Areas Related to Circles: questions with answers

Each cut along a diameter passes through the centre and separates the circle on both sides. So $4$ diameters produce $8$ sectors. To end up with $8$ equal slices, only $4$ cuts are needed. Reading the two counts as the same number makes $4$ look right in the first branch and $8$ in the second. The factor of two is the whole content of the question.

Section 11.1 names the two parts by their BOUNDARY, not by their size. Two radii and the arc between them enclose a SECTOR; a chord and the corresponding arc enclose a SEGMENT. Here the part is enclosed between a chord and the corresponding arc, so it is a segment of the circle. Giving the other name swaps the two definitions. The swap is not corrected by looking at how big the part is, because there are minor and major sectors just as there are minor and major segments. An arc is the curved edge alone and encloses no region, and a semicircle is one particular part rather than a kind of part.

A sector's area fraction equals its central angle divided by $360^\circ$. Thus $\theta = \frac{1}{ 10 } \times 360^\circ = 36^\circ$. Using $180^\circ$ for the whole circle gives $18^\circ$.

The horse must stay inside the field, so it can graze only within the corner, whose angle is $60^\circ$. The rope gives a radius of $6$ m. Every side not meeting the tether corner is at least $6$ m away in perpendicular distance. The perpendicular is the shortest distance from the corner to a side, so every point on any far side is at least $6$ m away. Thus no far side cuts off the grazing sector. The grazing area is $A=\frac{ 60^\circ }{360^\circ}\pi(6)^2=6\pi\ \text{m}^2$. The full-circle value $36\pi$ ignores that the horse must stay inside the field. The value $9\pi$ uses a $90^\circ$ angle instead of this field’s $60^\circ$ corner angle, and $18\pi$ uses a semicircle instead of the stated corner.

The whole disc has area $\pi \times 24^2 = 576\pi\ \text{cm}^2$, and the sector is the fraction $\frac{\theta}{360}$ of it. So $\frac{\theta}{360} = \frac{ 64 }{ 576 }$ and $\theta = 40^\circ$. Answering $360^\circ$ is the fraction dropped, read in reverse: it takes the given $64\pi$ for the WHOLE disc, and it cannot be right here because the disc measures $576\pi$. $80^\circ$ and $20^\circ$ put the factor 2 in the wrong place.

The hour hand goes once round in 12 HOURS, not one, so it turns $\frac{360^\circ}{720} = \frac{1}{2}^\circ$ a minute: in $50$ minutes it turns $\frac{ 50 }{2}^\circ = 25^\circ$. The hand's length, $14$ cm, is not needed for this angle -- it only matters if the question later asks for the area swept. Answering $300^\circ$ reads the rate off the MINUTE hand, twelve times too fast, and $50^\circ$ counts one degree a minute.

Read $\text{arc} = \frac{\theta}{360} \times 2\pi r$ backwards. The given number is a LENGTH, so it is compared with the CIRCUMFERENCE $2\pi \times 10$: $\frac{\theta}{360} = \frac{ 5\pi }{2\pi \times 10}$, giving $\theta = 90^\circ$. Comparing it with the AREA $\pi r^2$ instead divides by an extra factor of $\frac{r}{2}$ and gives $18^\circ$, also a whole number of degrees. Only the units reveal the mistake, since an area cannot equal a length. $180^\circ$ and $45^\circ$ mis-place the factor of 2 in $2\pi r$.

The two radii must come off before the arc can name an angle. The arc is $36 - 2 \times 7 = 22$ cm. Then $\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 7 = 22$ gives $\theta = 180^\circ$. Treating the whole perimeter as the arc gives $3240/11^\circ$ instead. That answer is not a whole number of degrees, which is the sign that the two radii were never removed. $360^\circ$ doubles the answer and $90^\circ$ guesses at the shape.

The perimeter of a sector counts the arc AND the two radii, so the two radii come off first: $\text{arc} = 72 - 2 \times 14 = 44$ cm. Reporting $72$ treats the whole boundary as the curved part. $58$ takes only one radius off, and $36$ halves the perimeter, which is right only when the two radii happen to equal the arc.

Drop the perpendicular from the centre to the chord. It bisects both the chord and the angle, leaving a right triangle with a HALF-angle of $\frac{ 90 }{2}^\circ$ at the centre. So half the chord is $r \sin \frac{ 90 }{2}^\circ$ and the whole chord is $2r \sin \frac{ 90 }{2} ^\circ = 10\sqrt{2}$ cm. At $60^\circ$ the two radii and the chord form an equilateral triangle. So the chord equals the radius itself. Answering $20$ uses $2r \sin 90^\circ$ with the whole angle instead of the half-angle. That length matches no side of this figure. $10\sqrt{3}$ is the other standard chord value, for a different angle.

The radius is $10$ cm. The perpendicular from the centre bisects the chord and the $60^\circ$ angle. This leaves a right-angled triangle with hypotenuse $10$ cm and a HALF-angle of $30^\circ$. So the distance is $10 \cos 30^\circ = 5\sqrt{3}$ cm. Answering $5$ uses $10 \cos 60^\circ$ with the whole angle -- and that number is in fact half the CHORD, not the distance to it. $10\sqrt{3}$ and $10$ drop the halving altogether.

The major segment is everything the chord leaves on the far side. So it is the disc, $12474\ \text{cm}^2$, less the MINOR SEGMENT, not less the minor sector. Taking away the minor sector leaves the major sector, $8316\ \text{cm}^2$. But the triangle between the two radii and the chord is still missing from it. That triangle lies on the major side of the chord, so it belongs to the answer. So the major segment is $8316 + \frac{3969}{4}\sqrt{3}\ \text{cm}^2$, larger than the major sector by exactly the triangle. $4158 - \frac{3969}{4}\sqrt{3}$ is the minor segment, the region on the other side of the chord.

The region is the same fraction $\frac{ 60 }{360}$ of the RING that a sector is of a disc, so it is $\frac{ 60 }{360} \times \pi(10^2 - 4^2) = 14\pi\ \text{cm}^2$. Both steps have to be taken: the squares are subtracted, then the fraction of the turn is applied. Answering $6\pi$ squares the width $6$ cm instead of subtracting the squares, and $84\pi$ is the whole ring, with the angle forgotten.

The ring is what is left when the inner disc is taken out of the outer one. Its area is $\pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \frac{22}{7}(21^2 - 14^2) = 770\ \text{cm}^2$. The subtraction happens between the SQUARES, after each disc is measured. Answering $154$ subtracts the radii first and squares the width, $\frac{22}{7}(21 - 14)^2$. That is the area of a circle of radius $7$ cm, a disc that has nothing to do with the ring, since $(R - r)^2$ is never $R^2 - r^2$. $1386$ is the outer disc alone and $2002$ adds the two discs.

The two sectors on one pair of radii fill the disc between them. So the major sector stands on $360^\circ - 30^\circ = 330$. Its area is $\frac{ 330 }{360} \times \frac{22}{7} \times 42^2 = 5082\ \text{cm}^2$. The same number comes from the disc, $5544\ \text{cm}^2$, less the minor sector. Answering $462$ puts the GIVEN angle into the formula, which measures the minor sector -- the smaller of the two, so it cannot be the major one. $2772$ takes the major part to be simply half the disc.