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The perimeter of a circle is 44 cm. What is its radius?
Answer: Let the radius be $r$.
Perimeter (Circumference) $= 2\pi r = 44$.
Using $\pi \approx \frac{22}{7}$:
$2 \times \frac{22}{7} \times r = 44$
$\frac{44}{7} \times r = 44$
$r = 7$ cm.
Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.
Answer: Circumference $C = 2\pi r$. Using $\pi \approx \frac{22}{7}$:
(i) $C = 2 \times \frac{22}{7} \times 7 = 44$ cm $\rightarrow$ 44.0 cm.
(ii) $C = 2 \times \frac{22}{7} \times 10 = \frac{440}{7} \approx 62.857$ $\rightarrow$ 62.9 cm.
(iii) $C = 2 \times \frac{22}{7} \times 12 = \frac{528}{7} \approx 75.428$ $\rightarrow$ 75.4 cm.
Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is $60^\circ$, and (ii) the radius is 6.3 m and the angle at the centre is $120^\circ$.
Answer: Arc length $l = 2\pi r \times \frac{\theta}{360}$.
(i) $l = 2 \times \frac{22}{7} \times 3.5 \times \frac{60}{360} = 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{1}{6} = 22 \times \frac{1}{6} = \frac{11}{3} \approx$ 3.67 cm.
(ii) $l = 2 \times \frac{22}{7} \times 6.3 \times \frac{120}{360} = 2 \times \frac{22}{7} \times \frac{63}{10} \times \frac{1}{3} = 2 \times 22 \times \frac{9}{10} \times \frac{1}{3} = 2 \times 22 \times \frac{3}{10} = \frac{132}{10} =$ 13.2 m.
Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle $75^\circ$.
Answer: Perimeter of sector = Arc length + 2(Radius).
Arc length $l = 2\pi r \times \frac{\theta}{360} = 2 \times \frac{22}{7} \times 14 \times \frac{75}{360}$.
$l = 2 \times 22 \times 2 \times \frac{5}{24} = 88 \times \frac{5}{24} = \frac{11 \times 5}{3} = \frac{55}{3} \approx 18.33$ cm.
Perimeter $= \frac{55}{3} + 14 + 14 = 18.33 + 28 = $ 46.33 cm.
The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Answer: This is an isosceles trapezium. Let the parallel sides be $a = 20$ cm and $b = 40$ cm, and non-parallel sides be $c = 26$ cm.
Drop perpendiculars from the ends of the shorter parallel side to the longer parallel side. This forms a rectangle in the middle (width 20) and two identical right triangles on the ends.
Base of each right triangle = $\frac{40 - 20}{2} = 10$ cm.
Using Pythagoras to find height $h$: $h^2 + 10^2 = 26^2 \Rightarrow h^2 = 676 - 100 = 576 \Rightarrow h = 24$ cm.
Area of trapezium = $\frac{1}{2}(a + b)h = \frac{1}{2}(20 + 40) \times 24 = \frac{1}{2}(60) \times 24 = 30 \times 24 =$ 720 cm².
Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Answer: Let sides be $a=8$, $b=11$, $c$. Perimeter $= a+b+c = 32 \Rightarrow 8+11+c = 32 \Rightarrow c = 13$ cm.
Semi-perimeter $s = \frac{32}{2} = 16$ cm.
Using Heron's formula: Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16(16-8)(16-11)(16-13)}$
Area $= \sqrt{16 \times 8 \times 5 \times 3} = \sqrt{16 \times 4 \times 2 \times 15} = 4 \times 2 \sqrt{30} =$ $8\sqrt{30}$ cm².
The sides of a triangular plot are in the ratio 3:5:7; its perimeter is 300 m. Find its area.
Answer: Let sides be $3x, 5x, 7x$.
$3x + 5x + 7x = 300 \Rightarrow 15x = 300 \Rightarrow x = 20$.
Sides are $a=60$ m, $b=100$ m, $c=140$ m. Semi-perimeter $s = 150$ m.
Heron's Formula: Area $= \sqrt{150(150-60)(150-100)(150-140)}$
Area $= \sqrt{150 \times 90 \times 50 \times 10} = \sqrt{6,750,000} =$ $1500\sqrt{3}$ m².
One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.
Answer: Let the shorter diagonal be $d_1 = x$. Then the longer diagonal $d_2 = 2x$.
Area of a rhombus = $\frac{1}{2} \times d_1 \times d_2$.
$\frac{1}{2} \times x \times 2x = 128 \Rightarrow x^2 = 128$
$x = \sqrt{128} = $ $8\sqrt{2}$ cm.
Find the area of a sector of a circle with radius 7 cm if the angle of the sector is $60^\circ$.
Answer: Area of sector = $\pi r^2 \times \frac{\theta}{360}$.
Area $= \frac{22}{7} \times 7^2 \times \frac{60}{360} = \frac{22}{7} \times 49 \times \frac{1}{6} = 22 \times 7 \times \frac{1}{6} = \frac{154}{6} = \frac{77}{3} \approx$ 25.67 cm².
Find the area of a quadrant of a circle whose circumference is 44 cm.
Answer: Circumference $= 2\pi r = 44 \Rightarrow 2 \times \frac{22}{7} \times r = 44 \Rightarrow \frac{44}{7}r = 44 \Rightarrow r = 7$ cm.
A quadrant is a sector with angle $90^\circ$ (or $\frac{1}{4}$ of a circle).
Area of quadrant = $\frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 7^2 = \frac{1}{4} \times 154 = \frac{77}{2} =$ 38.5 cm².
The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Answer: In 60 minutes, the minute hand sweeps $360^\circ$.
In 10 minutes, it sweeps $\frac{10}{60} \times 360^\circ = 60^\circ$.
This forms a sector with $r = 7$ cm and $\theta = 60^\circ$.
Area $= \frac{60}{360} \times \frac{22}{7} \times 7^2 = \frac{1}{6} \times 154 = \frac{77}{3} \approx$ 25.67 cm².
An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Answer: Let sides be $a=15$, $b=15$, $c$. Perimeter $= 15+15+c = 40 \Rightarrow c = 10$ cm.
Base $c = 10$. Drop a perpendicular height $h$ to bisect the base into two 5 cm segments.
By Pythagoras: $h^2 + 5^2 = 15^2 \Rightarrow h^2 + 25 = 225 \Rightarrow h^2 = 200 \Rightarrow h = 10\sqrt{2}$.
Area $= \frac{1}{2} \times \text{base} \times h = \frac{1}{2} \times 10 \times 10\sqrt{2} =$ $50\sqrt{2}$ cm².
An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?
Answer: Area = $\frac{1}{2} \times \text{base} \times h = 60 \Rightarrow \frac{1}{2} \times 10 \times h = 60 \Rightarrow 5h = 60 \Rightarrow h = 12$ cm.
The height bisects the base into 5 cm segments. Let equal sides be $a$.
By Pythagoras: $a^2 = h^2 + 5^2 = 12^2 + 5^2 = 144 + 25 = 169$.
$a = \sqrt{169} =$ 13 cm.
If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.
Answer: Distance in 1 rotation = Circumference $= \pi \times d = \frac{22}{7} \times 60 \approx 188.57$ cm.
Distance in 100 rotations = $188.57 \times 100 = 18857$ cm.
Distance in meters = $18857 / 100 =$ 188.57 m (or exactly $\frac{1320}{7}$ m).
Find the area of a quadrant of a circle whose circumference is 66 cm.
Answer: Circumference $= 2\pi r = 66 \Rightarrow 2 \times \frac{22}{7} \times r = 66 \Rightarrow \frac{44}{7}r = 66 \Rightarrow r = \frac{66 \times 7}{44} = \frac{3 \times 7}{2} = 10.5$ cm.
Area of quadrant = $\frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (10.5)^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{441}{4} = \frac{1}{4} \times 22 \times \frac{63}{4} = \frac{1386}{16} =$ 86.625 cm².
Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.
Answer:
(i) Algebra: A kite's diagonals are perpendicular. Let diagonals be $d_1$ and $d_2$. The diagonals split the kite into four right triangles. Let $d_1$ be split into segments $x$ and $y$ ($x+y=d_1$), and $d_2$ is bisected into $d_2/2$ and $d_2/2$.
Area = $\frac{1}{2}(x)(d_2/2) + \frac{1}{2}(x)(d_2/2) + \frac{1}{2}(y)(d_2/2) + \frac{1}{2}(y)(d_2/2)$
Area = $\frac{1}{2}(d_2/2)[x + x + y + y] = \frac{1}{4}d_2[2(x+y)] = \frac{1}{2}d_2(d_1)$.
(ii) Geometry: A kite can be divided by its main diagonal into two congruent triangles. If main diagonal is $d_1$, the height of each triangle is half the other diagonal ($d_2/2$).
Area of one triangle = $\frac{1}{2} \times d_1 \times (d_2/2)$.
Total Area = $2 \times \left[ \frac{1}{2} \times d_1 \times (d_2/2) \right] = \frac{1}{2} \times d_1 \times d_2$.
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