Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum / PECTA 2025 Syllabus
📖 What's Inside: This exercise covers interior and exterior angles of polygons, area formulas for geometric shapes, and the principles of tessellation. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Chapter 9: Similar Figures
1
Solve the following problems related to polygons.
(i) What is the sum of the interior angles of a decagon (10-sided polygon)?
\(Number\ of\ sides\ of\ decagon = n = 10\)
\[\begin{aligned} Sum\ of\ interior\ angles &= 180^\circ(n-2) \\ &= 180^\circ(10-2) \\ &= 180^\circ(8) \\ &= \boldsymbol{1440^\circ} \end{aligned}\](ii) Calculate the measure of each interior angle of a regular hexagon.
\(Number\ of\ sides\ of\ hexagon = n = 6\)
\[\begin{aligned} Each\ interior\ angle\ of\ hexagon &= \frac{180^\circ(n-2)}{n} \\[6pt] &= \frac{180^\circ(6-2)}{6} \\[6pt] &= \frac{180^\circ(4)}{6} \\[6pt] &= \frac{720^\circ}{6} \\[4pt] &= \boldsymbol{120^\circ} \end{aligned}\](iii) What is each exterior angle of a regular pentagon?
\(Number\ of\ sides\ of\ pentagon = n = 5\)
\[\begin{aligned} Each\ exterior\ angle\ of\ pentagon &= \frac{360^\circ}{n} \\[6pt] &= \frac{360^\circ}{5} \\[4pt] &= \boldsymbol{72^\circ} \end{aligned}\](iv) If the sum of the interior angles of a polygon is \(1260^\circ\), how many sides does the polygon have?
\(Number\ of\ sides\ of\ polygon = n = ?\)
\[\begin{aligned} Sum\ of\ interior\ angles &= 180^\circ(n-2) \\ 1260^\circ &= 180^\circ(n-2) \\ \frac{1260^\circ}{180^\circ} &= n-2 \\ 7 &= n-2 \\ 7+2 &= n \\ 9 &= n \\ \boldsymbol{n} &= \boldsymbol{9} \end{aligned}\]
2
In a parallelogram \(ABCD\), \(m\overline{AB} = 10\text{ cm}\), \(m\overline{AD} = 6\text{ cm}\) and \(m\angle BAD = 45^\circ\). Calculate the area of \(ABCD\).
\(m\overline{AB} = 10\text{ cm}\)
\(m\overline{AD} = 6\text{ cm}\)
\(m\angle BAD = 45^\circ\)
\(Area\ of\ parallelogram\ ABCD = ?\)
Consider a parallelogram \(ABCD\) and \(h\) is height between opposite sides.
In a right angle \(\triangle ADE\)
\[\begin{aligned} \sin m\angle A &= \frac{Perpendicular}{Hypotenuse} \\[6pt] \sin 45^\circ &= \frac{h}{6} \\[6pt] \frac{1}{\sqrt{2}} &= \frac{h}{6} \\[6pt] 6 \times \frac{1}{\sqrt{2}} &= h \\[4pt] h &= \frac{6}{\sqrt{2}} \text{ cm} \end{aligned}\]Now
\[\begin{aligned} Area\ of\ parallelogram\ ABCD &= Base \times Height \\ &= m\overline{AB} \times m\overline{DE} \\[4pt] &= 10 \times \frac{6}{\sqrt{2}} \\[6pt] &= \frac{60}{\sqrt{2}} \\[4pt] &= \boldsymbol{42.43\ cm^2} \end{aligned}\]
3
In a parallelogram \(ABCD\) if \(m\angle DAB = 70^\circ\), find the measures of all other angles in the parallelogram.
In parallelogram \(ABCD\)
\[m\angle A = m\angle C = 70^\circ \qquad (Opposite\ angle)\]Also
\[\begin{aligned} m\angle A + m\angle B &= 180^\circ \qquad (Adjacent\ angle) \\ 70^\circ + m\angle B &= 180^\circ \\ m\angle B &= 180^\circ - 70^\circ \\ m\angle B &= 110^\circ \end{aligned}\]Now
\[m\angle B = m\angle D = 110^\circ \qquad (Opposite\ angle)\]
\(m\angle A = m\angle C = 70^\circ\) and \(m\angle B = m\angle D = 110^\circ\)
4
A shape is created by cutting a square in half diagonally and then attaching a right-angled triangle to the hypotenuse of each half. Explain why this shape can tessellate and calculate the interior angle of the new shape.
Cutting a square diagonally creates two equal right-angled triangles with angles \(45^\circ\), \(45^\circ\) and \(90^\circ\).
A right-angled triangle is attached to the hypotenuse of each half to make a new shape. The shape can tessellate because its angles fit together to make \(360^\circ\), so there are no gaps or overlaps when repeated.
The exact interior angle of the new shape depends on the size and position of the attached right-angled triangles, so it cannot be found unless more information is given.
5
A tessellation is created by repeatedly reflecting a right-angled triangle with sides 3, 4, and 5 units. Find the minimum number of reflections needed to cover a square with an area of 3600 square units.
Given that right-angled triangle has sides 3, 4, and 5 units. So
\[\begin{aligned} Area\ of\ triangle &= \frac{1}{2} \times Base \times Height \\[6pt] &= \frac{1}{2} \times 3 \times 4 \\[4pt] &= \frac{12}{2} \\[4pt] &= 6\ square\ units \end{aligned}\]Now
\[\begin{aligned} Number\ of\ triangles &= \frac{Area\ of\ square}{Area\ of\ triangle} \\[6pt] &= \frac{3600}{6} \\[4pt] &= 600 \end{aligned}\]So 600 reflections are needed to cover the square.
6
A tessellation is created using regular hexagons, each with side length 5 cm. Find the total area of the tessellation if it consists of 25 hexagons, and find the total perimeter of the outer edge assuming it's a perfect hexagon.
\(Each\ side\ of\ hexagon = s = 5\ cm\)
\[\begin{aligned}
Area\ of\ 1\ regular\ hexagon &= A = \frac{3\sqrt{3}}{2} \cdot s^2 \\[6pt]
Area\ of\ 25\ regular\ hexagons &= A = 25 \times \frac{3\sqrt{3}}{2} \cdot s^2 \\[6pt]
A &= 25 \times \frac{3\sqrt{3}}{2} \cdot (5)^2 \\[6pt]
A &= \frac{1875\sqrt{3}}{2} \\[4pt]
A &= 1623.8\ cm^2
\end{aligned}\]
As the complete shape is also a perfect hexagon.
\[\begin{aligned} Area\ of\ regular\ hexagon &= \frac{3\sqrt{3}}{2} \cdot s^2 \\[6pt] 1623.8 &= \frac{3\sqrt{3}}{2} \cdot s^2 \\[6pt] \frac{1623.8 \times 2}{3\sqrt{3}} &= s^2 \\[6pt] 625 &= s^2 \\ s^2 &= 625 \\ \sqrt{s^2} &= \sqrt{625} \\ s &= 25\ cm \end{aligned}\]Now the total perimeter of outer edge of tessellation
\[\begin{aligned} Perimeter\ of\ hexagon &= 6 \times s \\ &= 6 \times 25 \\ &= \boldsymbol{125\ cm} \end{aligned}\]Total area = 1623.8 cm² | Total perimeter = 125 cm
7
A rectangular floor is 12 m by 15 m. How many square tiles, each 1 m by 1 m, are needed to cover the floor?
\[\begin{aligned}
Area\ of\ rectangular\ floor &= 12 \times 15 = 180\ m^2 \\
Area\ of\ each\ square\ tile &= 1 \times 1 = 1\ m^2
\end{aligned}\]
Now
\[\begin{aligned} Number\ of\ tiles &= \frac{Area\ of\ rectangular\ floor}{Area\ of\ each\ square\ tile} \\[6pt] &= \frac{180}{1} \\[4pt] &= \boldsymbol{180\ tiles} \end{aligned}\]180 tiles are needed.
8
A rectangular wall is 10 m tall and 120 m wide. How many gallons of paint are needed to cover the wall, if one gallon covers 35 m²?
\[\begin{aligned}
Area\ of\ rectangular\ wall &= 10 \times 120 = 1200\ m^2 \\
Area\ covered\ by\ one\ gallon &= 35\ m^2
\end{aligned}\]
Now
\[\begin{aligned} Number\ of\ gallons &= \frac{Area\ of\ rectangular\ wall}{Area\ covered\ by\ one\ gallon} \\[6pt] &= \frac{1200}{35} \\[4pt] &= 34.29\ gallons \\ &\approx \boldsymbol{35\ gallons} \end{aligned}\]Approximately 35 gallons of paint are needed.
9
A rectangular wall has a length of 10 m and a width of 4 metres. If 1 litre of paint covers 7 m², how many litres of paint are needed to cover the wall?
\[\begin{aligned}
Area\ of\ rectangular\ wall &= 10 \times 4 = 40\ m^2 \\
Area\ covered\ by\ one\ litre &= 7\ m^2
\end{aligned}\]
Now
\[\begin{aligned} Number\ of\ litres &= \frac{Area\ of\ rectangular\ wall}{Area\ covered\ by\ one\ litre} \\[6pt] &= \frac{40}{7} \\[4pt] &= 5.7\ litre \\ &\approx \boldsymbol{6\ litres} \end{aligned}\]Approximately 6 litres of paint are needed.
10
A window has a trapezoidal shape with parallel sides of 3 m and 1.5 m and a height of 2 m. Find the area of the window.
\(Length\ of\ first\ side = 3\ m\)
\(Length\ of\ second\ side = 1.5\ m\)
\(Height\ of\ trapezoidal = 2\ m\)
\[\begin{aligned}
Area\ of\ trapezoidal &= \frac{Sum\ of\ \parallel\ sides}{2} \times height \\[6pt]
&= \frac{3 + 1.5}{2} \times 2 \\[6pt]
&= \frac{4.5}{2} \times 2 \\[6pt]
&= \frac{9}{2} \\[4pt]
&= \boldsymbol{4.5\ m^2}
\end{aligned}\]
Area of the window = 4.5 m²
📈 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.
- Areas of Similar Triangles: Ratio of areas = (scale factor)².
- Applications: Shadow problems, map scales, model making, indirect measurement.