Mastery Guide: Geometry and Polygons - Demonstrative Geometry, Similarity of Figures, Regular Polygons & Scaling
Mastery Guide: Geometry and Polygons
Single National Curriculum (SNC) • Demonstrative Geometry, Similarity of 2D/3D Figures, Regular Polygons, Area/Volume Scaling, & Geometric Loci
📖 1. Unit Overview & Target Learning Outcomes
Geometry is the mathematical study of shapes, sizes, relative positions of figures, and properties of space. In Chapter 9, we advance from intuitive drawing to Demonstrative Geometry (formal deductive proof), explore the laws of Geometric Similarity across 2D shapes and 3D solids, calculate angle and perimeter/area metrics for Regular Polygons, and understand 2D motion and loci constraints.
🎯 Target Learning Outcomes:
- Demonstrative Proof Architecture: Differentiate between inductive reasoning (pattern observation) and deductive reasoning (logical deduction). Define statements, axioms, postulates, conjectures, theorems, corollaries, and converse statements. Master the 5-step proof structure (Figure, Given, To Prove, Construction, Proof).
- Similarity of Polygons: Apply criteria for polygon similarity (equal corresponding angles and proportional corresponding sides). Utilize Thales' Theorem (Basic Proportionality Theorem) and similarity postulates ($AA$, $SAS$, $SSS$) to solve missing lengths.
- Scaling Ratios of Area and Volume: Master the fundamental scaling laws: Linear Scale Factor $k = \frac{l_1}{l_2}$, Area Ratio $\frac{A_1}{A_2} = k^2 = \left(\frac{l_1}{l_2}\right)^2$, Volume Ratio $\frac{V_1}{V_2} = k^3 = \left(\frac{l_1}{l_2}\right)^3$, and Mass Ratio $\frac{m_1}{m_2} = k^3$ for identical material density.
- Regular Polygons & Angle Geometry: Calculate sum of interior angles $S = (n-2) \times 180^\circ$, each interior angle $\theta_i = \frac{(n-2) \times 180^\circ}{n}$, each exterior angle $\theta_e = \frac{360^\circ}{n}$, and total diagonals $d = \frac{n(n-3)}{2}$. Compute regular polygon area $A = \frac{1}{2} P a$ using perimeter $P$ and apothem $a$.
- Geometric Loci & Intersections: Understand locus as a set of points satisfying specific constraints (equidistant from a point $\to$ circle; equidistant from a line $\to$ pair of parallel lines; equidistant from two points $\to$ perpendicular bisector; equidistant from two intersecting lines $\to$ pair of angle bisectors).
💡 2. Kid-Friendly Tips for Success & Memory Hooks
Remember the Power Rule! Length scales as $k^1$, Area/Surface Area scales as $k^2$, and Volume/Mass scales as $k^3$. If side doubles ($\times 2$), Area quadruples ($\times 4$), and Volume octuples ($\times 8$)!
No matter how many sides a convex polygon has (triangle, 100-gon, or million-gon), if you walk around the perimeter, you make exactly ONE full spin ($360^\circ$). Thus, for a regular polygon: $\text{Exterior} = \frac{360^\circ}{n}$.
Axiom: Universal self-evident general truth (e.g. $a=b \implies b=a$).
Postulate: Self-evident truth specifically for GEOMETRY (e.g. line through two points).
Theorem: Must be logically PROVED!
The apothem ($a$) of a regular polygon is the perpendicular distance from the center to any side. The area is simply the sum of $n$ congruent triangles: $A = n \times \left(\frac{1}{2} s a\right) = \frac{1}{2} P a$.
🌍 3. Real-World Connections
🏗️ Architecture & Scale Models
Architects build scale models ($1:100$) of skyscrapers and bridges. If a model uses $1\text{ L}$ of paint, the actual building will require $1 \times 100^2 = 10,000\text{ L}$!
🔭 Optics & Astronomy
Telescopes and camera lenses use similar triangles formed by ray diagrams to determine the actual size and distance of distant celestial bodies or landscape landmarks from their small projected focal images.
🐝 Honeycombs & Polygon Tessellations
Bees build regular hexagonal cells because hexagons tessellate a 2D plane with minimum perimeter per unit area, maximizing honey storage while saving wax.
📡 GPS & Cellular Triangulation
GPS systems calculate location by intersecting geometric loci. Each satellite creates a spherical locus of distance; intersecting 3 or 4 loci pinpoints the receiver's precise coordinates on Earth.
🌟 4. Section-by-Section Explanations & Visual Models
4.1 Fundamentals of Demonstrative Geometry
Demonstrative geometry is the study of geometric figures using systematic logical deduction from accepted first principles.
| Term | Definition & Characteristics | Example |
|---|---|---|
| Axiom | A self-evident universal truth assumed without formal proof in general mathematics. | If $a=b$, then $a+c = b+c$. The whole is greater than its part. |
| Postulate | A self-evident geometric assumption accepted without proof. | A straight line may be drawn from any one point to any other point. |
| Conjecture | A mathematical statement believed to be true based on observations or patterns, but not yet formally proved. | Goldbach's Conjecture: Every even integer $>2$ is the sum of two primes. |
| Theorem | A statement that has been formally proven to be true using a chain of deductive reasoning. | The sum of interior angles of a triangle is $180^\circ$. |
| Corollary | A direct mathematical consequence that follows immediately from a proven theorem. | Each angle of an equilateral triangle is $60^\circ$. |
4.2 Similarity of Figures & Scaling Ratios
Two polygons are similar ($\sim$) if and only if:
- Their corresponding interior angles are equal: $\angle A = \angle A', \angle B = \angle B', \dots$
- Their corresponding sides are proportional: $\frac{A'B'}{AB} = \frac{B'C'}{BC} = \dots = k$ (scale factor).
4.3 Properties of Regular Polygons
A regular polygon is equilateral (all sides equal) and equiangular (all interior angles equal).
4.4 Geometric Locus & Intersecting Loci
A locus (plural: loci) is the set or path of all points in a plane that satisfy a given geometrical condition.
🎯 5. Unit Synthesis Summary
Chapter 9 builds the rigorous bridge between deductive geometric reasoning and algebraic computation. Starting with the distinction between empirical conjectures and formal deductive proofs (utilizing axioms and postulates), the chapter establishes similarity criteria for polygons ($AA$, $SAS$, $SSS$) and develops the powerful scaling power laws where lengths scale by $k$, surface areas by $k^2$, and 3D volumes/masses by $k^3$. For regular $n$-gons, interior angle sums $(n-2)\times 180^\circ$, exterior angles $\frac{360^\circ}{n}$, and apothem-based areas $A = \frac{1}{2}Pa$ allow direct calculation of multi-sided figures. Finally, the study of geometric loci unifies motion and geometric constraints into exact path descriptions (circles, parallel lines, perpendicular bisectors, and angle bisectors).
📝 Complete Solved Textbook Exercises & Examination Question Bank
Below is the exhaustive, step-by-step solution manual for every single textbook problem, example exercise, and review problem in Chapter 9, aligned strictly with FBISE scoring guidelines.
Exercise 9.1 • Step-by-Step Complete Solutions
What is the difference between axiom and conjecture?
Step 2: Define Conjecture: A conjecture is a statement formed based on observed patterns or empirical evidence that is believed to be true, but has not yet been formally proved or disproved by deductive logic.
Final Answer: An axiom is an accepted self-evident mathematical fact without proof, whereas a conjecture is an unproven proposition supported only by empirical evidence/patterns.
Which of the following are mathematical statements? Difference of 19 and 12 is 7.
Step 2: Evaluate truth value: $19 - 12 = 7$ is true.
Final Answer: It is a mathematical statement (True).
Which of the following are mathematical statements? -2 + 7 - 3 = 2
Step 2: Compare with RHS: $2 = 2$ (True).
Final Answer: It is a mathematical statement (True).
Which of the following are mathematical statements? 34 + 16 != 50
Step 2: Statement asserts $50 \neq 50$, which is definitively false.
Final Answer: It is a mathematical statement (False).
Which of the following are mathematical statements? a + b = 9
Step 2: Without values, it cannot be classified as definitively true or false; it is an open algebraic sentence.
Final Answer: Not a mathematical statement (Open sentence).
Which of the following are mathematical statements? (a + b)^2 = a^2 + 2ab + b^2
Step 2: Since it holds true for all real values of $a$ and $b$, it is a universally true statement (identity).
Final Answer: It is a mathematical statement (True Identity).
Which of the following are mathematical statements? 2 + 2 * 2 = 6
Step 2: LHS equals RHS ($6 = 6$).
Final Answer: It is a mathematical statement (True).
Which of the following are mathematical statements? The product of x and y is smaller than 5.
Step 2: Since the truth depends on undetermined variables $x$ and $y$, it cannot be assigned a fixed truth value.
Final Answer: Not a mathematical statement (Open sentence).
Which of the following are mathematical statements? If x is real then either x < 0 or x > 0 or x = 0.
Step 2: This is the foundational Law of Trichotomy for real numbers.
Final Answer: It is a mathematical statement (True).
Which of the following are mathematical statements? If a > b and b > c then a < c
Step 2: The given sentence asserts $a < c$, which is a contradiction (false).
Final Answer: It is a mathematical statement (False).
Which of the following are mathematical statements? xy + z = 12
Final Answer: Not a mathematical statement (Open sentence).
Which of the following are mathematical statements? s - t = 4 if s = 4 and t = 0
Step 2: LHS matches RHS ($4 = 4$).
Final Answer: It is a mathematical statement (True).
The sum of $a$ and $b$ is equal to 0. Is this sentence a mathematical statement? If not, how can we make it a mathematical statement?
Step 2: Conversion to mathematical statement:
- Method 1 (Assign specific values): 'The sum of $3$ and $-3$ is equal to $0$' (True statement).
- Method 2 (Universal quantification): 'For all real numbers $a$, the sum of $a$ and $-a$ is equal to $0$' (True statement).
Final Answer: No, it is an open sentence. It becomes a mathematical statement when specific numerical values are given to variables $a$ and $b$ (e.g., $s=4, t=0$) or when universal quantifiers are attached.
Prove $(x + 1)^2 + 5 = x^2 + 2x + 6$ by taking $x = 2, 5$ and $10$.
LHS = $(2 + 1)^2 + 5 = 3^2 + 5 = 9 + 5 = 14$
RHS = $2^2 + 2(2) + 6 = 4 + 4 + 6 = 14 \implies \text{LHS} = \text{RHS} = 14$.
Step 2: For $x = 5$:
LHS = $(5 + 1)^2 + 5 = 6^2 + 5 = 36 + 5 = 41$
RHS = $5^2 + 2(5) + 6 = 25 + 10 + 6 = 41 \implies \text{LHS} = \text{RHS} = 41$.
Step 3: For $x = 10$:
LHS = $(10 + 1)^2 + 5 = 11^2 + 5 = 121 + 5 = 126$
RHS = $10^2 + 2(10) + 6 = 100 + 20 + 6 = 126 \implies \text{LHS} = \text{RHS} = 126$.
Final Answer: $(x+1)^2 + 5 = x^2 + 2x + 6$ is verified for all given values $x=2, 5, 10$.
Find the next number in the pattern using conjecture: $1, 3, 7, 15, 31, \underline{\hspace{1cm}}$. State the conjecture used.
$3 - 1 = 2 = 2^1$
$7 - 3 = 4 = 2^2$
$15 - 7 = 8 = 2^3$
$31 - 15 = 16 = 2^4$
Step 2: State the Conjecture: The difference added to each consecutive term doubles each time (i.e. Add $2^5 = 32$), or rule $T_n = 2^n - 1$.
Step 3: Calculate next term: $31 + 32 = 63$ (or $2^6 - 1 = 64 - 1 = 63$).
Final Answer: Next number is 63. Conjecture: $T_n = 2T_{n-1} + 1$ (or adding successive powers of 2).
Which of the following are axioms? How many of the axioms are postulates? If a = b then b = a
Step 2: Not specific to geometry $\implies$ Axiom, not a postulate.
Final Answer: Axiom (0 postulates).
Which of the following are axioms? How many of the axioms are postulates? 2 plus 2 make 4.
Final Answer: Axiom (not a postulate).
Which of the following are axioms? How many of the axioms are postulates? One and only one line can pass through two points.
Final Answer: Axiom and a Postulate.
Which of the following are axioms? How many of the axioms are postulates? If two sides of a triangle are equal then opposite angles are also equal.
Final Answer: Theorem (Not an axiom).
Which of the following are axioms? How many of the axioms are postulates? Product of two negative real numbers is always greater than zero.
Final Answer: Axiom (not a postulate).
Which of the following are axioms? How many of the axioms are postulates? All right angles are equal to one another.
Final Answer: Axiom and a Postulate.
Which of the following are axioms? How many of the axioms are postulates? The whole is greater than its part.
Final Answer: Axiom (General, not purely geometric).
Which of the following are axioms? How many of the axioms are postulates? If a > b and c > d then a + c > b + d.
Final Answer: Axiom.
Which of the following are axioms? How many of the axioms are postulates? It is possible to extend a line segment continuously in both directions.
Final Answer: Axiom and a Postulate.
Which of the following are axioms? How many of the axioms are postulates? When we add three consecutive even numbers, their sum is even.
Final Answer: Theorem (Not an axiom).
Explain all the steps of geometrical proof.
Step 2: Given (Hypothesis): Mathematical translation of the initial conditions/facts given in the proposition expressed using the letters of the figure.
Step 3: Required to Prove (To Prove / Conclusion): The exact geometric result or property that must be established deductively.
Step 4: Construction: Any auxiliary lines, rays, or perpendiculars added to the figure to facilitate the proof (drawn with dashed lines).
Step 5: Proof (Statements and Reasons): A systematic two-column logical progression consisting of sequential geometric statements accompanied by valid reasons (axioms, postulates, definitions, or previously proved theorems).
Final Answer: The essential steps are: Figure, Given, To Prove, Construction, and Proof (Statements & Reasons).
Exercise 9.2 • Step-by-Step Complete Solutions
Which of the following pairs of figures are similar? Two right-angled triangles with angles 30°, 60°, 90°.
Step 2: By $AA$ (Angle-Angle) similarity criterion, triangles with identical corresponding angles are similar.
Final Answer: Similar.
Which of the following pairs of figures are similar? Equilateral triangle of side 2 and equilateral triangle of side 3.
Step 2: Corresponding sides ratio is constant: $\frac{3}{2} = 1.5$.
Final Answer: Similar.
Which of the following pairs of figures are similar? Rectangle of base 4 and parallelogram of base 4, slant side 3.
Step 2: A general non-rectangular parallelogram has non-right angles. Since corresponding angles are not equal, they cannot be similar.
Final Answer: Not Similar.
Which of the following pairs of figures are similar? Triangles with angle 120° and adjacent sides (6, 4.5) and (4, 3).
Step 2: The included angle between these sides is $120^\circ$ in both triangles.
Step 3: By $SAS$ (Side-Angle-Side) similarity criterion, the triangles are similar.
Final Answer: Similar.
Which of the following pairs of figures are similar? Triangles with sides (2, 3, 4) and (4, 6, 8).
Step 2: Since all 3 pairs of corresponding sides are in the same constant ratio $k = 2$, they are similar by $SSS$ criterion.
Final Answer: Similar.
Which of the following pairs of figures are similar? Rectangles of dimensions 12x16 and 6x8.
Step 2: Check side ratios: $\frac{12}{6} = 2$ and $\frac{16}{8} = 2$. Ratios are equal.
Final Answer: Similar.
Which of the following pairs of figures are similar? Triangles with 30° included between sides (5, 6) and sides (7.5, 9).
Step 2: Included angle is $30^\circ$ in both.
Final Answer: Similar.
Find the unknown quantities in the following similar figures. Right triangles: Triangle 1 (base 8, hyp x, perp y), Triangle 2 (base 4, hyp 5, perp 3).
Step 2: Calculate hypotenuse $x = 5 \times k = 5 \times 2 = 10\text{ cm}$.
Step 3: Calculate perpendicular $y = 3 \times k = 3 \times 2 = 6\text{ cm}$.
Final Answer: x = 10 cm, y = 6 cm.
Find the unknown quantities in the following similar figures. Triangles with sides (3, 5) and (6, 10) with angle 30° and angle a.
Step 2: In similar figures, corresponding angles are strictly equal: $a = 30^\circ$.
Final Answer: a = 30°.
Find the unknown quantities in the following similar figures. Rectangles of size 5x2 and x*3.
Step 2: Solve for $x$: $x = 5 \times 1.5 = 7.5\text{ cm}$.
Final Answer: x = 7.5 cm.
Find the unknown quantities in the following similar figures. Trapezoids: Top 3, Base 4, Height 1.8; Top 6, Base z, Height y.
Step 2: Unknown base $z = 4 \times 2 = 8\text{ cm}$.
Step 3: Unknown height $y = 1.8 \times 2 = 3.6\text{ cm}$.
Final Answer: z = 8 cm, y = 3.6 cm.
Find the unknown quantities in the following similar figures. Right triangles: (a, 2.1, 3.2) and (8, b, 6.4).
Step 2: Hypotenuse $a = \frac{8}{k} = \frac{8}{2} = 4\text{ cm}$.
Step 3: Base $b = 2.1 \times k = 2.1 \times 2 = 4.2\text{ cm}$.
Final Answer: a = 4 cm, b = 4.2 cm.
Find the unknown quantities in the following similar figures. Right triangles: (a, 3, 5) and (8, b, 10).
Step 2: Base $a = \frac{8}{2} = 4\text{ cm}$ (or $\sqrt{5^2 - 3^2} = 4$).
Step 3: Height $b = 3 \times 2 = 6\text{ cm}$.
Final Answer: a = 4 cm, b = 6 cm.
Find the unknown quantities in the following similar figures. Triangles LMN and PON: LN = 18, MN = a; PN = 12, ON = 8.
Step 2: Solve for side $a = ON \times k = 8 \times 1.5 = 12\text{ cm}$.
Final Answer: a = 12 cm.
Two Triangles $ABC$ and $DEF$ are similar and $\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = 2$. If $AB = 6\text{ cm}, BC = 8\text{ cm}, CA = 4\text{ cm}$, find the lengths of sides of $\triangle DEF$.
Step 2: Using ratio $\frac{BC}{EF} = 2 \implies EF = \frac{BC}{2} = \frac{8}{2} = 4\text{ cm}$.
Step 3: Using ratio $\frac{CA}{FD} = 2 \implies FD = \frac{CA}{2} = \frac{4}{2} = 2\text{ cm}$.
Final Answer: DE = 3 cm, EF = 4 cm, FD = 2 cm.
While going back to home from school, Wajid noted that his shadow was $\frac{2}{3}$ of his height. Find the height of a pole nearby having shadow of $12\text{ m}$.
Step 2: For the pole with height $H$ and shadow $12\text{ m}$:
$$\frac{12}{H} = \frac{2}{3}$$
Step 3: Cross multiply to solve for $H$:
$$2H = 12 \times 3 = 36 \implies H = \frac{36}{2} = 18\text{ m}$$
Final Answer: The height of the pole is 18 m.
Find the length of the larger rope of the hanging bridge, where $x = y$, with left pillar height 6 m, span 12 m, and right pillar height 4 m, span 8 m, smaller rope 10 m.
Step 2: Calculate scale factor from the spans (or pillar heights):
$$k = \frac{12\text{ m}}{8\text{ m}} = \frac{6\text{ m}}{4\text{ m}} = 1.5$$
Step 3: Calculate length of larger rope $L$ using smaller rope length $10\text{ m}$:
$$L = 10 \times 1.5 = 15\text{ m}$$
Final Answer: The length of the larger rope is 15 m.
In the figure, $OA$ is object lying in front of a convex lens at a distance of $10\text{ cm}$. Find the distance of image from the lens if its size is twice that of the object.
Step 2: Given magnification factor $M = 2$ and $d_o = 10\text{ cm}$:
$$\frac{d_i}{10} = 2$$
Step 3: Solve for $d_i$: $d_i = 10 \times 2 = 20\text{ cm}$.
Final Answer: The distance of the image from the lens is 20 cm.
In the figure, an electricity tower is seen through the telescope. Find height of tower if height of its image is $0.5\text{ m}$, distance inside telescope is $5\text{ m}$, and distance to tower is $100\text{ m}$.
Step 2: Substitute given values: $\frac{H}{0.5} = \frac{100}{5} = 20$.
Step 3: Solve for $H$: $H = 20 \times 0.5 = 10\text{ m}$.
Final Answer: The height of the electricity tower is 10 m.
Find the values of unknown quantities in the following figures. UV || YZ with XU = a, UY = 2, XV = 6, VZ = 4.
Step 2: Substitute values: $\frac{a}{2} = \frac{6}{4} = 1.5$.
Step 3: Solve for $a$: $a = 2 \times 1.5 = 3\text{ cm}$.
Final Answer: a = 3 cm.
Find the values of unknown quantities in the following figures. BC || DE with AD = 2, DB = 5, AE = x, EC = 6, DE = 1.5, BC = y.
Step 2: Find $y$ using similar triangles $\triangle ADE \sim \triangle ABC$: $\frac{BC}{DE} = \frac{AB}{AD} = \frac{2+5}{2} = \frac{7}{2} = 3.5$.
Step 3: Solve for $y$: $y = 1.5 \times 3.5 = 5.25\text{ cm}$.
Final Answer: x = 2.4 cm, y = 5.25 cm.
Find the values of unknown quantities in the following figures. AB || DE with CD = a, DA = 12, CE = 20, EB = 15.
Step 2: Substitute: $\frac{a}{12} = \frac{20}{15} = \frac{4}{3}$.
Step 3: Solve for $a$: $a = 12 \times \frac{4}{3} = 16\text{ cm}$.
Final Answer: a = 16 cm.
Find the values of unknown quantities in the following figures. AC || DE with AD = 3, DB = y, CE = 6, EB = 9, AC = 12, DE = x.
Step 2: Total length $CB = 6 + 9 = 15\text{ cm}$.
Step 3: From similar triangles $\triangle BDE \sim \triangle BAC$: $\frac{DE}{AC} = \frac{EB}{CB} \implies \frac{x}{12} = \frac{9}{15} = 0.6 \implies x = 12 \times 0.6 = 7.2\text{ cm}$.
Final Answer: x = 7.2 cm, y = 4.5 cm.
In the following figure, $\angle A = 40^\circ, \angle B = 90^\circ, \angle ADE = 90^\circ$. Find the measure of $\angle C$ and $\angle AED$. Is $ED \parallel CB$?
Step 2: In right-angled $\triangle ADE$: $\angle AED = 180^\circ - (90^\circ + 40^\circ) = 50^\circ$.
Step 3: Verify parallel condition: Since corresponding angles $\angle AED = \angle C = 50^\circ$ (and $\angle ADE = \angle B = 90^\circ$), the line segments $ED$ and $CB$ are parallel.
Final Answer: m∠C = 50°, m∠AED = 50°. Yes, ED || CB.
In the figure, $\triangle XYZ$ is an equilateral triangle and $LM \parallel YZ$. Find measure of $a$ and $b$. What type of triangle is $XLM$ with respect to sides and angles?
Step 2: Since $LM \parallel YZ$, corresponding angles are equal: $a = \angle Y = 60^\circ$ and $b = \angle Z = 60^\circ$.
Step 3: In $\triangle XLM$, all three angles are $60^\circ$ ($\angle X = a = b = 60^\circ$). Thus, all sides are equal ($XL = LM = MX$).
Final Answer: a = 60°, b = 60°. Triangle XLM is an equilateral (and equiangular) triangle.
Find the height of shorter tree if longer one is $12\text{ m}$ high from the ground level, where observer is at a distance of $25\text{ m}$ from the shorter tree and distance between both trees is $5\text{ m}$.
Step 2: Total distance from observer to longer tree $d_2 = 25 + 5 = 30\text{ m}$.
Step 3: Form similar right triangles equation: $\frac{h_{\text{short}}}{h_{\text{long}}} = \frac{d_1}{d_2} \implies \frac{h}{12} = \frac{25}{30} = \frac{5}{6}$.
Step 4: Solve for $h$: $h = 12 \times \frac{5}{6} = 10\text{ m}$.
Final Answer: The height of the shorter tree is 10 m.
Exercise 9.3 • Step-by-Step Complete Solutions
Following pairs of shapes are similar. Find unknown area in each case. Right triangles: side 4 with A1 = ?, and side 6 with A2 = 36 cm².
Step 2: Substitute: $\frac{A_1}{36} = \left(\frac{4}{6}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.
Step 3: Solve: $A_1 = 36 \times \frac{4}{9} = 16\text{ cm}^2$.
Final Answer: A1 = 16 cm².
Following pairs of shapes are similar. Find unknown area in each case. Rectangles: length 9 with A1 = 162 cm², and length 4 with A2 = ?.
Step 2: Solve: $A_2 = 162 \times \frac{16}{81} = 2 \times 16 = 32\text{ cm}^2$.
Final Answer: A2 = 32 cm².
Following pairs of shapes are similar. Find unknown area in each case. Trapezoids: top 3 with A1 = ?, and top 6 with A2 = 996 cm².
Step 2: Solve: $A_1 = \frac{996}{4} = 249\text{ cm}^2$.
Final Answer: A1 = 249 cm².
Following pairs of shapes are similar. Find unknown area in each case. Circles: radius 3.6 with A1 = 180 cm², and radius 1.8 with A2 = ?.
Step 2: Solve: $A_2 = \frac{180}{4} = 45\text{ cm}^2$.
Final Answer: A2 = 45 cm².
Following pairs of shapes are similar. Find unknown length $x$ in each case. Triangles: base 2 with A1 = 12 cm², and base x with A2 = 108 cm².
Step 2: Take square root: $\frac{x}{2} = 3 \implies x = 6\text{ cm}$.
Final Answer: x = 6 cm.
Following pairs of shapes are similar. Find unknown length $x$ in each case. Parallelograms: side x with A1 = 50 cm², and side 10 with A2 = 200 cm².
Step 2: Take square root: $\frac{10}{x} = 2 \implies x = \frac{10}{2} = 5\text{ cm}$.
Final Answer: x = 5 cm.
Following pairs of shapes are similar. Find unknown length $x$ in each case. Cones: slant x with A1 = 900 cm², and slant 10 with A2 = 100 cm².
Step 2: Take square root: $\frac{x}{10} = 3 \implies x = 30\text{ cm}$.
Final Answer: x = 30 cm.
Following pairs of shapes are similar. Find unknown length $x$ in each case. Sectors: radius 16 with A1 = 120 cm², and radius x with A2 = 180 cm².
Step 2: Take square root: $\frac{x}{16} = \sqrt{1.5} \approx 1.2247$.
Step 3: Solve: $x = 16 \times 1.2247 \approx 19.6\text{ cm}$.
Final Answer: x = 19.6 cm.
Radii of two spheres are $6\text{ cm}$ and $8\text{ cm}$ respectively. Find: (i) the ratio of areas of both spheres. (ii) area of larger sphere if area of smaller sphere is $360\text{ cm}^2$. (iii) area of smaller sphere if area of larger sphere is $1600\text{ cm}^2$.
Step 2: (ii) Given $A_1 = 360\text{ cm}^2$:
$$A_2 = 360 \times \frac{16}{9} = 40 \times 16 = 640\text{ cm}^2$$
Step 3: (iii) Given $A_2 = 1600\text{ cm}^2$:
$$A_1 = 1600 \times \frac{9}{16} = 100 \times 9 = 900\text{ cm}^2$$
Final Answer: (i) 9:16, (ii) 640 cm², (iii) 900 cm².
Ratio of areas of two regular pentagons is $16 : 25$. Find the ratio of sides of pentagons. Also find length of side of second pentagon if length of side of first pentagon is $8\text{ cm}$.
$$\frac{s_1}{s_2} = \sqrt{\frac{A_1}{A_2}} = \sqrt{\frac{16}{25}} = \frac{4}{5} = 4:5$$
Step 2: Given $s_1 = 8\text{ cm}$, solve for $s_2$:
$$\frac{8}{s_2} = \frac{4}{5} \implies s_2 = \frac{8 \times 5}{4} = 10\text{ cm}$$
Final Answer: Ratio of sides is 4 : 5; Side of second pentagon is 10 cm.
In the figure, $AB \parallel YZ$. If areas of triangles $XAB$ and $XYZ$ are in the ratio $25 : 36$, find $\frac{XB}{XZ}$ and $\frac{AB}{YZ}$. Are the ratios equal?
Step 2: Ratio of corresponding linear sides equals the square root of area ratio:
$$\frac{XB}{XZ} = \sqrt{\frac{25}{36}} = \frac{5}{6}$$
$$\frac{AB}{YZ} = \sqrt{\frac{25}{36}} = \frac{5}{6}$$
Step 3: Yes, the ratios are equal because corresponding sides of similar triangles are in a constant proportion.
Final Answer: XB/XZ = 5/6, AB/YZ = 5/6. The ratios are equal.
In a map, length of a $10\text{ m}$ wall is shown by $5\text{ cm}$. If area of wall shown on the map is $1400\text{ cm}^2$, find the area of actual wall.
Step 2: Determine linear scale factor $k = \frac{\text{Map Length}}{\text{Actual Length}} = \frac{5\text{ cm}}{1000\text{ cm}} = \frac{1}{200}$.
Step 3: Apply area scaling: $\frac{\text{Map Area}}{\text{Actual Area}} = k^2 = \left(\frac{1}{200}\right)^2 = \frac{1}{40000}$.
Step 4: Calculate actual area: $\text{Actual Area} = 1400 \times 40000 = 56,000,000\text{ cm}^2 = \frac{56,000,000}{10000} = 5600\text{ m}^2$.
Final Answer: The actual area of the wall is 5600 m².
Two cuboids are similar. Height of smaller cuboid is one-third of bigger one. (i) Find the ratio of surface area of larger cuboid to that of smaller one. (ii) Find the surface area of bigger cuboid if surface area of smaller cuboid is $350\text{ cm}^2$.
Ratio of surface areas: $\frac{A_{\text{large}}}{A_{\text{small}}} = k^2 = 3^2 = 9 = 9:1$.
Step 2: (ii) Given $A_{\text{small}} = 350\text{ cm}^2$:
$$A_{\text{large}} = 350 \times 9 = 3150\text{ cm}^2$$
Final Answer: (i) 9 : 1, (ii) 3150 cm².
In the figure, $BC \parallel DE$, $AC = 8\text{ cm}, CE = 5\text{ cm}$. Find: (i) $\frac{BC}{DE}$ and $\frac{AB}{AD}$ (ii) $\frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle ADE}$ (iii) the area of $\triangle ABC$ if area of $\triangle ADE$ is $507\text{ cm}^2$. (iv) area of quadrilateral $BDEC$. What type of quadrilateral is it?
Since $BC \parallel DE$, $\triangle ABC \sim \triangle ADE$ with scale factor $k = \frac{AC}{AE} = \frac{8}{13}$.
Thus, $\frac{BC}{DE} = \frac{8}{13}$ and $\frac{AB}{AD} = \frac{8}{13}$.
Step 2: (ii) Area ratio: $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle ADE)} = \left(\frac{8}{13}\right)^2 = \frac{64}{169}$.
Step 3: (iii) Given $\text{Area}(\triangle ADE) = 507\text{ cm}^2$:
$$\text{Area}(\triangle ABC) = 507 \times \frac{64}{169} = 3 \times 64 = 192\text{ cm}^2$$
Step 4: (iv) $\text{Area}(BDEC) = \text{Area}(\triangle ADE) - \text{Area}(\triangle ABC) = 507 - 192 = 315\text{ cm}^2$.
Since $BC \parallel DE$ and non-parallel sides $BD, CE$ exist, quadrilateral $BDEC$ is a Trapezium (Trapezoid).
Final Answer: (i) 8/13, 8/13; (ii) 64/169; (iii) 192 cm²; (iv) Area = 315 cm², Type = Trapezium.
Exercise 9.4 • Step-by-Step Complete Solutions
Determine whether the solids are similar or not. Cuboid 1 (3x5) and Cuboid 2 (6x10).
Step 2: Since all corresponding linear dimensions are in constant ratio $2$, the cuboids are similar.
Final Answer: Similar.
Determine whether the solids are similar or not. Cylinder 1 (D=24, H=12) and Cylinder 2 (d=14, h=8).
Step 2: Height ratio: $\frac{12}{8} = 1.5$.
Step 3: Since $1.714 \neq 1.5$, the cylinders are not in proportion.
Final Answer: Not Similar.
Determine whether the solids are similar or not. Pyramids: Base 5x5, H=6, Slant=6.5; Base 10x10, H=12, Slant=13.
Step 2: Height ratio: $\frac{12}{6} = 2$.
Step 3: Slant height ratio: $\frac{13}{6.5} = 2$. All ratios equal $2$.
Final Answer: Similar.
Determine whether the solids are similar or not. Cones: Radius 9, Slant 15, H=12; Radius 21, Slant 29, H=20.
Step 2: Height ratio: $\frac{20}{12} = \frac{5}{3} \approx 1.667$.
Step 3: Ratios are unequal.
Final Answer: Not Similar.
Solids are similar. Find the values of unknowns. Also find the ratios of volume of solids. Cylinders: D=100m, H=25m; h=5m, d=?.
Step 2: Unknown diameter $d = \frac{D}{k} = \frac{100}{5} = 20\text{ m}$.
Step 3: Volume ratio $\frac{V_1}{V_2} = k^3 = 5^3 = 125:1$.
Final Answer: d = 20 m, Volume Ratio = 125 : 1.
Solids are similar. Find the values of unknowns. Also find the ratios of volume of solids. Triangular prisms: sides (5, 12, 13) with length 6; and sides (7.5, b, c) with length h.
Step 2: Unknowns: $b = 12 \times 1.5 = 18\text{ m}$, $c = 13 \times 1.5 = 19.5\text{ m}$, $h = 6 \times 1.5 = 9\text{ m}$.
Step 3: Volume ratio $\frac{V_2}{V_1} = k^3 = (1.5)^3 = \frac{27}{8} = 27:8$.
Final Answer: b = 18 m, c = 19.5 m, h = 9 m, Volume Ratio = 27 : 8.
Solids are similar. Find the unknown volume. Pyramids: base 21 mm with V1 = 9000 mm³; base 7 mm with V2 = ?.
Step 2: Substitute: $\frac{V_2}{9000} = \left(\frac{7}{21}\right)^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$.
Step 3: Solve: $V_2 = \frac{9000}{27} = 333.33\text{ mm}^3$.
Final Answer: V2 = 333.33 mm³.
Solids are similar. Find the unknown volume. Cylinders: height 10 ft with V1 = ?; height 12 ft with V2 = 360 ft³.
Step 2: Solve: $V_1 = 360 \times \frac{125}{216} = \frac{5 \times 125}{3} = 208.33\text{ ft}^3$.
Final Answer: V1 = 208.33 ft³.
Solids are similar. Find the unknown volume. Cones: height 6 cm with V1 = ?; height 15 cm with V2 = 460 cm³.
Step 2: Solve: $V_1 = 460 \times \frac{8}{125} = 29.44\text{ cm}^3$.
Final Answer: V1 = 29.44 cm³.
Solids are similar. Find the unknown volume. Cuboids: height 12 m with V1 = 288 m³; height 9 m with V2 = ?.
Step 2: Solve: $V_2 = 288 \times \frac{27}{64} = 4.5 \times 27 = 121.5\text{ m}^3$.
Final Answer: V2 = 121.5 m³.
Find the ratio of scale factors of the following pairs of similar solids: Cubes with volumes $V_1 = 512\text{ m}^3$ and $V_2 = 1728\text{ m}^3$.
$$k = \sqrt[3]{\frac{V_1}{V_2}} = \sqrt[3]{\frac{512}{1728}}$$
Step 2: Simplify cube roots: $\sqrt[3]{512} = 8$ and $\sqrt[3]{1728} = 12$.
Step 3: Reduce fraction: $\frac{8}{12} = \frac{2}{3} = 2:3$.
Final Answer: The ratio of scale factors is 2 : 3.
Two swimming pools are similar with a scale factor of $4 : 5$. The amount of chlorine mixture to be added is proportional to the volume of water in the pool. If three cups of chlorine mixture are needed for the smaller pool, how much of the chlorine mixture is needed for the larger pool?
Step 2: Since chlorine is proportional to volume, chlorine needed for larger pool $C_2$:
$$C_2 = 3 \times \frac{125}{64} = \frac{375}{64} \approx 5.86\text{ cups}$$
Final Answer: The larger pool requires 5.86 cups of chlorine mixture.
A model bus is built with a scale of $1 : 10$. The model bus has a volume of $30\text{ m}^3$ (or $0.03\text{ m}^3$). What is the volume of the actual bus?
Step 2: Volume scaling ratio is $k^3 = 10^3 = 1000$.
Step 3: Actual volume $V_{\text{actual}} = V_{\text{model}} \times 1000 = 30 \times 1000 = 30,000\text{ m}^3$.
Final Answer: The volume of the actual bus is 30,000 m³.
Solid A is similar to solid B with a scale factor of $2 : 3$. Find the surface area and volume of solid B if surface area and volume of solid A are $130\pi\text{ cm}^2$ and $280\pi\text{ cm}^3$ respectively.
Step 2: Surface area of B: $S_B = S_A \times k^2 = 130\pi \times \left(\frac{3}{2}\right)^2 = 130\pi \times \frac{9}{4} = 292.5\pi\text{ cm}^2 \approx 918.9\text{ cm}^2$.
Step 3: Volume of B: $V_B = V_A \times k^3 = 280\pi \times \left(\frac{3}{2}\right)^3 = 280\pi \times \frac{27}{8} = 35 \times 27\pi = 945\pi\text{ cm}^3 \approx 2968.8\text{ cm}^3$.
Final Answer: Surface Area of B = 292.5π cm²; Volume of B = 945π cm³.
Solid I is similar to Solid II. Find the scale factor of solid I to solid II if $V_1 = 8\pi\text{ ft}^3$ and $V_2 = 125\pi\text{ ft}^3$.
$$k = \sqrt[3]{\frac{V_1}{V_2}} = \sqrt[3]{\frac{8\pi}{125\pi}} = \sqrt[3]{\frac{8}{125}} = \frac{2}{5} = 2:5$$
Final Answer: The scale factor of solid I to solid II is 2 : 5.
Solid A and solid B are mathematically similar. The volume of solid A is $32\text{ cm}^3$. The volume of solid B is $108\text{ cm}^3$. The height of solid A is $10\text{ cm}$. Find the height of solid B.
Step 2: Calculate height of solid B: $h_B = h_A \times k = 10 \times 1.5 = 15\text{ cm}$.
Final Answer: The height of solid B is 15 cm.
P and Q are two similar solids. Solid P has surface area and volume $108\text{ cm}^2$ and $135\text{ cm}^3$ respectively. Find volume of solid Q if Q has surface area $300\text{ cm}^2$.
$$k = \sqrt{\frac{A_Q}{A_P}} = \sqrt{\frac{300}{108}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$$
Step 2: Calculate volume of solid Q using $V_Q = V_P \times k^3$:
$$V_Q = 135 \times \left(\frac{5}{3}\right)^3 = 135 \times \frac{125}{27} = 5 \times 125 = 625\text{ cm}^3$$
Final Answer: The volume of solid Q is 625 cm³.
X and Y are two similar cylinders such that $\text{base area of X} : \text{base area of Y} = 16 : 25$. Find: (i) ratio of heights of both cylinders. (ii) Ratio of areas of curved surface of cylinders. (iii) Ratio of volumes of cylinders.
Ratio of heights = $4 : 5$.
Step 2: (ii) Ratio of curved surface areas equals the ratio of areas: $k^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25} = 16 : 25$.
Step 3: (iii) Ratio of volumes: $k^3 = \left(\frac{4}{5}\right)^3 = \frac{64}{125} = 64 : 125$.
Final Answer: (i) 4:5, (ii) 16:25, (iii) 64:125.
The volume of one right circular cone is 8 times the other one. If the radius of larger cone is $12\text{ cm}$, find the radius of the smaller one.
Step 2: Radius of smaller cone $r = \frac{R}{k} = \frac{12}{2} = 6\text{ cm}$.
Final Answer: The radius of the smaller cone is 6 cm.
Masses of two similar objects are $8\text{ kg}$ and $27\text{ kg}$ respectively. If the height of first object is $2\text{ m}$, what is the height of second object?
Step 2: Find linear scale factor $k = \sqrt[3]{\frac{m_2}{m_1}} = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} = 1.5$.
Step 3: Calculate height of second object: $h_2 = h_1 \times k = 2 \times 1.5 = 3\text{ m}$.
Final Answer: The height of the second object is 3 m.
Exercise 9.5 • Step-by-Step Complete Solutions
Find the number of sides of a regular polygon if each exterior angle = 45°.
Step 2: Substitute: $n = \frac{360^\circ}{45^\circ} = 8$.
Final Answer: 8 sides (Octagon).
Find the number of sides of a regular polygon if each exterior angle = 60°.
Final Answer: 6 sides (Hexagon).
Find the number of sides of a regular polygon if each exterior angle = 120°.
Final Answer: 3 sides (Equilateral triangle).
Find the number of sides of a regular polygon if each exterior angle = 40°.
Final Answer: 9 sides (Nonagon).
Draw a regular pentagon whose exterior angles are $p, q, r, s, t$ and each interior angle is $k$. What is the measure of: (a) each interior angle (b) each exterior angle
$$k = \frac{(n-2) \times 180^\circ}{n} = \frac{(5-2) \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ$$
Step 2: (b) Each exterior angle:
$$p = q = r = s = t = \frac{360^\circ}{n} = \frac{360^\circ}{5} = 72^\circ$$
(Check: $108^\circ + 72^\circ = 180^\circ$).
Final Answer: (a) k = 108°, (b) p = q = r = s = t = 72°.
Each interior angle of a polygon is five times the exterior angle of the polygon. Find the number of sides.
Step 2: Since interior and exterior angles form a linear pair:
$$\theta_i + \theta_e = 180^\circ \implies 5\theta_e + \theta_e = 180^\circ \implies 6\theta_e = 180^\circ \implies \theta_e = 30^\circ$$
Step 3: Calculate number of sides $n = \frac{360^\circ}{\theta_e} = \frac{360^\circ}{30^\circ} = 12$.
Final Answer: The polygon has 12 sides (Dodecagon).
Find the minimum interior angles and maximum exterior angles possible in a regular polygon. Give reasons to support your answer.
Step 2: As $n$ increases, the interior angle $\theta_i = \frac{(n-2)180^\circ}{n}$ increases monotonically, so the minimum interior angle occurs at $n = 3$:
$$\theta_{i,\text{min}} = \frac{(3-2) \times 180^\circ}{3} = 60^\circ$$
Step 3: Since $\theta_e = 180^\circ - \theta_i$, the maximum exterior angle occurs when interior angle is minimum:
$$\theta_{e,\text{max}} = 180^\circ - 60^\circ = 120^\circ$$
Final Answer: Minimum interior angle is 60° and maximum exterior angle is 120°, occurring in an equilateral triangle.
Find the exterior angle of a regular polygon of 5 sides.
Final Answer: 72°.
Find the exterior angle of a regular polygon of 9 sides.
Final Answer: 40°.
Find the exterior angle of a regular polygon of 15 sides.
Final Answer: 24°.
Find the exterior angle of a regular polygon of 20 sides.
Final Answer: 18°.
The ratio between an exterior angle and the interior angle of a regular polygon is $1 : 2$. Find: (a) the measure of each exterior angle. (b) the measure of each interior angle. (c) the number of sides in the polygon.
$$x + 2x = 180^\circ \implies 3x = 180^\circ \implies x = 60^\circ$$
Step 2: (a) Each exterior angle $\theta_e = 60^\circ$.
Step 3: (b) Each interior angle $\theta_i = 2(60^\circ) = 120^\circ$.
Step 4: (c) Number of sides $n = \frac{360^\circ}{60^\circ} = 6$ (Regular Hexagon).
Final Answer: (a) 60°, (b) 120°, (c) 6 sides.
Is it possible to have a regular polygon each of whose exterior angle is $50^\circ$? Give reason to support your answer.
Step 2: Substitute $\theta_e = 50^\circ$:
$$n = \frac{360^\circ}{50^\circ} = 7.2$$
Step 3: A polygon must have an integer number of sides ($n \in \mathbb{N}, n \ge 3$). Since $7.2$ is not an integer, such a regular polygon cannot exist.
Final Answer: No, it is not possible because 360° is not evenly divisible by 50° (n = 7.2 is not an integer).
Name the polygon whose sum of interior angles is equal to the sum of its exterior angles.
Step 2: Sum of interior angles is $(n-2) \times 180^\circ$.
Step 3: Equate both sums:
$$(n - 2) \times 180^\circ = 360^\circ \implies n - 2 = \frac{360}{180} = 2 \implies n = 4$$
Final Answer: The polygon is a Quadrilateral (n = 4).
The sum of all the interior angles of a regular polygon is four times the sum of its exterior angles. Identify the polygon.
Step 2: Sum of interior angles $= 4 \times 360^\circ = 1440^\circ$.
Step 3: Use interior sum formula: $(n - 2) \times 180^\circ = 1440^\circ$.
$$n - 2 = \frac{1440}{180} = 8 \implies n = 10$$
Final Answer: The polygon is a Decagon (10 sides).
An exterior angle of a regular polygon is $12^\circ$. What is the sum of all the interior angles?
Step 2: Calculate sum of interior angles:
$$S = (n - 2) \times 180^\circ = (30 - 2) \times 180^\circ = 28 \times 180^\circ = 5040^\circ$$
Final Answer: The sum of all interior angles is 5040°.
Prove that each interior angle and its corresponding exterior angle in any polygon are supplementary.
Step 2: The interior angle and the exterior angle adjacent to it lie on this straight line, forming a linear pair of angles.
Step 3: By the Linear Pair Axiom, the sum of angles forming a linear pair is $180^\circ$:
$$\theta_{\text{interior}} + \theta_{\text{exterior}} = 180^\circ$$
Final Answer: Since their sum is 180°, they are supplementary.
Find the number of sides in a regular polygon when the measure of each exterior angle is $72^\circ$.
Step 2: Substitute $\theta_e = 72^\circ$:
$$n = \frac{360^\circ}{72^\circ} = 5$$
Final Answer: The polygon has 5 sides (Regular Pentagon).
The exterior angles of a pentagon are $(y + 5)^\circ, (2y + 3)^\circ, (3y + 2)^\circ, (4y + 1)^\circ$ and $(5y + 4)^\circ$ respectively. Find the measure of each angle.
Step 2: Add the five given algebraic expressions:
$$(y + 5) + (2y + 3) + (3y + 2) + (4y + 1) + (5y + 4) = 360$$
$$(1 + 2 + 3 + 4 + 5)y + (5 + 3 + 2 + 1 + 4) = 360$$
$$15y + 15 = 360 \implies 15y = 345 \implies y = 23^\circ$$
Step 3: Calculate each individual angle:
- 1st Angle: $23 + 5 = 28^\circ$
- 2nd Angle: $2(23) + 3 = 46 + 3 = 49^\circ$
- 3rd Angle: $3(23) + 2 = 69 + 2 = 71^\circ$
- 4th Angle: $4(23) + 1 = 92 + 1 = 93^\circ$
- 5th Angle: $5(23) + 4 = 115 + 4 = 119^\circ$
(Check: $28 + 49 + 71 + 93 + 119 = 360^\circ$).
Final Answer: y = 23°. The angles are 28°, 49°, 71°, 93°, and 119°.
A convex polygon has 14 diagonals. Find the number of sides of the polygon.
Step 2: Set equal to 14:
$$\frac{n(n - 3)}{2} = 14 \implies n(n - 3) = 28$$
$$n^2 - 3n - 28 = 0$$
Step 3: Factorize quadratic: $(n - 7)(n + 4) = 0$.
Since $n > 0$, $n = 7$.
Final Answer: The polygon has 7 sides (Heptagon).
Find the sum of all the interior angles of a polygon having 13 sides.
Step 2: Substitute $n = 13$:
$$S = (13 - 2) \times 180^\circ = 11 \times 180^\circ = 1980^\circ$$
Final Answer: The sum of all interior angles is 1980°.
The sum of all the interior angles of a polygon is $2880^\circ$. How many sides does the polygon have?
Step 2: Solve for $n - 2$:
$$n - 2 = \frac{2880}{180} = 16$$
Step 3: $n = 16 + 2 = 18$.
Final Answer: The polygon has 18 sides.
Exercise 9.6 • Step-by-Step Complete Solutions
A lawn is in the shape of equilateral triangle. Find the perimeter of lawn if length of one side is $5\text{ m}$. Also find the cost of boundary wall of lawn @ Rs. 220 per metre.
Step 2: Calculate total cost: $\text{Cost} = 15\text{ m} \times \text{Rs. } 220 = \text{Rs. } 3300$.
Final Answer: Perimeter = 15 m, Total Cost = Rs. 3300.
Cricket ground in a village is in the shape of parallelogram. One side of ground is $65\text{ m}$ long and distance between parallel sides having length $65\text{ m}$ is $42\text{ m}$. Find the cost of planting grass @ Rs. 10 per square metre.
Step 2: Calculate total cost: $\text{Cost} = 2730\text{ m}^2 \times \text{Rs. } 10/\text{m}^2 = \text{Rs. } 27,300$.
Final Answer: Area = 2730 m², Cost of planting grass = Rs. 27,300.
Base of a minaret of a Masjid is built in the shape of regular pentagon as shown in the figure (side $6\text{ m}$, slant to vertex $5\text{ m}$, half-side $3\text{ m}$, apothem $a = \sqrt{5^2 - 3^2} = 4\text{ m}$). Find the perimeter and area of base of minaret.
Step 2: Find apothem $a$: In right triangle with hypotenuse $5\text{ m}$ and base $3\text{ m}$:
$$a = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ m}$$
Step 3: Calculate area: $A = \frac{1}{2} P a = \frac{1}{2} \times 30 \times 4 = 60\text{ m}^2$.
Final Answer: Perimeter = 30 m, Area = 60 m².
A plot in the shopping area is in the shape of regular hexagon. One side of the plot is $10\text{ m}$ long. Find: (i) the cost of fencing the plot @ Rs. 160 per metre. (ii) area of the plot. (iii) the cost of filling the plot @ Rs. 500 per $\text{m}^2$.
$$\text{Cost of Fencing} = 60 \times 160 = \text{Rs. } 9600$$
Step 2: (ii) A regular hexagon consists of 6 equilateral triangles of side $s = 10\text{ m}$:
$$\text{Area} = 6 \times \left(\frac{\sqrt{3}}{4} s^2\right) = 6 \times \frac{\sqrt{3}}{4} \times 100 = 150\sqrt{3} \approx 259.81\text{ m}^2$$
Step 3: (iii) Cost of filling plot:
$$\text{Cost of Filling} = 259.81 \times 500 = \text{Rs. } 129,905$$
Final Answer: (i) Rs. 9600, (ii) 259.81 m², (iii) Rs. 129,905.
A room is in the shape of square having perimeter of 48 feet. Find: (i) the cost of carpeting the floor @ Rs. 350 per $\text{ft}^2$. (ii) the inner area of each wall if room is 10 feet high. (iii) the cost of painting inner sides of the 4 walls @ Rs. 100 per $\text{ft}^2$.
Step 2: (i) Floor area $= 12 \times 12 = 144\text{ ft}^2$.
$$\text{Cost of Carpeting} = 144 \times 350 = \text{Rs. } 50,400$$
Step 3: (ii) Area of one wall $= \text{Length} \times \text{Height} = 12 \times 10 = 120\text{ ft}^2$.
Step 4: (iii) Total area of 4 walls $= 4 \times 120 = 480\text{ ft}^2$.
$$\text{Cost of Painting} = 480 \times 100 = \text{Rs. } 48,000$$
Final Answer: (i) Rs. 50,400, (ii) 120 ft², (iii) Rs. 48,000.
A tile is in the shape of regular hexagon. Each side of the tile is one foot long. Find the perimeter of four tiles joined together as shown in the figure (forming a 4-hexagon cluster sharing 5 interior edges).
Step 2: Since each outer side is 1 foot long:
$$\text{Perimeter} = 14 \times 1\text{ ft} = 14\text{ ft}$$
Final Answer: The perimeter of the joined tiles is 14 feet.
Exercise 9.7 • Step-by-Step Complete Solutions
Draw $AB = 6\text{ cm}$. Bisect $AB$ at $O$ and draw a locus of point $P$ equidistant from $O$ and above $AB$. What could be the name of locus?
Step 2: The locus of all points at a constant distance $r = OP$ from a fixed point $O$ lying in the half-plane above $AB$ forms a semicircular arc.
Final Answer: The locus is a semicircle with centre O.
Draw coordinate axes. Take two points $A$ and $B$ on $x$-axis and $y$-axis respectively at a distance of $4.5\text{ cm}$ each. Draw a locus of points from $A$ to $B$ equidistant from origin. What is specific name of this locus? How many such loci can be drawn around the origin?
Step 2: By choosing different radial distances $r > 0$, infinitely many concentric circular loci can be drawn around the origin.
Final Answer: It is a circular arc of radius 4.5 cm; Infinitely many such concentric circular loci can be drawn.
Draw a horizontal line $l$. (i) Take a point $T$ above $l$ at a distance of $3\text{ cm}$ and draw a locus of points through $T$ parallel to $l$. (ii) Now take a point $Q$ below $l$ at a distance of $3.2\text{ cm}$ and draw a locus of points through $Q$ parallel to $l$. (iii) What is distance between both loci?
Step 2: (ii) The locus of points at fixed distance $3.2\text{ cm}$ below $l$ is a straight line $l_2 \parallel l$.
Step 3: (iii) Total perpendicular separation between loci $l_1$ and $l_2 = 3\text{ cm} + 3.2\text{ cm} = 6.2\text{ cm}$.
Final Answer: (i) Line parallel to l, (ii) Line parallel to l, (iii) 6.2 cm.
Diagram shows a circle with centre $P$. $X$ and $Y$ are two points on the circumference of the circle. (i) Draw locus of points which are equidistant from $X$ and $Y$ through $P$. (ii) Take another point $Z$ on the locus outside the circle and draw another circle of radius $PZ$. (iii) What is the relation of this circle with given circle?
Step 2: (iii) Since both circles have the same center $P$, they are concentric circles.
Final Answer: (i) Perpendicular bisector of chord XY, (iii) Concentric circles.
Draw a line $AB$. Let $P$ and $Q$ be two points not on $AB$ but coplanar with $AB$. Draw the locus of points from $P$ to $Q$. (i) What is the name of that locus? (ii) Is there any point on the locus $PQ$ if extended which lies on line $AB$? (iii) How can we take points $P$ and $Q$ such that no point of the above locus lies in line $AB$? (iv) How can we take points $P$ and $Q$ such that every point of line $AB$ may lie on locus?
Step 2: (ii) Yes, if lines $PQ$ and $AB$ are non-parallel coplanar lines, extending line $PQ$ will intersect line $AB$ at a unique point.
Step 3: (iii) To ensure no intersection, choose points $P$ and $Q$ such that line $PQ$ is parallel to line $AB$ ($PQ \parallel AB$).
Step 4: (iv) To ensure all points lie on the locus, choose $P$ and $Q$ to lie directly on the line $AB$ itself (coincident lines).
Final Answer: (i) Line segment PQ, (ii) Yes (intersection point), (iii) Choose PQ || AB, (iv) Choose P and Q on line AB.
Draw an equilateral triangle $PQR$ of suitable measurement. (i) Draw right bisectors of any two sides and locate a point $A$ where both bisectors meet. (ii) Draw angle bisectors of any two vertices and locate a point $B$ where both bisectors meet. (iii) What is the relation between locus of $A$ and $B$?
Step 2: (ii) The intersection of angle bisectors of vertices is the incentre $B$.
Step 3: (iii) In an equilateral triangle, the circumcentre, incentre, centroid, and orthocentre all coincide at the exact same single point ($A \equiv B$).
Final Answer: (i) Circumcentre A, (ii) Incentre B, (iii) Points A and B coincide.
Figure shows an isosceles triangle $ABC$ with $AB = AC$. Prove that the locus of bisector of angle $A$ is right bisector of side $BC$.
Step 2: Compare $\triangle ABD$ and $\triangle ACD$:
- $AB = AC$ (Given, isosceles triangle)
- $\angle BAD = \angle CAD$ ($AD$ is angle bisector)
- $AD = AD$ (Common side)
By $SAS$ Congruence Postulate, $\triangle ABD \cong \triangle ACD$.
Step 3: From congruence:
- $BD = CD$ ($D$ is midpoint of $BC$)
- $\angle ADB = \angle ADC$. Since $\angle ADB + \angle ADC = 180^\circ$, $\angle ADB = 90^\circ$.
Thus, $AD$ is perpendicular to $BC$ and bisects it, meaning $AD$ is the right bisector of side $BC$.
Final Answer: The bisector of vertex angle A is the perpendicular bisector of base BC.
Draw three non-collinear points in the plane. Find the locus of the points which are equidistant form these three points. How many such points exist?
Step 2: The locus of points equidistant from $A$ and $B$ is the perpendicular bisector of $AB$.
Step 3: The locus of points equidistant from $B$ and $C$ is the perpendicular bisector of $BC$.
Step 4: These two lines intersect at exactly one unique point $O$ (the circumcentre), which is equidistant from all three points ($OA = OB = OC$).
Final Answer: The locus is the circumcentre; exactly 1 unique point exists.
Take two lines $AB$ and $CD$ inclined at $60^\circ$ intersecting at $O$. (i) Draw a locus of points which are equidistant from both lines. (ii) Draw bisector of $60^\circ$. (iii) What is relation between locus of points equidistant from lines and angle bisector? (iv) Draw bisector of adjacent angle at $O$. Find the relation between both angle bisectors.
Step 2: (ii) The bisector of the $60^\circ$ angle divides it into two equal angles of $30^\circ$.
Step 3: (iv) The adjacent supplementary angle is $180^\circ - 60^\circ = 120^\circ$. Its bisector creates angles of $60^\circ$.
The angle between the two bisectors is:
$$\theta = 30^\circ + 60^\circ = 90^\circ$$
Thus, the internal and external angle bisectors are perpendicular to each other.
Final Answer: (i) Pair of angle bisectors; (iii) Locus is the angle bisector; (iv) The angle bisectors are perpendicular (90°).
Review Exercise 9 • Step-by-Step Complete Solutions
Which of the following is polygon?
A polygon is a 2D closed figure made of straight line segments. A quadrilateral is a 4-sided polygon.
Final Answer: Option (C)
Which of the following is regular polygon?
A regular polygon must be both equilateral and equiangular. A square has all 4 sides equal and all 4 angles equal to 90°.
Final Answer: Option (D)
If two triangles are similar, their corresponding sides are.
By definition of similarity, corresponding sides are in a constant ratio (proportional).
Final Answer: Option (A)
What is the sum of interior angles for an irregular hexagon?
Sum of interior angles of ANY hexagon (regular or irregular) is (6 - 2) * 180° = 4 * 180° = 720°.
Final Answer: Option (B)
What is the sum of interior angles for a regular 12 sided polygon?
S = (12 - 2) * 180° = 10 * 180° = 1800°.
Final Answer: Option (A)
How many sides a regular polygon has if its exterior angle is 15°?
n = 360° / 15° = 24 sides.
Final Answer: Option (C)
In the figure, AB = 21 cm and P divides AB in the ratio 3 : 4. What is length of AP?
AP = 21 * (3 / (3 + 4)) = 21 * (3/7) = 9 cm.
Final Answer: Option (D)
In the figure, a/b = c/d. Which one is true?
By Converse of Basic Proportionality Theorem, if AD/DB = AE/EC (a/b = c/d), then DE || BC.
Final Answer: Option (D)
In the figure if x = y. Then the value of b is:
By angle bisector theorem: a/c = b/d => b = ad/c.
Final Answer: Option (B)
In the figure, triangle ABC is equilateral and DE || BC, then triangle ADE is
Since ABC is equilateral (angles 60°) and DE || BC, triangle ADE has all angles 60°, so it is equilateral.
Final Answer: Option (D)
In the figure, GH || EF. Then a : b = ?
By similar triangles DGH ~ DEF, a/b = GH/EF = DG/DE.
Final Answer: Option (A)
What is the sum of all the exterior angles of a 13-sided polygon whose one interior angle is equal to x°?
The sum of exterior angles of ANY convex polygon is always identically 360°.
Final Answer: Option (B)
Which polygon has both its interior and exterior angles the same?
In a regular square (4-gon), interior angle = (4-2)*180/4 = 90° and exterior angle = 360/4 = 90°.
Final Answer: Option (C)
The formation or expression of an opinion or theory without sufficient evidence for proof is known as:
A conjecture is an unproven proposition based on observation without complete proof.
Final Answer: Option (B)
A mathematical statement that is proved true based on already accepted statements is called:
A theorem is a statement established by deductive proof.
Final Answer: Option (D)
A mathematical statement that is assumed to be true without proof is called:
An axiom is an accepted self-evident mathematical fact without proof.
Final Answer: Option (A)
Two solids with equal ratios of corresponding linear measures:
Solids with equal ratios of corresponding linear dimensions are mathematically similar.
Final Answer: Option (A)
Find the exterior angle of a polygon with 6 sides.
Step 2: Substitute $n = 6$: $\theta_e = \frac{360^\circ}{6} = 60^\circ$.
Final Answer: The exterior angle is 60°.
Is it possible to have a polygon in which sum of interior angles is 9 right angles?
Step 2: Set equal to $(n-2) \times 180^\circ$:
$$(n - 2) \times 180^\circ = 810^\circ \implies n - 2 = \frac{810}{180} = 4.5 \implies n = 6.5$$
Step 3: Since $n$ must be a positive integer $\ge 3$, such a polygon cannot exist.
Final Answer: No, it is not possible (n = 6.5 is not an integer).
Is it possible to have a polygon whose sum of interior angles is $7200^\circ$?
Step 2: Solve for $n - 2$:
$$n - 2 = \frac{7200}{180} = 40 \implies n = 42$$
Step 3: Since 42 is a positive integer, it is fully possible.
Final Answer: Yes, it is possible for a 42-sided polygon.
Find the measure of each angle of a regular nonagon.
Step 2: Use formula $\theta_i = \frac{(n-2) \times 180^\circ}{n}$:
$$\theta_i = \frac{(9 - 2) \times 180^\circ}{9} = \frac{7 \times 180^\circ}{9} = 7 \times 20^\circ = 140^\circ$$
Final Answer: Each angle of a regular nonagon is 140°.
In the figure, $XY$ is a concave mirror, $OA$ is object and $IB$ is its image. (i) Show that $\triangle OAP \sim \triangle IBP$. (ii) Find height of object if $IB = 2\text{ cm}, OP = 10\text{ cm}, IP = 4\text{ cm}$.
- $\angle AOP = \angle BIP = 90^\circ$ (Both object and image stand perpendicular to principal axis).
- $\angle APO = \angle BPI$ (Law of reflection: angle of incidence equals angle of reflection).
By $AA$ Similarity Criterion, $\triangle OAP \sim \triangle IBP$.
Step 2: (ii) From similarity, ratios of corresponding sides are equal:
$$\frac{OA}{IB} = \frac{OP}{IP} \implies \frac{OA}{2} = \frac{10}{4} = 2.5$$
$$OA = 2 \times 2.5 = 5\text{ cm}$$
Final Answer: (i) Proven by AA similarity; (ii) Height of object OA = 5 cm.
In the figure, $AB = 6\text{ cm}, BD = 9\text{ cm}$. Find the diameter of smaller circle if diameter of bigger one is $80\text{ cm}$.
Step 2: Total length $AD = AB + BD = 6 + 9 = 15\text{ cm}$.
Step 3: The lines from vertex $A$ tangent to circles form similar right triangles with the radii:
$$\frac{r}{R} = \frac{AB}{AD} \implies \frac{r}{40} = \frac{6}{15} = \frac{2}{5}$$
$$r = 40 \times \frac{2}{5} = 16\text{ cm}$$
Step 4: Diameter of smaller circle $d = 2r = 2 \times 16 = 32\text{ cm}$.
Final Answer: The diameter of the smaller circle is 32 cm.
Prove that if vertex angles of two isosceles triangles are equal then the two triangles are similar.
Step 2: In an isosceles triangle, base angles are equal:
$$\angle B = \angle C = \frac{180^\circ - \theta}{2}$$
$$\angle E = \angle F = \frac{180^\circ - \theta}{2}$$
Step 3: Thus $\angle A = \angle D$, $\angle B = \angle E$, and $\angle C = \angle F$.
Since all corresponding angles are equal, $\triangle ABC \sim \triangle DEF$ by $AAA$ similarity criterion.
Final Answer: Proven by AAA similarity.
In the figure, $BC \parallel DE$. $AB = 8\text{ cm}, BD = 4\text{ cm}, BC = x + 2, DE = 2x$. Find the length of $\overline{BC}$ and $\overline{DE}$.
Step 2: By similar triangles $\triangle ABC \sim \triangle ADE$:
$$\frac{BC}{DE} = \frac{AB}{AD} \implies \frac{x + 2}{2x} = \frac{8}{12} = \frac{2}{3}$$
Step 3: Cross-multiply and solve for $x$:
$$3(x + 2) = 2(2x) \implies 3x + 6 = 4x \implies x = 6$$
Step 4: Calculate lengths:
$$BC = x + 2 = 6 + 2 = 8\text{ cm}$$
$$DE = 2x = 2(6) = 12\text{ cm}$$
Final Answer: BC = 8 cm, DE = 12 cm.
Triangles $SQT$ and $PQR$ are similar. $SQ = 3\text{ cm}, SP = 6\text{ cm}$. Find the ratio of area of triangle $SQT$ to that of triangle $PQR$.
Step 2: Linear scale factor $k = \frac{SQ}{PQ} = \frac{3}{9} = \frac{1}{3}$.
Step 3: Ratio of areas:
$$\frac{\text{Area}(\triangle SQT)}{\text{Area}(\triangle PQR)} = k^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} = 1:9$$
Final Answer: The ratio of areas is 1 : 9.
The following pyramids are similar and larger pyramid has a surface area of $392\text{ cm}^2$ with base side $14\text{ cm}$, while smaller pyramid has base side $10\text{ cm}$. What is the surface area of smaller pyramid?
Step 2: Apply area scaling formula:
$$A_{\text{small}} = A_{\text{large}} \times k^2 = 392 \times \left(\frac{5}{7}\right)^2 = 392 \times \frac{25}{49}$$
Step 3: Solve: $A_{\text{small}} = 8 \times 25 = 200\text{ cm}^2$.
Final Answer: The surface area of the smaller pyramid is 200 cm².
The two cylinders are similar with radii $R = 3\text{ ft}$ and $r = 2\text{ ft}$. What is the volume of the larger cylinder if the volume of smaller cylinder is $40\text{ ft}^3$?
Step 2: Volume scaling formula:
$$V_{\text{large}} = V_{\text{small}} \times k^3 = 40 \times \left(\frac{3}{2}\right)^3 = 40 \times \frac{27}{8} = 5 \times 27 = 135\text{ ft}^3$$
Final Answer: The volume of the larger cylinder is 135 ft³.
The following pairs of solids are similar. Find the surface area of red solid. Similar cuboids: length 4 m with Surface area = 336 m²; and length 6 m.
Step 2: $A_2 = 336 \times \left(\frac{3}{2}\right)^2 = 336 \times \frac{9}{4} = 84 \times 9 = 756\text{ m}^2$.
Final Answer: Surface area = 756 m².
The following pairs of solids are similar. Find the surface area of red solid. Similar cones: slant 20 in with Surface area = 1800 in²; and slant 15 in.
Step 2: $A_2 = 1800 \times \left(\frac{3}{4}\right)^2 = 1800 \times \frac{9}{16} = 112.5 \times 9 = 1012.5\text{ in}^2$.
Final Answer: Surface area = 1012.5 in².
Extra Exercise • Step-by-Step Complete Solutions
: A statement that can be deduced directly from a previously proven theorem without a separate lengthy proof is called a:
Final Answer: Option (C)
: If the linear dimensions of two similar polygons are in the ratio $3 : 7$, then the ratio of their areas is:
Final Answer: Option (C)
: If two similar solid cylinders have volumes in the ratio $1 : 64$, what is the ratio of their heights?
Final Answer: Option (A)
: How many diagonals does a regular octagon (8 sides) have?
Final Answer: Option (B)
: What is the sum of exterior angles of a 50-sided convex polygon?
Final Answer: Option (B)
: The apothem of a regular polygon is identical to the:
Final Answer: Option (B)
: The locus of points equidistant from two intersecting lines consists of:
Final Answer: Option (C)
: State True or False: (i) All regular hexagons are similar. (ii) The converse of a true theorem is always true.
Step 2: (ii) The converse of a true theorem is not necessarily true (e.g., 'If a shape is a square, it has 4 right angles' is true, but 'If a shape has 4 right angles, it is a square' is false, as it could be a rectangle) (False).
Final Answer: (i) True, (ii) False.
: Match the geometric term in Column A with its exact formula / property in Column B: Column A: 1. Sum of interior angles of n-gon 2. Each exterior angle of regular n-gon 3. Area of regular polygon 4. Number of diagonals in n-gon 5. Volume ratio of similar solids Column B: A. 360° / n B. (n - 2) * 180° C. (l1 / l2)³ D. (1/2) * P * a E. n(n - 3) / 2
Step 2: 2. Each exterior angle $= \frac{360^\circ}{n}$ (A)
Step 3: 3. Area of regular polygon $= \frac{1}{2} P a$ (D)
Step 4: 4. Number of diagonals $= \frac{n(n-3)}{2}$ (E)
Step 5: 5. Volume ratio $= \left(\frac{l_1}{l_2}\right)^3$ (C)
Final Answer: 1-B, 2-A, 3-D, 4-E, 5-C.
More Chapter Notes for Class 9 (FBISE)
MathematicsTest Your Knowledge on Chapter 9: Mastery Guide: Geometry and Polygons - Demonstrative Geometry, Similarity of Figures, Regular Polygons & Scaling
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