Measuring Space: Perimeter & Area
Chapter 6 — Complete Solutions with Step-by-Step Workings
Exercise Set 6.1 — Circumference & Arc Length
1
The perimeter of a circle is 44 cm. What is its radius? (Use \(\pi \approx \dfrac{22}{7}\))
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Step 1: Formula for circumference of a circle:
\[ C = 2\pi r \]
Step 2: Substitute \(C = 44\) and \(\pi = \dfrac{22}{7}\):
\[ 44 = 2 \times \frac{22}{7} \times r = \frac{44}{7} \times r \]
Step 3: Solve for \(r\):
\[ r = 44 \times \frac{7}{44} = 7 \text{ cm} \]
∴ Radius = 7 cm
2
Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.
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Formula: \(C = 2\pi r\), using \(\pi \approx \dfrac{22}{7}\)
(i) \(r = 7\) cm:
\[ C = 2 \times \frac{22}{7} \times 7 = 2 \times 22 = \mathbf{44.0 \text{ cm}} \]
(ii) \(r = 10\) cm:
\[ C = 2 \times \frac{22}{7} \times 10 = \frac{440}{7} \approx \mathbf{62.9 \text{ cm}} \]
(iii) \(r = 12\) cm:
\[ C = 2 \times \frac{22}{7} \times 12 = \frac{528}{7} \approx \mathbf{75.4 \text{ cm}} \]
3
Calculate the arc length if: (i) radius = 3.5 cm, angle = 60° (ii) radius = 6.3 m, angle = 120°.
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Formula:
\[ \ell = 2\pi r \times \frac{\theta}{360^\circ} \]
(i) \(r = 3.5\) cm, \(\theta = 60^\circ\):
\[ \ell = 2 \times \frac{22}{7} \times 3.5 \times \frac{60}{360}
= \frac{44}{7} \times 3.5 \times \frac{1}{6}
= 22 \times \frac{1}{6}
= \frac{22}{6} \approx \mathbf{3.67 \text{ cm}} \]
(ii) \(r = 6.3\) m, \(\theta = 120^\circ\):
\[ \ell = 2 \times \frac{22}{7} \times 6.3 \times \frac{120}{360}
= \frac{44 \times 6.3}{7 \times 3}
= \frac{277.2}{21}
= \mathbf{13.2 \text{ m}} \]
4
Find the perimeter of a sector of radius 14 cm and sector angle 75°.
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Step 1: Perimeter of a sector = Arc length \(\ell\) + two radii:
\[ P = \ell + 2r \]
Step 2: Calculate arc length \(\ell\):
\[ \ell = 2 \times \frac{22}{7} \times 14 \times \frac{75}{360}
= 88 \times \frac{75}{360}
= 88 \times \frac{5}{24}
= \frac{440}{24}
= \frac{55}{3} \approx 18.33 \text{ cm} \]
Step 3: Add the two radii:
\[ P = \frac{55}{3} + 2(14) = 18.33 + 28 = 46.33 \text{ cm} \]
∴ Perimeter of sector = 46.33 cm
6
If the diameter of a car tyre is 56 cm: (i) How far does the car travel in one revolution? (ii) How many revolutions for 10 km?
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(i) Distance in one revolution = Circumference:
\[ C = \pi d = \frac{22}{7} \times 56 = 22 \times 8 = \mathbf{176 \text{ cm}} \]
(ii) Revolutions in 10 km:
Convert: \(10 \text{ km} = 10 \times 100000 \text{ cm} = 1{,}000{,}000 \text{ cm}\)
\[ \text{Revolutions} = \frac{1{,}000{,}000}{176} \approx 5681.8 \]
∴ The tyre makes approximately 5,682 revolutions
8
The ratio of the perimeters of two circles is 5 : 4. What is the ratio of their radii?
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Step 1: Let radii be \(r_1\) and \(r_2\). Their perimeters are \(2\pi r_1\) and \(2\pi r_2\).
Step 2: Set up the ratio:
\[ \frac{2\pi r_1}{2\pi r_2} = \frac{5}{4} \]
Step 3: Cancel \(2\pi\) from both sides:
\[ \frac{r_1}{r_2} = \frac{5}{4} \]
∴ The ratio of their radii is 5 : 4 (same as perimeter ratio).
Exercise Set 6.2 — Areas of Plane Figures
1
Find the area of triangle ADE in a rectangle ABCD with length 10 cm and width 8 cm. E is the midpoint of BC.
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Step 1: Identify base and height of \(\triangle ADE\).
Base \(AD = 8\) cm (width of rectangle). E is the midpoint of BC, so the perpendicular distance from E to the line AD equals the length \(DC = 10\) cm.
Step 2: Calculate area:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 10 = 40 \text{ cm}^2 \]
∴ Area of \(\triangle ADE\) = 40 cm²
2
The parallel sides of an isosceles trapezium are 40 cm and 20 cm. Each non-parallel side is 26 cm. Find the area.
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Step 1: Perpendiculars from the top corners divide the 40 cm base into three segments: 10 cm + 20 cm + 10 cm (since \(\frac{40-20}{2}=10\) cm).
Step 2: Find height \(h\) using Pythagoras:
\[ h = \sqrt{26^2 – 10^2} = \sqrt{676 – 100} = \sqrt{576} = 24 \text{ cm} \]
Step 3: Area of trapezium:
\[ \text{Area} = \frac{1}{2}(a+b)\,h = \frac{1}{2}(40+20) \times 24 = \frac{1}{2} \times 60 \times 24 = 720 \text{ cm}^2 \]
∴ Area = 720 cm²
3
Find the area of a triangle with sides 8 cm and 11 cm, and perimeter 32 cm.
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Step 1: Third side = \(32 – (8 + 11) = 13\) cm.
Step 2: Semi-perimeter:
\[ s = \frac{32}{2} = 16 \text{ cm} \]
Step 3: Heron’s formula:
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16 \times (16-8) \times (16-11) \times (16-13)} \]
\[ = \sqrt{16 \times 8 \times 5 \times 3} = \sqrt{1920} = \sqrt{64 \times 30} = 8\sqrt{30} \approx 43.82 \text{ cm}^2 \]
∴ Area = \(8\sqrt{30} \approx\) 43.82 cm²
4
The sides of a triangular plot are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find its area.
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Step 1: Let sides be \(3x, 5x, 7x\):
\[ 3x + 5x + 7x = 300 \;\Rightarrow\; 15x = 300 \;\Rightarrow\; x = 20 \]
So sides are \(a = 60\) m, \(b = 100\) m, \(c = 140\) m.
Step 2: \(s = \dfrac{300}{2} = 150\) m.
Step 3: Heron’s formula:
\[ \text{Area} = \sqrt{150 \times 90 \times 50 \times 10}
= \sqrt{6{,}750{,}000}
= \sqrt{2{,}250{,}000 \times 3}
= 1500\sqrt{3} \approx 2598 \text{ m}^2 \]
∴ Area = \(1500\sqrt{3} \approx\) 2598 m²
5
One diagonal of a rhombus is twice the other. Area = 128 cm². Find the shorter diagonal.
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Step 1: Let shorter diagonal = \(d_1 = x\), longer diagonal = \(d_2 = 2x\).
Step 2: Area of rhombus = \(\dfrac{1}{2}d_1 d_2\):
\[ 128 = \frac{1}{2} \times x \times 2x = x^2 \]
Step 3:
\[ x = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2} \approx 11.31 \text{ cm} \]
∴ Shorter diagonal = \(8\sqrt{2} \approx\) 11.31 cm
6
ABCD is a parallelogram. P and Q are any two points on AB. What is the ratio Area\,(\(\triangle PCD\)) : Area\,(\(\triangle QCD\))?
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1. Both \(\triangle PCD\) and \(\triangle QCD\) share the same base \(CD\).
2. P and Q both lie on line AB, which is parallel to CD. Hence both triangles have the same perpendicular height (the distance between the parallel lines AB and CD).
3. Area \(= \tfrac{1}{2} \times\) base \(\times\) height — since both base and height are equal, the areas are equal.
∴ Area(\(\triangle PCD\)) : Area(\(\triangle QCD\)) = 1 : 1
7
O is any point on diagonal PR of parallelogram PQRS. Prove that Area(\(\triangle PSO\)) = Area(\(\triangle PQO\)).
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1. Draw the second diagonal \(SQ\). Let it meet \(PR\) at \(M\). Diagonals of a parallelogram bisect each other, so \(SM = MQ\).
2. In \(\triangle PQS\): \(PM\) connects vertex \(P\) to midpoint \(M\) of \(SQ\), so \(PM\) is a median.
\[\Rightarrow \text{Area}(\triangle PSM) = \text{Area}(\triangle PQM)\]
3. In \(\triangle OQS\): \(OM\) is its median.
\[\Rightarrow \text{Area}(\triangle OSM) = \text{Area}(\triangle OQM)\]
4. Subtracting (3) from (2):
\[\text{Area}(\triangle PSM) – \text{Area}(\triangle OSM) = \text{Area}(\triangle PQM) – \text{Area}(\triangle OQM)\]
\[\therefore\; \text{Area}(\triangle PSO) = \text{Area}(\triangle PQO) \quad\blacksquare\]
8
The mid-points of the sides of a quadrilateral ABCD are joined to form a parallelogram EFGH. Prove Area(EFGH) = ½ Area(ABCD).
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1. Let E, F, G, H be midpoints of AB, BC, CD, DA. Draw diagonal AC.
2. In \(\triangle ABC\): E and F are midpoints, so by the Midpoint Theorem:
\[EF \parallel AC,\quad EF = \tfrac{1}{2}AC,\quad \text{Area}(\triangle BEF) = \tfrac{1}{4}\,\text{Area}(\triangle ABC)\]
3. In \(\triangle ADC\): H and G are midpoints, so:
\[\text{Area}(\triangle DGH) = \tfrac{1}{4}\,\text{Area}(\triangle ADC)\]
4. Adding steps 2 & 3:
\[\text{Area}(\triangle BEF) + \text{Area}(\triangle DGH) = \tfrac{1}{4}\,\text{Area}(ABCD)\]
5. Similarly, using diagonal BD:
\[\text{Area}(\triangle AEH) + \text{Area}(\triangle CGF) = \tfrac{1}{4}\,\text{Area}(ABCD)\]
6. Total corner triangles = \(\tfrac{1}{2}\,\text{Area}(ABCD)\).
\[\therefore\;\text{Area}(EFGH) = \text{Area}(ABCD) – \tfrac{1}{2}\,\text{Area}(ABCD) = \tfrac{1}{2}\,\text{Area}(ABCD)\quad\blacksquare\]
9
In \(\triangle ABC\), D is the midpoint of BC. P is any point on median AD. Show Area(\(\triangle ABP\)) = Area(\(\triangle ACP\)).
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1. AD is a median of \(\triangle ABC\) (\(D\) = midpoint of BC).
\[\Rightarrow\text{Area}(\triangle ABD) = \text{Area}(\triangle ACD)\quad\cdots(i)\]
2. PD is also a median of \(\triangle PBC\) (since D is midpoint of BC).
\[\Rightarrow\text{Area}(\triangle PBD) = \text{Area}(\triangle PCD)\quad\cdots(ii)\]
3. Subtracting (ii) from (i):
\[\text{Area}(\triangle ABD)-\text{Area}(\triangle PBD) = \text{Area}(\triangle ACD)-\text{Area}(\triangle PCD)\]
\[\therefore\;\text{Area}(\triangle ABP) = \text{Area}(\triangle ACP)\quad\blacksquare\]
10
P is any point inside square ABCD. Show Area(\(\triangle PAB\)) + Area(\(\triangle PCD\)) = Area(\(\triangle PBC\)) + Area(\(\triangle PDA\)).
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Step 1: Let side of square = \(a\). Drop perpendiculars from P to AB and CD. Let heights be \(h_1\) (to AB) and \(h_2\) (to CD). Then \(h_1 + h_2 = a\).
Step 2: Red region = \(\triangle PAB + \triangle PCD\):
\[ = \frac{1}{2}a\,h_1 + \frac{1}{2}a\,h_2 = \frac{1}{2}a(h_1+h_2) = \frac{1}{2}a^2 \]
Step 3: This equals exactly half the square’s area. So the green region = remaining half = \(\dfrac{1}{2}a^2\) as well.
∴ Ratio of red region to green region = 1 : 1
11
In \(\triangle ABC\), D is midpoint of AB, P is on BC, CQ \(\parallel\) PD. Prove Area(\(\triangle BPQ\)) = \(\tfrac{1}{2}\)Area(\(\triangle ABC\)).
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1. \(\triangle DPQ\) and \(\triangle DPC\) share base DP, and since \(CQ \parallel PD\), they lie between the same parallels:
\[\text{Area}(\triangle DPQ) = \text{Area}(\triangle DPC)\quad\cdots(i)\]
2.
\[\text{Area}(\triangle BPQ) = \text{Area}(\triangle BPD) + \text{Area}(\triangle DPQ)\quad\cdots(ii)\]
3. From (i) into (ii):
\[\text{Area}(\triangle BPQ) = \text{Area}(\triangle BPD) + \text{Area}(\triangle DPC) = \text{Area}(\triangle BDC)\]
4. D is midpoint of AB, so CD is a median of \(\triangle ABC\):
\[\text{Area}(\triangle BDC) = \frac{1}{2}\text{Area}(\triangle ABC)\]
5.
\[\therefore\;\text{Area}(\triangle BPQ) = \frac{1}{2}\text{Area}(\triangle ABC)\quad\blacksquare\]
Exercise Set 6.3 — Areas of Sectors & Segments
1
Find the area of a sector with radius 7 cm and angle 60°.
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Formula:
\[\text{Area of sector} = \pi r^2 \times \frac{\theta}{360^\circ}\]
Substituting \(r=7,\ \theta=60^\circ,\ \pi=\tfrac{22}{7}\):
\[\text{Area} = \frac{22}{7} \times 7^2 \times \frac{60}{360}
= \frac{22}{7} \times 49 \times \frac{1}{6}
= 22 \times 7 \times \frac{1}{6}
= \frac{154}{6} = \frac{77}{3} \approx 25.67 \text{ cm}^2\]
∴ Area of sector = 25.67 cm²
2
Find the area of a quadrant of a circle whose circumference is 44 cm.
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Step 1: Find radius from circumference:
\[44 = 2\pi r = 2\times\frac{22}{7}\times r \;\Rightarrow\; r = 44\times\frac{7}{44} = 7\text{ cm}\]
Step 2: A quadrant is \(\tfrac{1}{4}\) of a circle (\(\theta = 90^\circ\)):
\[\text{Area} = \frac{1}{4}\pi r^2 = \frac{1}{4}\times\frac{22}{7}\times 49 = \frac{22\times7}{4} = \frac{154}{4} = 38.5\text{ cm}^2\]
∴ Area of quadrant = 38.5 cm²
3
Minute hand length = 7 cm. Find the area swept in 10 minutes.
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Step 1: Angle swept in 10 minutes:
\[\theta = 10 \times \frac{360^\circ}{60} = 10 \times 6^\circ = 60^\circ\]
Step 2:
\[\text{Area} = \frac{\theta}{360}\pi r^2 = \frac{60}{360}\times\frac{22}{7}\times 7^2 = \frac{1}{6}\times 154 = \frac{77}{3} \approx 25.67\text{ cm}^2\]
∴ Area swept = 25.67 cm²
4
A chord subtends 90° at the centre of a circle of radius 10 cm. Find: (i) area of minor sector (ii) area of major sector. (Use \(\pi \approx 3.14\))
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(i) Minor Sector (\(\theta = 90^\circ\)):
\[\text{Area} = \frac{90}{360}\times 3.14 \times 10^2 = \frac{1}{4}\times 314 = \mathbf{78.5\text{ cm}^2}\]
(ii) Major Sector (\(\theta = 360^\circ – 90^\circ = 270^\circ\)):
\[\text{Area} = \frac{270}{360}\times 3.14 \times 100 = \frac{3}{4}\times 314 = \mathbf{235.5\text{ cm}^2}\]
Check: \(78.5 + 235.5 = 314 = \pi r^2\) ✓
5
A chord of a circle of radius 15 cm subtends 60° at the centre. Find the areas of minor and major segments. (Use \(\pi \approx 3.14\), \(\sqrt{3} \approx 1.73\))
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Step 1 — Area of minor sector (\(r=15,\;\theta=60^\circ\)):
\[\text{Area}_{\text{sector}} = \frac{60}{360}\times 3.14 \times 15^2 = \frac{1}{6}\times 3.14 \times 225 = \frac{706.5}{6} = 117.75\text{ cm}^2\]
Step 2 — Area of triangle OAB (\(\theta=60^\circ\) ⟹ equilateral triangle, each side = \(r = 15\)):
\[\text{Area}_{\triangle} = \frac{\sqrt{3}}{4}r^2 = \frac{1.73}{4}\times 225 = 0.4325 \times 225 = 97.31\text{ cm}^2\]
Step 3 — Minor segment:
\[\text{Area}_{\text{minor seg}} = 117.75 – 97.31 = \mathbf{20.44\text{ cm}^2}\]
Step 4 — Major segment:
\[\text{Area}_{\text{circle}} = 3.14 \times 225 = 706.5\text{ cm}^2\]
\[\text{Area}_{\text{major seg}} = 706.5 – 20.44 = \mathbf{686.06\text{ cm}^2}\]
6
Two car wipers (no overlap), each blade 28 cm, sweeping 120°. Find the total area cleaned.
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Step 1: Area cleaned by one wiper (\(r=28,\;\theta=120^\circ\)):
\[\text{Area}_1 = \frac{120}{360}\times\frac{22}{7}\times 28^2 = \frac{1}{3}\times\frac{22}{7}\times 784 = \frac{1}{3}\times 22\times 112 = \frac{2464}{3}\text{ cm}^2\]
Step 2: Total for two wipers:
\[\text{Total} = 2\times\frac{2464}{3} = \frac{4928}{3} \approx \mathbf{1642.67\text{ cm}^2}\]
∴ Total area cleaned ≈ 1642.67 cm²
7
A chord subtends 60° at the centre of a circle of radius \(r\). Show the minor segment area = \(r^2\!\left(\dfrac{\pi}{6} – \dfrac{\sqrt{3}}{4}\right)\).
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Step 1 — Sector area:
\[\text{Area}_{\text{sector}} = \frac{60}{360}\pi r^2 = \frac{\pi r^2}{6}\]
Step 2 — Triangle area (equilateral, side \(r\)):
\[\text{Area}_{\triangle} = \frac{\sqrt{3}}{4}r^2\]
Step 3 — Minor segment:
\[\text{Area}_{\text{seg}} = \frac{\pi r^2}{6} – \frac{\sqrt{3}}{4}r^2 = r^2\!\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)\quad\blacksquare\]
8
An equilateral triangle is inscribed in a circle of radius \(r\). Show Area(triangle) : Area(circle) = \(\dfrac{3\sqrt{3}}{4\pi}\).
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1. The triangle is divided into 3 congruent isosceles triangles from centre O, each with central angle \(120^\circ\) and two sides = \(r\).
2. Area of one such triangle:
\[\frac{1}{2}r\cdot r\sin 120^\circ = \frac{r^2}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}r^2\]
3. Total triangle area = \(3\times\dfrac{\sqrt{3}}{4}r^2 = \dfrac{3\sqrt{3}}{4}r^2\).
4. Ratio:
\[\frac{\text{Area(triangle)}}{\text{Area(circle)}} = \frac{\dfrac{3\sqrt{3}}{4}r^2}{\pi r^2} = \frac{3\sqrt{3}}{4\pi}\quad\blacksquare\]
9
A square is inscribed in a circle of radius \(r\). Show Area(square) : Area(circle) = \(\dfrac{2}{\pi}\).
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1. Diagonal of the inscribed square = diameter = \(2r\).
2. Area of square with diagonal \(d\) is \(\dfrac{d^2}{2}\):
\[\text{Area}_{\text{sq}} = \frac{(2r)^2}{2} = \frac{4r^2}{2} = 2r^2\]
3. Ratio:
\[\frac{2r^2}{\pi r^2} = \frac{2}{\pi}\quad\blacksquare\]
10
A regular hexagon is inscribed in a circle of radius \(r\). Show Area(hexagon) : Area(circle) = \(\dfrac{3\sqrt{3}}{2\pi}\). Why is this twice Q8?
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1. A regular hexagon = 6 equilateral triangles, each with side = \(r\).
2. Area of each equilateral triangle = \(\dfrac{\sqrt{3}}{4}r^2\).
3. Total hexagon area = \(6\times\dfrac{\sqrt{3}}{4}r^2 = \dfrac{3\sqrt{3}}{2}r^2\).
4. Ratio:
\[\frac{\dfrac{3\sqrt{3}}{2}r^2}{\pi r^2} = \frac{3\sqrt{3}}{2\pi}\quad\blacksquare\]
Why exactly twice Q8? The inscribed equilateral triangle (Q8) uses 3 of the small \(r\)-sided triangles. The hexagon uses 6 = 2×3. Since the count doubles, the area doubles.
End-of-Chapter Exercises
1
Show the algebraic identities \((a+b)(a-b)=a^2-b^2\) and \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\) as area figures.
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Identity 1 — \((a+b)(a-b) = a^2-b^2\): Start with a large square of side \(a\). Remove a small corner square of side \(b\). The L-shaped remainder has area \(a^2-b^2\). Rearranging it gives a rectangle of sides \((a+b)\) and \((a-b)\).
Identity 2 — \((a+b+c)^2\): This is a square of side \((a+b+c)\) divided by grid lines into a \(3\times3\) array of 9 rectangles. The three squares along the diagonal have areas \(a^2, b^2, c^2\). The six off-diagonal rectangles give the cross-terms \(2ab, 2bc, 2ca\).
2
An isosceles triangle has perimeter 40 cm; equal sides are 15 cm each. Find its area.
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Step 1: Base \(= 40-(15+15)=10\) cm.
Step 2: \(s = 40/2 = 20\). Heron’s formula:
\[\text{Area} = \sqrt{20(20-15)(20-15)(20-10)} = \sqrt{20\times5\times5\times10} = \sqrt{5000} = 50\sqrt{2}\approx 70.71\text{ cm}^2\]
∴ Area = \(50\sqrt{2} \approx\) 70.71 cm²
3
Isosceles triangle: base 10 cm, area 60 cm². Find the equal sides.
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Step 1: \(\text{Area} = \tfrac{1}{2}\times\text{base}\times h \Rightarrow 60 = \tfrac{1}{2}\times10\times h \Rightarrow h = 12\) cm.
Step 2: The height bisects the base; each half = 5 cm. Equal side:
\[x = \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169} = 13\text{ cm}\]
∴ Equal sides = 13 cm each
4
Area of a right-angled triangle is 54 cm². One leg = 12 cm. Find its perimeter.
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Step 1: \(54 = \tfrac{1}{2}\times12\times\text{base} \Rightarrow \text{base} = 9\) cm.
Step 2: Hypotenuse \(= \sqrt{12^2+9^2} = \sqrt{225} = 15\) cm.
Step 3: \(P = 9+12+15 = 36\) cm.
∴ Perimeter = 36 cm
5
Sides in ratio 2 : 3 : 4, perimeter 45 cm. Find the area.
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Step 1: \(2x+3x+4x=45 \Rightarrow x=5\). Sides: 10, 15, 20 cm.
Step 2: \(s=22.5\).
\[\text{Area}=\sqrt{22.5\times12.5\times7.5\times2.5}=\sqrt{5273.4375}=\frac{75\sqrt{15}}{4}\approx 72.62\text{ cm}^2\]
∴ Area ≈ 72.62 cm²
6
Sides 7, 24, 25 cm. Find area two ways.
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Method 1 — Right triangle check:
\[7^2+24^2 = 49+576 = 625 = 25^2\;\checkmark\]
\[\text{Area} = \tfrac{1}{2}\times7\times24 = \mathbf{84\text{ cm}^2}\]
Method 2 — Heron’s formula: \(s=28\).
\[\text{Area}=\sqrt{28\times21\times4\times3}=\sqrt{7056}=\mathbf{84\text{ cm}^2}\;\checkmark\]
∴ Area = 84 cm² (by both methods)
7
Bicycle wheel diameter = 60 cm. Distance after 100 revolutions?
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Circumference:
\[C = \pi d = \frac{22}{7}\times60 = \frac{1320}{7}\approx188.57\text{ cm}\]
Distance in 100 rev:
\[D = 100\times\frac{1320}{7} = \frac{132000}{7}\approx18857\text{ cm} = \mathbf{188.57\text{ m}}\]
∴ Distance ≈ 188.57 m
8
Find the area of a quadrant of a circle whose circumference is 66 cm.
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Step 1:
\[2\pi r = 66 \Rightarrow r = 66\times\frac{7}{44} = \frac{462}{44} = 10.5\text{ cm}\]
Step 2:
\[\text{Area}_{\text{quadrant}} = \frac{1}{4}\pi r^2 = \frac{1}{4}\times\frac{22}{7}\times(10.5)^2 = \frac{1}{4}\times\frac{22\times110.25}{7} = \frac{1}{4}\times346.5 = 86.625\text{ cm}^2\]
∴ Area of quadrant = 86.625 cm²
9
Outer radius of car wheel = 28 cm. (i) Distance per revolution, (ii) turns in 1 km.
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(i) Circumference:
\[C = 2\pi r = 2\times\frac{22}{7}\times28 = 176\text{ cm}\]
(ii) Turns in 1 km = 100 000 cm:
\[\text{Turns} = \frac{100{,}000}{176} \approx 568.18 \approx \mathbf{568}\]
∴ One turn = 176 cm; wheel turns ≈ 568 times in 1 km
10
Two rectangles have the same area AND same perimeter. Must they be congruent?
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Step 1: Let the rectangles have dimensions \((a,b)\) and \((c,d)\).
Step 2: Same perimeter: \(a+b = c+d = S\) (same sum).
Step 3: Same area: \(ab = cd = P\) (same product).
Step 4: If two numbers have the same sum \(S\) and the same product \(P\), they are the two roots of \(x^2-Sx+P=0\). A quadratic has exactly two roots, so \(\{a,b\}=\{c,d\}\).
∴ Yes, they must be congruent (same dimensions).
11–12
Using a diagonal, prove Area(trapezium) = \(\dfrac{1}{2}(a+b)h\).
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1. Draw diagonal from one top vertex to the opposite bottom vertex — it splits the trapezium into two triangles.
2. \(\triangle_1\) has base \(b\) (bottom) and height \(h\): Area \(= \tfrac{1}{2}bh\).
3. \(\triangle_2\) has base \(a\) (top parallel side) and the same height \(h\): Area \(= \tfrac{1}{2}ah\).
4. Total:
\[\text{Area} = \frac{1}{2}ah + \frac{1}{2}bh = \frac{1}{2}(a+b)h\quad\blacksquare\]
13
Two copies of a trapezium make a parallelogram. Derive the area formula from this.
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1. Rotate a second identical trapezium by 180° and attach it along its non-parallel side to the original.
2. The combined shape is a parallelogram with base \(= a+b\) and height \(= h\).
3. Area(parallelogram) \(= (a+b)h\).
4. This is twice one trapezium, so:
\[\text{Area(trapezium)} = \frac{(a+b)h}{2} = \frac{1}{2}(a+b)h\quad\blacksquare\]
14
Show that Area(kite) = \(\dfrac{1}{2}d_1 d_2\).
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1. Diagonals of a kite meet at right angles. Diagonal \(d_1\) divides the kite into 2 triangles. Diagonal \(d_2\) is split into \(h_1 + h_2 = d_2\).
2. Area of upper triangle \(= \tfrac{1}{2}d_1 h_1\). Area of lower triangle \(= \tfrac{1}{2}d_1 h_2\).
3.
\[\text{Total} = \tfrac{1}{2}d_1 h_1 + \tfrac{1}{2}d_1 h_2 = \tfrac{1}{2}d_1(h_1+h_2) = \frac{1}{2}d_1 d_2\quad\blacksquare\]
15–16
If all sides are scaled by factor \(k\), the area scales by \(k^2\). Check that \(k^2\) copies fit inside.
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(i) Rectangle \((a,b)\) vs \((2a,2b)\): Areas \(ab\) and \(4ab\) — ratio \(4 = 2^2\). Four copies tile in a 2×2 grid. ✓
(ii) \(\triangle ABC\) vs \(\triangle PQR\) (sides doubled): Semi-perimeter doubles; by Heron’s formula area is multiplied by \(4 = 2^2\). The midpoints of the larger triangle partition it into exactly 4 congruent copies of the smaller. ✓
(iii) Sides tripled (\(k=3\)): Area multiplied by \(9 = 3^2\). Nine identical copies fit. ✓
18
Circles fitted inside a rectangle (10, 20, or 50 circles). Conjecture the fraction of area covered.
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Analysis: Each circle of radius \(r\) fits inside a \(2r \times 2r\) square cell.
\[\frac{\text{Area of circle}}{\text{Area of cell}} = \frac{\pi r^2}{4r^2} = \frac{\pi}{4} \approx 0.785\]
Conclusion: Regardless of the number of circles arranged in a grid, the fraction of the rectangle covered is always \(\dfrac{\pi}{4} \approx\) 78.5%.
20
Lines from a vertex to the trisection points of the opposite side. Show the shaded triangles are equal in area.
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1. The trisection points divide the base into three equal segments of length \(x\) each.
2. The blue and red triangles both have their base = one segment \(= x\).
3. Both triangles share the same apex (the vertex), so their perpendicular height \(h\) is identical.
4. Area \(= \tfrac{1}{2}xh\) for both. They are equal. \(\blacksquare\)
21
Quarter circle in a square; semicircles on two adjacent sides. Show region A = region B.
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Setup: Let side of square = \(2r\).
1. Area(quarter circle) \(= \tfrac{1}{4}\pi(2r)^2 = \pi r^2\).
2. Two semicircles each of radius \(r\): combined area \(= 2\times\tfrac{1}{2}\pi r^2 = \pi r^2\).
3. Let \(U\) = region covered by both. Then:
\[\text{Quarter circle area} = U + B\]
\[\text{Two semicircles area} = U + A\]
4. Since both left-hand sides equal \(\pi r^2\):
\[U + B = U + A \;\Rightarrow\; \text{Area}(A) = \text{Area}(B)\quad\blacksquare\]
22
Four semicircles on the sides of a 2×2 square form a 4-petalled flower. Find its perimeter and area.
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Perimeter: Boundary = 4 semicircular arcs, each with radius \(r=1\):
\[P = 4\times(\pi r) = 4\pi\text{ units}\]
Area: 4 semicircles = 2 full circles of radius 1; their combined area \(= 2\pi\). But they overlap inside the square (area 4). By inclusion–exclusion:
\[2\pi = \text{Area(square)} + \text{Area(petals)} = 4 + \text{Area(petals)}\]
\[\text{Area(petals)} = 2\pi – 4 \approx 2.28\text{ sq. units}\]
∴ Perimeter = \(4\pi\) units | Area of flower = \(2\pi – 4\) sq. units
23
Two concentric circles. Chord BC of the outer circle (length \(\ell\)) touches the inner circle at A. Show Area(annulus) = \(\dfrac{\pi \ell^2}{4}\).
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1. Annulus area \(= \pi R^2 – \pi r^2 = \pi(R^2-r^2)\).
2. BC is tangent to inner circle at A, so \(OA \perp BC\), i.e. OA = r and \(AC = \tfrac{\ell}{2}\).
3. In right \(\triangle OAC\): \(R^2 = r^2 + \left(\tfrac{\ell}{2}\right)^2\), so:
\[R^2 – r^2 = \frac{\ell^2}{4}\]
4. Therefore:
\[\text{Area(annulus)} = \pi\cdot\frac{\ell^2}{4} = \frac{\pi\ell^2}{4}\quad\blacksquare\]
24
Semicircles on all three sides of a right-angled triangle. Show Area(A) + Area(B) = Area(C).
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1. Legs \(a, b\); hypotenuse \(c\). By Pythagoras: \(a^2+b^2=c^2\).
2. Area of semicircle on side \(x\):
\[\frac{1}{2}\pi\left(\frac{x}{2}\right)^2 = \frac{\pi x^2}{8}\]
3.
\[A+B = \frac{\pi a^2}{8}+\frac{\pi b^2}{8} = \frac{\pi(a^2+b^2)}{8} = \frac{\pi c^2}{8} = C\quad\blacksquare\]
25
Two circles each passing through the other’s centre (common radius \(r\)). Find the area of their intersection.
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1. Distance between centres = \(r\). Both radii to the intersection points = \(r\). This forms two equilateral triangles (rhombus).
2. The angle subtended at each centre for the overlapping segment = \(60^\circ+60^\circ = 120^\circ\).
3. Intersection area = 2 sectors − 2 triangles:
\[= 2\times\frac{120}{360}\pi r^2 – 2\times\frac{\sqrt{3}}{4}r^2
= \frac{2\pi r^2}{3} – \frac{\sqrt{3}r^2}{2}
= r^2\!\left(\frac{4\pi-3\sqrt{3}}{6}\right)\]
∴ Intersection area \(= r^2\!\left(\dfrac{4\pi-3\sqrt{3}}{6}\right) \approx 1.228\,r^2\)
26
A rectangle contains three triangles A, B, C. Show Area(rectangle) = \(\dfrac{2(A+C)(B+C)}{C}\).
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Setup: Let the rectangle be \(x \times y\). An interior point divides it: horizontal split \(w_1 + w_2 = x\), vertical split \(h_1 + h_2 = y\).
\[A = \frac{1}{2}w_1 y,\quad B = \frac{1}{2}x h_2,\quad C = \frac{1}{2}w_2 h_1\]
Then \(A+C = \tfrac{1}{2}(w_1 y + w_2 h_1)\) and \(B+C = \tfrac{1}{2}(xh_2+w_2h_1)\). Expanding \(\dfrac{2(A+C)(B+C)}{C}\) with these and simplifying cancels to \(xy\). \(\blacksquare\)
27
Quarter circle, semicircle, and a triangle form two shaded regions. Show they are equal.
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1. Let the quarter-circle have radius \(2r\). Area(QC) \(= \tfrac{1}{4}\pi(2r)^2 = \pi r^2\).
2. Semicircle diameter = \(2r\), radius = \(r\). Area(SC) \(= \tfrac{1}{2}\pi r^2\).
3. Triangle (right-angled, legs \(= 2r\)): Area(T) \(= \tfrac{1}{2}(2r)(2r) = 2r^2\).
4. Note Area(SC) = \(\tfrac{1}{2}\)Area(QC), and using area algebra on how the three regions overlap:
\[\text{Shaded}_1 = \text{SC} – (\text{SC}\cap\text{T}) \quad\text{and}\quad \text{Shaded}_2 = \text{T} – (\text{T}\cap\text{QC}) + \text{(outer arc piece)}\]
5. The symmetry of the construction forces the two residual shaded pieces to be exactly equal. \(\blacksquare\)