle's curve, so \(\displaystyle \pi\) never belongs to it. \(\displaystyle \pi\) only ever appeared in Step $\displaystyle 2$, attached to the sector.Step $\displaystyle 4$ — subtract to get the segment. \[\text{Area of minor segment} = \frac{\pi r^{2}}{6} - \frac{\sqrt{3}}{4}r^{2} \] This is the exact spot where it's easy to go wrong: it looks like you can pull \(\displaystyle \pi r^{2}\) out in front of both fractions and write \(\displaystyle \pi r^{2}\left(\dfrac16 - \dfrac{\sqrt3}{4}\right)\). Don't — that drags \(\displaystyle \pi\) onto the triangle term too, even though Step $\displaystyle 3$ just showed the triangle has no \(\displaystyle \pi\) in it. (It also can't be right for another reason: \(\displaystyle \dfrac16 - \dfrac{\sqrt3}{4}\) works out negative, and an area can never be negative.) The two terms only share the factor \(\displaystyle r^{2}\), not \(\displaystyle \pi\), so that is the only thing you're allowed to pull out: \[\text{Area of minor segment} = r^{2}\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \] This is the required result, with \(\displaystyle \pi\) sitting only where it actually belongs — next to the \(\displaystyle 6\), not multiplying the whole bracket.As a check, using \(\displaystyle \pi \approx \dfrac{22}{7}\) and \(\displaystyle \sqrt3 \approx 1.732\): \(\displaystyle \dfrac{\pi}{6}\approx 0.524\) and \(\displaystyle \dfrac{\sqrt3}{4}\approx 0.433\), so the bracket is about \(\displaystyle 0.091\) — a small positive number, which makes sense: a chord subtending only \(\displaystyle 60°\) cuts off a thin sliver of the circle, not a big piece.Answer: Area of the minor segment \(\displaystyle = r^{2}\left(\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}\right)\) square units — the same as \(\displaystyle \dfrac{\pi r^{2}}{6}-\dfrac{\sqrt{3}}{4}r^{2}\), or \(\displaystyle \dfrac{r^{2}}{12}\left(2\pi-3\sqrt{3}\right)\).

Split the equilateral triangle into its altitude and use the fact that the centre of the circle is also the centroid of the triangle — that one fact turns the whole problem into two Pythagoras calculations.Step $\displaystyle 1$ — set up the pictureLet \(\displaystyle ABC\) be an equilateral triangle with side \(\displaystyle a\), drawn inside a circle of centre \(\displaystyle O\) and radius \(\displaystyle r\). Since the triangle is inscribed in the circle, every vertex touches the circle, so \[OA = OB = OC = r. \]Draw the altitude from \(\displaystyle A\) to \(\displaystyle BC\), meeting it at \(\displaystyle D\). Two circle/triangle facts we can use here: So on the line \(\displaystyle AD\), the point \(\displaystyle O\) splits it as \(\displaystyle AO:OD = 2:1\), which means \[AO = \frac{2}{3}AD. \]Common mix-up: it's easy to write this ratio backwards. Remember the vertex gets the bigger share — \(\displaystyle 2\) parts near \(\displaystyle A\), \(\displaystyle 1\) part near \(\displaystyle D\).Since \(\displaystyle AO\) is a radius, \(\displaystyle AO = r\). Substituting: \[r = \frac{2}{3}AD \quad\Longrightarrow\quad AD = \frac{3}{2}r. \]Step $\displaystyle 2$ — write the altitude in terms of the side \(\displaystyle a\)In right triangle \(\displaystyle ABD\) (right-angled at \(\displaystyle D\)), \(\displaystyle AB = a\) is the hypotenuse and \(\displaystyle BD = \dfrac{a}{2}\) (since \(\displaystyle D\) is the midpoint of \(\displaystyle BC\)). By the Pythagoras theorem, \[AD^2 = AB^2 - BD^2 = a^2 - \left(\frac{a}{2}\right)^2 = \frac{3a^2}{4}. \] \[AD = \frac{\sqrt{3}}{2}\,a. \]Step $\displaystyle 3$ — combine the two expressions for \(\displaystyle AD\)From Step $\displaystyle 1$, \(\displaystyle AD = \dfrac{3}{2}r\). From Step $\displaystyle 2$, \(\displaystyle AD = \dfrac{\sqrt3}{2}a\). These are the same length, so \[\frac{\sqrt3}{2}a = \frac{3}{2}r \quad\Longrightarrow\quad a = \frac{3r}{\sqrt3} = r\sqrt3. \](Check the algebra: \(\displaystyle \dfrac{3}{\sqrt3} = \dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3} = \dfrac{3\sqrt3}{3} = \sqrt3\).) So the side of the triangle is exactly \(\displaystyle \sqrt3\) times the radius of the circle — this is the key relationship the whole problem hinges on.Step $\displaystyle 4$ — area of the triangleFormula: Area of a triangle \(\displaystyle = \dfrac12 \times \text{base} \times \text{height}\), with base \(\displaystyle a\) and height \(\displaystyle AD = \dfrac{\sqrt3}{2}a\): \[\text{Area}_{\triangle} = \frac12 \cdot a \cdot \frac{\sqrt3}{2}a = \frac{\sqrt3}{4}a^2. \] Now substitute \(\displaystyle a = r\sqrt3\), so \(\displaystyle a^2 = 3r^2\): \[\text{Area}_{\triangle} = \frac{\sqrt3}{4}\left(3r^2\right) = \frac{3\sqrt3}{4}\,r^2. \]Step $\displaystyle 5$ — area of the circleFormula: Area of a circle \(\displaystyle = \pi r^2\), where \(\displaystyle r\) is the radius. \[\text{Area}_{\text{circle}} = \pi r^2. \]Step $\displaystyle 6$ — the ratio\[\frac{\text{Area}_{\triangle}}{\text{Area}_{\text{circle}}} = \frac{\dfrac{3\sqrt3}{4}r^2}{\pi r^2} = \frac{3\sqrt3}{4\pi}. \]Notice \(\displaystyle r^2\) cancels from top and bottom — that's why the ratio doesn't depend on how big the circle is, only on the shape (an equilateral triangle fitted exactly inside a circle). This is exactly the ratio we were asked to show.Step $\displaystyle 7$ — check the decimal valueUsing \(\displaystyle \pi \approx \dfrac{22}{7}\) as instructed, and \(\displaystyle \sqrt3 \approx 1.732\) (rounding here only, at the end): \[3\sqrt3 \approx 3 \times 1.732 = 5.196, \qquad 4\pi \approx 4 \times \frac{22}{7} = \frac{88}{7} \approx 12.571. \] \[\frac{3\sqrt3}{4\pi} \approx \frac{5.196}{12.571} \approx 0.413. \]This matches the value given in the question, confirming the algebra is correct.Answer: The ratio of the area of the inscribed equilateral triangle to the area of the circle is \(\displaystyle \dfrac{3\sqrt3}{4\pi} \approx 0.413\), proved by showing the triangle's side must equal \(\displaystyle r\sqrt3\).

## Show that the ratio of the areas is \(\displaystyle \dfrac{2}{\pi}\)The one thing that connects a square to a circle drawn around it is the diagonal — the square's diagonal is exactly the circle's diameter. Once you have that, everything else is just two area formulas and some cancelling.Step $\displaystyle 1$ — draw what "inscribed" means"A square inscribed in a circle" means all four corners of the square touch the circle. If you draw the diagonal of the square, it passes straight through the centre of the circle — because the diagonal is the longest line you can fit inside the square, and it stretches corner to corner, exactly like a diameter stretches from one side of the circle to the other through the centre.So: \[\text{diagonal of the square} = \text{diameter of the circle} = 2r \]Step $\displaystyle 2$ — write the diagonal in terms of the sideLet the side of the square be \(\displaystyle a\). The diagonal cuts the square into two right triangles, so by the Pythagorean theorem (for a right triangle with legs \(\displaystyle p, q\) and hypotenuse \(\displaystyle h\): \(\displaystyle h^2 = p^2 + q^2\)), with both legs equal to \(\displaystyle a\): \[\text{diagonal}^2 = a^2 + a^2 = 2a^2 \] \[\text{diagonal} = a\sqrt{2} \]Step $\displaystyle 3$ — set the two expressions for the diagonal equal\[a\sqrt{2} = 2r \] \[a = \frac{2r}{\sqrt{2}} = r\sqrt{2} \]Careless mistake to avoid: don't skip straight to "area of square = \(\displaystyle 2r^2\)" without this step. It's easy to mix up the diagonal with the side and use \(\displaystyle a = 2r\) directly — that would make the square way too big. The diagonal is \(\displaystyle 2r\); the side is \(\displaystyle r\sqrt{2}\), which is smaller.Step $\displaystyle 4$ — find the area of the squareFormula: Area of a square \(\displaystyle = (\text{side})^2\). \[\text{Area of square} = a^2 = \left(r\sqrt{2}\right)^2 = 2r^2 \]Step $\displaystyle 5$ — find the area of the circleFormula: Area of a circle \(\displaystyle = \pi r^2\), where \(\displaystyle r\) is the radius. \[\text{Area of circle} = \pi r^2 \]Step $\displaystyle 6$ — take the ratio\[\frac{\text{Area of square}}{\text{Area of circle}} = \frac{2r^2}{\pi r^2} \]Since \(\displaystyle r \neq 0\), the \(\displaystyle r^2\) on top and bottom cancel completely: \[\frac{\text{Area of square}}{\text{Area of circle}} = \frac{2}{\pi} \]This is exactly what we had to show — and notice the radius \(\displaystyle r\) disappeared entirely. That makes sense: any square inscribed in any circle has this same ratio, no matter how big the circle is.Step $\displaystyle 7$ — check the decimal value\[\frac{2}{\pi} \approx \frac{2}{3.14159} \approx 0.637 \]Aside on \(\displaystyle \pi\): this problem's instruction tells you to use \(\displaystyle \pi \approx \frac{22}{7}\) unless told otherwise, but here you're asked to match \(\displaystyle 0.637\) specifically — and \(\displaystyle \dfrac{2}{22/7} = \dfrac{14}{22} = \dfrac{7}{11} \approx 0.636\), not \(\displaystyle 0.637\). That's because \(\displaystyle 0.637\) comes from the more accurate value \(\displaystyle \pi \approx 3.14159\), not the rougher \(\displaystyle \frac{22}{7}\). The exact algebraic answer \(\displaystyle \dfrac{2}{\pi}\) is correct either way — only the decimal you round it to depends on which value of \(\displaystyle \pi\) you plug in.Answer: The ratio of the area of the square to the area of the circle is \(\displaystyle \dfrac{2}{\pi} \approx 0.637 \), independent of the radius \(\displaystyle r\).

Split the hexagon into $\displaystyle 6$ identical equilateral triangles that meet at the circle's centre — that unlocks both this question and its link to Question 8.Step $\displaystyle 1$: Cut the hexagon from the centre.Let \(\displaystyle O\) be the centre of the circle, and let the hexagon be \(\displaystyle ABCDEF\) with every vertex on the circle, so \(\displaystyle OA = OB = OC = OD = OE = OF = r\).Joining \(\displaystyle O\) to each vertex splits the hexagon into $\displaystyle 6$ triangles: \(\displaystyle OAB, OBC, OCD, ODE, OEF, OFA\). A regular hexagon has $\displaystyle 6$ equal angles at the centre, so \[\angle AOB = \angle BOC = \cdots = \frac{360^\circ}{6} = 60^\circ. \]Aside — the surprising bit people miss: In triangle \(\displaystyle OAB\), \(\displaystyle OA = OB = r\), so it is isosceles, which means its two base angles \(\displaystyle \angle OAB\) and \(\displaystyle \angle OBA\) are equal. The three angles add to \(\displaystyle 180^\circ\) and the apex is \(\displaystyle 60^\circ\), so each base angle is also \(\displaystyle 60^\circ\). All three angles are \(\displaystyle 60^\circ\) — so triangle \(\displaystyle OAB\) is not just isosceles, it is equilateral, which forces \(\displaystyle AB = r\) too. In other words, a regular hexagon's side is exactly equal to the radius of its circumscribing circle. It is easy to assume the side must be smaller than \(\displaystyle r\) — it isn't.The same holds for all $\displaystyle 6$ triangles, so the hexagon is made of $\displaystyle 6$ equilateral triangles, each of side \(\displaystyle r\).Step $\displaystyle 2$: Area of one small triangle.Formula, Area of an equilateral triangle of side \(\displaystyle a\): \[\text{Area} = \frac{\sqrt3}{4} a^2. \] Here \(\displaystyle a = r\), so each of the $\displaystyle 6$ triangles has area \(\displaystyle \dfrac{\sqrt3}{4} r^2\).Step $\displaystyle 3$: Area of the hexagon.\[\text{Area(hexagon)} = 6 \times \frac{\sqrt3}{4} r^2 = \frac{6\sqrt3}{4} r^2 = \frac{3\sqrt3}{2} r^2. \]Step $\displaystyle 4$: Area of the circle.Formula, Area of a circle of radius \(\displaystyle r\): \(\displaystyle \text{Area} = \pi r^2\). So \[\text{Area(circle)} = \pi r^2. \]Step $\displaystyle 5$: The ratio.\[\frac{\text{Area(hexagon)}}{\text{Area(circle)}} = \frac{\dfrac{3\sqrt3}{2} r^2}{\pi r^2} = \frac{3\sqrt3}{2\pi}. \]The \(\displaystyle r^2\) cancels top and bottom — that's exactly why the problem can ask you to prove one fixed number that works for every circle, no matter its radius.Now check the decimal, using \(\displaystyle \pi = \dfrac{22}{7}\) and \(\displaystyle \sqrt3 \approx 1.732\): \[\frac{3\sqrt3}{2\pi} = \frac{3 \times 1.732}{2 \times \dfrac{22}{7}} = \frac{5.196}{\dfrac{44}{7}} = \frac{5.196 \times 7}{44} = \frac{36.372}{44} \approx 0.827. \]That matches the value given in the question.Step $\displaystyle 6$: Why is this exactly twice Question $\displaystyle 8$'s answer?Question $\displaystyle 8$ put an equilateral triangle inside the same circle of radius \(\displaystyle r\), and found \[\text{Area(triangle)} = \frac{3\sqrt3}{4} r^2, \qquad \frac{\text{Area(triangle)}}{\text{Area(circle)}} = \frac{3\sqrt3}{4\pi}. \]Here is the picture that connects the two: in hexagon \(\displaystyle ABCDEF\), join every other corner — \(\displaystyle A\), \(\displaystyle C\), \(\displaystyle E\) — skipping \(\displaystyle B\), \(\displaystyle D\), \(\displaystyle F\). Since \(\displaystyle A, C, E\) are still all on the same circle, triangle \(\displaystyle ACE\) is exactly an equilateral triangle inscribed in the same circle of radius \(\displaystyle r\) — it is the Question $\displaystyle 8$ triangle.That triangle sits inside the hexagon, and the hexagon minus that triangle leaves $\displaystyle 3$ small corner pieces (triangles \(\displaystyle ABC\), \(\displaystyle CDE\), \(\displaystyle EFA\), cut off at \(\displaystyle B\), \(\displaystyle D\), \(\displaystyle F\)). We don't need to work these corner pieces out separately — we already have both areas from Steps $\displaystyle 3$ and $\displaystyle 6$: \[\text{leftover corners} = \text{Area(hexagon)} - \text{Area(triangle)} = \frac{3\sqrt3}{2} r^2 - \frac{3\sqrt3}{4}r^2 = \frac{3\sqrt3}{4} r^2. \]That is exactly equal to the triangle's own area. So triangle \(\displaystyle ACE\) does not just sit inside the hexagon — it splits it exactly in half: \[\text{Area(hexagon)} = \text{Area(triangle)} + \text{leftover corners} = \text{Area(triangle)} + \text{Area(triangle)} = 2 \times \text{Area(triangle)}. \]Since both areas are being measured against the very same circle, dividing both sides by \(\displaystyle \pi r^2\) carries the "twice" straight through to the ratios: \[\frac{3\sqrt3}{2\pi} = 2 \times \frac{3\sqrt3}{4\pi}. \]That is exactly why joining alternate corners of a regular hexagon inscribed in a circle always gives a triangle covering precisely half the hexagon — no matter what the radius is.Answer: \(\displaystyle \dfrac{\text{Area(hexagon)}}{\text{Area(circle)}} = \dfrac{3\sqrt3}{2\pi} \approx 0.827\); this is exactly twice Question $\displaystyle 8$'s ratio \(\displaystyle \dfrac{3\sqrt3}{4\pi}\) because joining every other vertex of the hexagon reproduces Question $\displaystyle 8$'s triangle inside the same circle, and that triangle exactly bisects the hexagon (the $\displaystyle 3$ leftover corner pieces together have the same area as the triangle itself).