Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum / PECTA 2025 Syllabus
📖 What's Inside: This review exercise covers all key concepts of Similar Figures including similarity of polygons, triangles, area ratios, and volume ratios. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Chapter 9: Similar Figures
1. Four options are given against each statement. Encircle the correct one.
| Sr. No. |
Questions | A | B | C | D |
|---|---|---|---|---|---|
| 1 | If two polygons are similar, then: | their corresponding angles are equal. | their areas are equal. | their volumes are equal. | their corresponding sides are equal. |
| 2 | The ratio of the areas of two similar polygons is: | equal to the ratio of their perimeters. | equal to the square of the ratio of their corresponding sides. | equal to the cube of the ratio of their corresponding sides. | equal to the sum of their corresponding sides. |
| 3 | If the volume of two similar solids is \(125\ cm^3\) and \(27\ cm^3\), the ratio of their corresponding heights is: | 3 : 5 | 5 : 3 | 25 : 9 | 9 : 25 |
| 4 | The exterior angle of a regular pentagon is: | \(40^\circ\) | \(45^\circ\) | \(60^\circ\) | \(72^\circ\) |
| 5 | A parallelogram has an area of \(64\ cm^2\) and a similar parallelogram has an area of \(144\ cm^2\). If a side of the smaller parallelogram is \(8\ cm\), the corresponding side of the larger parallelogram is: | \(10\ cm\) | \(12\ cm\) | \(18\ cm\) | \(16\ cm\) |
| 6 | The total number of diagonals in a polygon with 9 sides is: | 18 | 21 | 25 | 27 |
| 7 | Two spheres are similar, and their radii are in the ratio 4 : 5. If the surface area of the larger sphere is \(500\pi\ cm^2\), what is the surface area of the smaller sphere? | \(256\pi\ cm^2\) | \(320\pi\ cm^2\) | \(400\pi\ cm^2\) | \(405\pi\ cm^2\) |
| 8 | A regular polygon has an exterior angle of \(30^\circ\). How many diagonals does the polygon have? | 54 | 90 | 72 | 108 |
| 9 | In a regular hexagon, the ratio of the length of a diagonal to the side length is: | \(\sqrt{3}:1\) | \(2:1\) | \(3:2\) | \(2:3\) |
| 10 | A regular polygon has an interior angle of \(165^\circ\). How many sides does it have? | 15 | 16 | 20 | 24 |
Solution of MCQs
| 1 | Similar polygons have corresponding angles equal and corresponding sides proportional. |
| 2 | The area ratio of similar polygons is the square of the ratio of their corresponding sides. |
| 3 | \[\begin{aligned} \frac{V_1}{V_2} &= \left(\frac{l_1}{l_2}\right)^3 \\ \Rightarrow \frac{125}{27} &= \left(\frac{l_1}{l_2}\right)^3 \\ \sqrt[3]{\frac{125}{27}} &= \sqrt[3]{\left(\frac{l_1}{l_2}\right)^3} \\ \frac{5}{3} &= \frac{l_1}{l_2} \\ \frac{l_1}{l_2} &= \frac{5}{3} \end{aligned}\] |
| 4 | Exterior angle of pentagon is \(= \dfrac{360^\circ}{5} = 72^\circ\) |
| 5 | Ratio of areas of similar figures \(\dfrac{A_1}{A_2} = \left(\dfrac{l_1}{l_2}\right)^2\) \[\begin{aligned} \frac{64}{144} &= \left(\frac{8}{l_2}\right)^2 \\ \sqrt{\frac{64}{144}} &= \sqrt{\left(\frac{8}{l_2}\right)^2} \\ \frac{8}{12} &= \frac{8}{l_2} \\ l_2 \times 8 &= 12 \times 8 \\ l_2 &= \frac{96}{8} \\ l_2 &= 12\ cm \end{aligned}\] |
| 6 | The total number of diagonals in a polygon with 9 sides \(= \dfrac{n(n-3)}{2} = \dfrac{9(9-3)}{2} = \dfrac{54}{2} = 27\) |
| 7 | Ratio of areas of similar figures is \(\dfrac{A_1}{A_2} = \left(\dfrac{l_1}{l_2}\right)^2\) \[\begin{aligned} \frac{A_1}{500\pi} &= \left(\frac{4}{5}\right)^2 \\ \frac{A_1}{500\pi} &= \frac{16}{25} \\ A_1 \times 25 &= 16 \times 500\pi \\ A_1 &= \frac{16 \times 500\pi}{25} \\ A_1 &= 320\pi\ cm^2 \end{aligned}\] |
| 8 | Exterior angle of a regular polygon is given: \(Exterior\ Angle = \dfrac{360^\circ}{n}\) \[\begin{aligned} 30^\circ &= \frac{360^\circ}{n} \\ n &= \frac{360^\circ}{30^\circ} \\ no.\ of\ sides &= 12 \end{aligned}\] Now the total number of diagonals in a polygon with 12 sides \(= \dfrac{n(n-3)}{2} = \dfrac{12(12-3)}{2} = \dfrac{108}{2} = 54\) |
| 9 | In a regular hexagon, the longest diagonal connects opposite vertices and is twice the side length, so the ratio is \(2:1\). |
| 10 | Exterior angle \(= 180^\circ - 165^\circ = 15^\circ\) \[\begin{aligned} Exterior\ Angle &= \frac{360^\circ}{n} \\ 15^\circ &= \frac{360^\circ}{n} \\ n &= \frac{360^\circ}{15^\circ} \\ no.\ of\ sides &= 24 \end{aligned}\] |
Short & Long Questions
2If the sum of the interior angles of a polygon is \(1080^\circ\), how many sides does the polygon have?
\(Sum\ of\ the\ interior\ angles\ of\ a\ polygon = 1080^\circ\)
\(Number\ of\ sides = n = ?\)
We know that
\[\begin{aligned} Sum\ of\ interior\ angles &= 180^\circ(n-2) \\ 1080^\circ &= 180^\circ(n-2) \\ \frac{1080^\circ}{180^\circ} &= n-2 \\ 6 &= n-2 \\ 6+2 &= n \\ 8 &= n \\ \boldsymbol{n} &= \boldsymbol{8} \end{aligned}\]Thus, the polygon has 8 sides.
3Two similar bottles are such that one is twice as high as the other. What is the ratio of their surface areas and their capacities?
Let
\(Height\ of\ second\ bottle = h_2 = h\)
\(Height\ of\ first\ bottle = h_1 = 2h\)
If \(A_1\) and \(A_2\) be the surface areas of first and second bottles respectively. Using formula of area for similar solids
\[\begin{aligned} \frac{A_1}{A_2} &= \left(\frac{h_1}{h_2}\right)^2 \\ \frac{A_1}{A_2} &= \left(\frac{2h}{h}\right)^2 \\ \frac{A_1}{A_2} &= \left(\frac{2}{1}\right)^2 \\ \frac{A_1}{A_2} &= \frac{4}{1} \\ \boldsymbol{A_1 : A_2} &= \boldsymbol{4:1} \end{aligned}\]Let \(C_1\) and \(C_2\) be the capacities (volumes) of these bottles. Using formula of volumes for similar solids.
\[\begin{aligned} \frac{C_1}{C_2} &= \left(\frac{h_1}{h_2}\right)^3 \\ \frac{C_1}{C_2} &= \left(\frac{2h}{h}\right)^3 \\ \frac{C_1}{C_2} &= \left(\frac{2}{1}\right)^3 \\ \frac{C_1}{C_2} &= \frac{8}{1} \\ \boldsymbol{C_1 : C_2} &= \boldsymbol{8:1} \end{aligned}\]Ratio of surface areas \(= \mathbf{4:1}\) | Ratio of capacities \(= \mathbf{8:1}\)
4Each dimension of a model car is \(\dfrac{1}{10}\) of the corresponding car dimension. Find the ratio of:
\[Linear\ scale\ factor = k = \frac{Model\ dimension}{Actual\ dimension} = \frac{1}{10}\]
(a) the areas of their windscreens
Let \(A_1\) and \(A_2\) be the areas of model and actual car's windscreens. Using formula of areas for similar solids.
\[\begin{aligned} \frac{A_1}{A_2} &= k^2 \\ \frac{A_1}{A_2} &= \left(\frac{1}{10}\right)^2 \\ \frac{A_1}{A_2} &= \frac{1}{100} \\ \boldsymbol{A_1 : A_2} &= \boldsymbol{1:100} \end{aligned}\](b) the capacities of their boots
Let \(C_1\) and \(C_2\) be the capacities (volumes) of their boots. Using formula of volumes for similar solids.
\[\begin{aligned} \frac{C_1}{C_2} &= k^3 \\ \frac{C_1}{C_2} &= \left(\frac{1}{10}\right)^3 \\ \frac{C_1}{C_2} &= \frac{1}{1000} \\ \boldsymbol{C_1 : C_2} &= \boldsymbol{1:1000} \end{aligned}\](c) the widths of the cars
Let \(W_1\) and \(W_2\) be the widths of model and actual car
\[\begin{aligned} \frac{W_1}{W_2} &= k \\ \frac{W_1}{W_2} &= \frac{1}{10} \\ \boldsymbol{W_1 : W_2} &= \boldsymbol{1:10} \end{aligned}\](d) the number of wheels they have
\[\begin{aligned} No.\ of\ wheels\ of\ model\ car &= 4 \\ No.\ of\ wheels\ of\ actual\ car &= 4 \\ Ratio\ of\ no.\ of\ wheels &= 4:4 \\ &= \boldsymbol{1:1} \end{aligned}\]
5Three similar jugs have heights 8 cm, 12 cm and 16 cm. If the smallest jug holds \(\dfrac{1}{2}\) litre, find the capacities of the other two.
\(Height\ of\ first\ jug = h_1 = 8\ cm\)
\(Height\ of\ second\ jug = h_2 = 12\ cm\)
\(Height\ of\ third\ jug = h_3 = 16\ cm\)
\(Volume\ of\ first\ jug = V_1 = \dfrac{1}{2}\ litre\)
\(Volume\ of\ second\ jug = V_2 = ?\)
\(Volume\ of\ third\ jug = V_3 = ?\)
For volume of second jug \(V_2\) using formula of volumes for similar solids.
\[\begin{aligned} \frac{V_1}{V_2} &= \left(\frac{h_1}{h_2}\right)^3 \\ \frac{\frac{1}{2}}{V_2} &= \left(\frac{8}{12}\right)^3 \\ \frac{1}{2V_2} &= \frac{512}{1728} \\ (1)(1728) &= (512)(2V_2) \\ 1728 &= 1024V_2 \\ V_2 &= \frac{1728}{1024} \\ V_2 &= 1.69 \\ \boldsymbol{V_2} &= \boldsymbol{1.69\ litre} \end{aligned}\]For volume of third jug \(V_3\) using formula of volumes for similar solids.
\[\begin{aligned} \frac{V_1}{V_3} &= \left(\frac{h_1}{h_3}\right)^3 \\ \frac{\frac{1}{2}}{V_3} &= \left(\frac{8}{16}\right)^3 \\ \frac{1}{2V_3} &= \frac{512}{4096} \\ (1)(4096) &= (512)(2V_3) \\ 4096 &= 1024V_3 \\ V_3 &= \frac{4096}{1024} \\ \boldsymbol{V_3} &= \boldsymbol{4\ litre} \end{aligned}\]Capacity of second jug \(= \mathbf{1.69\ litre}\) | Capacity of third jug \(= \mathbf{4\ litres}\)
6Three similar drinking glasses have heights 7.5 cm, 9 cm and 10.5 cm. If the tallest glass holds 343 millilitres, find the capacities of the other two.
\(Height\ of\ first\ glass = h_1 = 7.5\ cm\)
\(Height\ of\ second\ glass = h_2 = 9\ cm\)
\(Height\ of\ third\ glass = h_3 = 10.5\ cm\)
\(Volume\ (capacity)\ of\ third\ glass = V_3 = 343\ ml\)
\(Volume\ of\ first\ glass = V_1 = ?\)
\(Volume\ of\ second\ glass = V_2 = ?\)
For capacity of first glass \(V_1\) using formula of volumes for similar solids.
\[\begin{aligned} \frac{V_1}{V_3} &= \left(\frac{h_1}{h_3}\right)^3 \\ \frac{V_1}{343} &= \left(\frac{7.5}{10.5}\right)^3 \\ \frac{V_1}{343} &= \frac{421.875}{1157.625} \\ (V_1)(1157.625) &= (421.875)(343) \\ 1157.625V_1 &= 144703.125 \\ V_1 &= \frac{144703.125}{1157.625} \\ \boldsymbol{V_1} &= \boldsymbol{125\ ml} \end{aligned}\]For capacity of second glass \(V_2\) using formula of volumes for similar solids.
\[\begin{aligned} \frac{V_2}{V_3} &= \left(\frac{h_2}{h_3}\right)^3 \\ \frac{V_2}{343} &= \left(\frac{9}{10.5}\right)^3 \\ \frac{V_2}{343} &= \frac{729}{1157.625} \\ (V_2)(1157.625) &= (729)(343) \\ 1157.625V_2 &= 250047 \\ V_2 &= \frac{250047}{1157.625} \\ \boldsymbol{V_2} &= \boldsymbol{216\ ml} \end{aligned}\]Capacity of first glass \(= \mathbf{125\ ml}\) | Capacity of second glass \(= \mathbf{216\ ml}\)
7A toy manufacturer produces model cars similar to actual cars. The ratio of door area (model to actual) is \(1 : 2500\). Find:
Ratio of areas of doors of model and actual cars: \(A_1 : A_2 = 1 : 2500\)
(a) the ratio of their lengths
Let \(l_1\) and \(l_2\) be the lengths of model and actual cars respectively. So,
\[\begin{aligned} \frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\ \frac{1}{2500} &= \left(\frac{l_1}{l_2}\right)^2 \\ \sqrt{\left(\frac{l_1}{l_2}\right)^2} &= \sqrt{\frac{1}{2500}} \\ \frac{l_1}{l_2} &= \frac{1}{50} \\ \boldsymbol{l_1 : l_2} &= \boldsymbol{1:50} \end{aligned}\](b) the ratio of the capacities of their petrol tanks
Let \(V_1\) and \(V_2\) be the capacities (volumes) of petrol tanks. Using formula of volumes for similar solids.
\[\begin{aligned} \frac{V_1}{V_2} &= \left(\frac{l_1}{l_2}\right)^3 \\ \frac{V_1}{V_2} &= \left(\frac{1}{50}\right)^3 \\ \frac{V_1}{V_2} &= \frac{1}{125000} \\ \boldsymbol{V_1 : V_2} &= \boldsymbol{1:125000} \end{aligned}\](c) the width of the model, if the actual car is 150 cm wide
Let \(W_1\) and \(W_2 = 150\ cm\) be the widths of model and actual car
\[\begin{aligned} \frac{W_1}{W_2} &= \frac{l_1}{l_2} \\ \frac{W_1}{150} &= \frac{1}{50} \\ W_1 &= \frac{1}{50} \times 150 \\ \boldsymbol{W_1} &= \boldsymbol{3\ cm} \end{aligned}\](d) the area of the rear window of the actual car, if the area of the rear window of the model is \(3\ cm^2\)
Let \(A_1 = 3\ cm^2\) and \(A_2\) be the areas of windows of model and actual cars respectively. So,
\[\begin{aligned} \frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\ \frac{3}{A_2} &= \left(\frac{1}{50}\right)^2 \\ \frac{3}{A_2} &= \frac{1}{2500} \\ \frac{A_2}{3} &= \frac{2500}{1} \\ A_2 &= \frac{2500}{1} \times 3 \\ \boldsymbol{A_2} &= \boldsymbol{7500\ cm^2} \end{aligned}\]8The ratio of the areas of two similar labels on two similar jars of coffee is \(144:169\). Find the ratio of:
Ratio of areas of two similar jars: \(A_1 : A_2 = 144 : 169\)
(a) the heights of the two jars
Let \(h_1\) and \(h_2\) be the heights of two jars respectively
\[\begin{aligned} \frac{A_1}{A_2} &= \left(\frac{h_1}{h_2}\right)^2 \\ \left(\frac{h_1}{h_2}\right)^2 &= \frac{144}{169} \\ \sqrt{\left(\frac{h_1}{h_2}\right)^2} &= \sqrt{\frac{144}{169}} \\ \frac{h_1}{h_2} &= \frac{12}{13} \\ \boldsymbol{h_1 : h_2} &= \boldsymbol{12:13} \end{aligned}\](b) their capacities
Let \(V_1\) and \(V_2\) be the capacities (volumes) of jars. Using formula of volumes for similar solids.
\[\begin{aligned} \frac{V_1}{V_2} &= \left(\frac{h_1}{h_2}\right)^3 \\ \frac{V_1}{V_2} &= \left(\frac{12}{13}\right)^3 \\ \frac{V_1}{V_2} &= \frac{1728}{2197} \\ \boldsymbol{V_1 : V_2} &= \boldsymbol{1728:2197} \end{aligned}\]Height ratio \(= \mathbf{12:13}\) | Capacity ratio \(= \mathbf{1728:2197}\)
9A tessellation of tiles on a floor uses a repeating pattern of 1 regular hexagon, 6 squares and 6 equilateral triangles. Find the total area of a single pattern with side length \(\dfrac{1}{2}\) metre of each polygon.
Given 1 regular hexagon, 6 equilateral triangles, and 6 squares.
\(Side\ length\ of\ each\ polygon = s = \dfrac{1}{2}\ m\)
Now
\[\begin{aligned} Area\ of\ hexagon &= \frac{3\sqrt{3}}{2} \cdot s^2 \\ &= \frac{3\sqrt{3}}{2} \cdot \left(\frac{1}{2}\right)^2 \\ &= \frac{3\sqrt{3}}{8} \\ \boldsymbol{A_{hexagon}} &= \boldsymbol{0.65\ m^2} \end{aligned}\] \[\begin{aligned} Area\ of\ equilateral\ triangle &= \frac{\sqrt{3}}{4} \cdot s^2 \\ &= \frac{\sqrt{3}}{4} \cdot \left(\frac{1}{2}\right)^2 \\ &= \frac{\sqrt{3}}{16} \\ \boldsymbol{A_{triangle}} &= \boldsymbol{0.108\ m^2} \end{aligned}\] \[\begin{aligned} Area\ of\ square &= s^2 \\ &= \left(\frac{1}{2}\right)^2 \\ &= \frac{1}{4} \\ \boldsymbol{A_{square}} &= \boldsymbol{0.25\ m^2} \end{aligned}\] \[\begin{aligned} Total\ area &= A_{hexagon} + 6A_{triangle} + 6A_{square} \\ &= 0.65 + 6(0.108) + 6(0.25) \\ &= \boldsymbol{2.8\ m^2} \end{aligned}\]Thus, \(\mathbf{2.8\ m^2}\) is the total area of a single pattern.
📈 Key Concepts – Similar Figures
- Similar Figures: Same shape, different size. Corresponding angles are equal, corresponding sides are proportional.
- Scale Factor: Ratio of corresponding sides. Used to enlarge or reduce figures.
- Triangle Similarity Conditions: AAA, SSS, SAS.
- Area Ratio: Ratio of areas = (scale factor)².
- Volume Ratio: Ratio of volumes = (scale factor)³.
- Applications: Shadow problems, map scales, model making, indirect measurement.