Focus Keyword: NCERT Solutions Class 9 Math Chapter 6 Measuring Space Perimeter and Area
Secondary Keywords & LSI: Class 9 Maths Chapter 6 solutions, Heron’s formula Class 9 questions and answers, measuring space perimeter and area Class 9, area of triangles and quadrilaterals Class 9, CBSE Class 9 Maths exam preparation
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H1 Title: NCERT Solutions for Class 9 Math Chapter 6: Measuring Space: Perimeter and Area (Complete Guide)
Navigating through the CBSE Class 9 Mathematics curriculum requires an in-depth understanding of two-dimensional boundary metrics, spatial enclosure concepts, triangular decompositions, and Heron’s formula. Chapter 6 of Class 9 Mathematics, “Measuring Space: Perimeter and Area”, forms the foundation of architectural land surveying, civil structural design, geospatial mapping, and material optimization. It investigates the quantitative distinction between one-dimensional boundary perimeter and two-dimensional enclosed area; explores the derivation and algebraic mechanics of Heron’s semi-perimeter formula for scalene, isosceles, and equilateral triangles; and details the calculation of composite polygonal regions such as trapeziums, rhombuses, and irregular quadrilaterals. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from standard triangular field evaluations and cost estimations to composite park layouts and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic four-step mathematical workflows (Given Data $\rightarrow$ Formula Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Answer with Units), and clear inline geometric diagrams have been highlighted to ensure students secure maximum marks in their CBSE examinations.
Chapter 6: Measuring Space: Perimeter and Area
Master Chapter Summary & Formula Blueprint
In the rationalised curriculum, this chapter equips students with analytical tools to determine the area of any triangular or polygonal region without requiring the vertical altitude to be known in advance.
| Geometric Entity | Core Mathematical Definition | Standard Governing Formula | Key Operational Constraint | CBSE Marks Weightage |
|---|---|---|---|---|
| Perimeter ($P$) | Total linear boundary length of a 2D closed figure | $P = a + b + c + \dots$ | Linear units ($\text{cm}, \text{m}, \text{km}$) | 1 Mark |
| Right-Angled Triangle | Triangle with one perpendicular ($90^\circ$) corner | $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ | Base and height must be perpendicular | 1 to 2 Marks |
| Equilateral Triangle | Triangle with all three sides equal ($a = b = c$) | $\text{Area} = \frac{\sqrt{3}}{4}a^2$ | Derived directly from Heron’s formula | 2 to 3 Marks |
| Semi-Perimeter ($s$) | Half of the total perimeter of a triangle | $s = \frac{a + b + c}{2}$ | Crucial intermediate step for Heron’s formula | 1 Mark (Step Mark) |
| Heron’s Formula | General formula for area of any triangle | $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ | Requires all three side lengths ($a, b, c$) | 3 to 4 Marks |
| Rhombus Area | Quadrilateral with four equal sides | $\text{Area} = \frac{1}{2} \times d_1 \times d_2 = 2 \times \text{Area}(\triangle)$ | Diagonals bisect at right angles | 3 to 4 Marks |
| Trapezium Area | Quadrilateral with one pair of parallel sides | $\text{Area} = \frac{1}{2}(a + b) \times h$ | Height $h$ found via triangular decomposition | 4 to 5 Marks |
🧠 Examiner’s Secret: In CBSE evaluations, students often lose a half-mark for omitting the calculation of the semi-perimeter $s$. Never write the final square root expression directly; always state $s = \frac{a+b+c}{2}$ clearly, compute the individual factors $(s-a)$, $(s-b)$, and $(s-c)$ separately, and then substitute them into $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
Foundational Geometric Concepts and Derivations
Perimeter vs. Area
Perimeter is the total one-dimensional continuous boundary length enclosing a closed coplanar figure, whereas area is the total two-dimensional surface space bounded within that perimeter.
Perimeter is measured in linear units such as metres ($\text{m}$) or centimetres ($\text{cm}$). Area represents the number of unit squares that fit within the boundary and is measured in square units such as square metres ($\text{m}^2$) or square centimetres ($\text{cm}^2$).
Heron’s Formula
Heron’s formula states that the area of a triangle whose sides have lengths $a$, $b$, and $c$ is given by the square root of the continued product of its semi-perimeter and the differences between the semi-perimeter and each side. A B C Side c Side b Side a h
Formulated by Heron of Alexandria in approximately 60 CE:
$$\text{Semi-Perimeter: } s = \frac{a + b + c}{2}$$
$$\text{Area: } \Delta = \sqrt{s(s – a)(s – b)(s – c)}$$
This formula is especially useful when the vertical altitude ($h$) of a triangle is difficult or impossible to measure directly, requiring only the three side lengths.
Decomposition Method for Trapeziums and Quadrilaterals
To determine the area of a trapezium whose four sides are known:
- Divide the trapezium into a parallelogram and a triangle by drawing a line from one upper vertex parallel to the opposite non-parallel leg.
- The sides of the newly formed triangle are completely known.
- Compute the area of this triangle using Heron’s formula.
- Equate this area to $\frac{1}{2} \times \text{base} \times h$ to solve for the vertical height $h$.
- Use this height $h$ to calculate the area of the parallelogram ($\text{base} \times h$) or apply the trapezium formula $\frac{1}{2}(\text{sum of parallel sides}) \times h$.
💡 Did You Know?: Heron’s formula can also be expressed without the semi-perimeter:
$$\Delta = \frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)} = \frac{1}{4}\sqrt{4a^2b^2 – (a^2 + b^2 – c^2)^2}$$
This form connects Heron’s geometric formula directly to the Law of Cosines.
[👉 Also Read: Class 9 Math Chapter 5 I’m Up and Down and Round and Round NCERT Solutions]
Step-by-Step Solutions: Core Textbook Exercises and Applied Problems
Question 1. [Traffic Signal Board & Equilateral Triangle]
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘$a$’. Find the area of the signal board, using Heron’s formula. If its perimeter is $180\text{ cm}$, what will be the area of the signal board?
Answer:
Part I: Derivation of Equilateral Triangle Formula Using Heron’s Formula
- Step 1: Identify Given Data
Sides of the equilateral triangle: $a = b = c$. - Step 2: Calculate Semi-Perimeter ($s$)
$$s = \frac{a + a + a}{2} = \frac{3a}{2}$$ - Step 3: Calculate Differences
$$s – a = \frac{3a}{2} – a = \frac{a}{2}$$
$$s – b = \frac{3a}{2} – a = \frac{a}{2}$$
$$s – c = \frac{3a}{2} – a = \frac{a}{2}$$ - Step 4: Substitute into Heron’s Formula
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)}$$
$$\Delta = \sqrt{\left(\frac{3a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)} = \sqrt{\frac{3a^4}{16}} = \frac{\sqrt{3}}{4}a^2$$
Part II: Numerical Calculation for Perimeter = $180\text{ cm}$
- Step 1: Find Side Length $a$
$$\text{Perimeter} = 3a = 180\text{ cm} \implies a = \frac{180}{3} = 60\text{ cm}$$ - Step 2: Compute Area
$$\Delta = \frac{\sqrt{3}}{4}(60)^2 = \frac{\sqrt{3}}{4} \times 3600 = 900\sqrt{3}\text{ cm}^2$$
If evaluating with $\sqrt{3} \approx 1.732$:
$$\Delta = 900 \times 1.732 = 1558.8\text{ cm}^2$$
Final Answer:
The general formula is $\frac{\sqrt{3}}{4}a^2$, and the area of the board is $900\sqrt{3}\text{ cm}^2$ (or approximately $1558.8\text{ cm}^2$).
Question 2. [Commercial Advertisement on Flyover Wall]
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122\text{ m}$, $22\text{ m}$ and $120\text{ m}$ (see figure). The advertisements yield an earning of ₹$5000\text{ per m}^2\text{ per year}$. A company hired one of its walls for $3\text{ months}$. How much rent did it pay?
Answer:
Step 1: Identify Given Dimensions
- Sides of triangular wall: $a = 122\text{ m}$, $b = 22\text{ m}$, $c = 120\text{ m}$.
- Rate of rent = ₹$5000\text{ per m}^2\text{ per year}$.
- Duration = $3\text{ months} = \frac{3}{12}\text{ year} = \frac{1}{4}\text{ year}$.
Step 2: Calculate Semi-Perimeter ($s$)
$$s = \frac{a + b + c}{2} = \frac{122 + 22 + 120}{2} = \frac{264}{2} = 132\text{ m}$$
Step 3: Calculate Differences
$$s – a = 132 – 122 = 10\text{ m}$$
$$s – b = 132 – 22 = 110\text{ m}$$
$$s – c = 132 – 120 = 12\text{ m}$$
Step 4: Compute Area Using Heron’s Formula
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)}$$
$$\Delta = \sqrt{132 \times 10 \times 110 \times 12}$$
Factorise into prime/composite factors for clean square-root extraction:
- $132 = 12 \times 11$
- $110 = 11 \times 10$
$$\Delta = \sqrt{(12 \times 11) \times 10 \times (11 \times 10) \times 12}$$
$$\Delta = \sqrt{12^2 \times 11^2 \times 10^2} = 12 \times 11 \times 10 = 1320\text{ m}^2$$
(Verification: $120^2 + 22^2 = 14400 + 484 = 14884 = 122^2$. The wall forms a right-angled triangle: $\frac{1}{2} \times 120 \times 22 = 1320\text{ m}^2$).
Step 5: Compute Total Rent Paid
$$\text{Rent} = \text{Area} \times \text{Annual Rate} \times \text{Time in Years}$$
$$\text{Rent} = 1320 \times 5000 \times \frac{3}{12} = 1320 \times 1250 = ₹16,50,000$$
Final Answer:
The company paid a total rent of ₹$16,50,000$.
Question 3. [Park Slide Wall Painting]
There is a slide in a park. One of its side walls has been painted in some colour with a message ‘KEEP THE PARK GREEN AND CLEAN’. If the sides of the wall are $15\text{ m}$, $11\text{ m}$ and $6\text{ m}$, find the area painted in colour.
Answer:
Step 1: Identify Given Data
Sides of the triangular wall: $a = 15\text{ m}$, $b = 11\text{ m}$, $c = 6\text{ m}$.
Step 2: Calculate Semi-Perimeter ($s$)
$$s = \frac{a + b + c}{2} = \frac{15 + 11 + 6}{2} = \frac{32}{2} = 16\text{ m}$$
Step 3: Calculate Differences
$$s – a = 16 – 15 = 1\text{ m}$$
$$s – b = 16 – 11 = 5\text{ m}$$
$$s – c = 16 – 6 = 10\text{ m}$$
Step 4: Compute Area Using Heron’s Formula
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)}$$
$$\Delta = \sqrt{16 \times 1 \times 5 \times 10}$$
$$\Delta = \sqrt{16 \times 50} = \sqrt{16 \times 25 \times 2} = 4 \times 5 \times \sqrt{2} = 20\sqrt{2}\text{ m}^2$$
Substitute $\sqrt{2} \approx 1.414$:
$$\Delta = 20 \times 1.414 = 28.28\text{ m}^2$$
Final Answer:
The area painted in colour is $20\sqrt{2}\text{ m}^2$ (or approximately $28.28\text{ m}^2$).
Question 4. [Triangle with Two Known Sides and Perimeter]
Find the area of a triangle two sides of which are $18\text{ cm}$ and $10\text{ cm}$ and the perimeter is $42\text{ cm}$.
Answer:
Step 1: Find the Unknown Third Side ($c$)
- Let $a = 18\text{ cm}$, $b = 10\text{ cm}$.
- Perimeter $P = a + b + c = 42\text{ cm}$.
$$18 + 10 + c = 42 \implies 28 + c = 42 \implies c = 42 – 28 = 14\text{ cm}$$
Step 2: Calculate Semi-Perimeter ($s$)
$$s = \frac{P}{2} = \frac{42}{2} = 21\text{ cm}$$
Step 3: Calculate Differences
$$s – a = 21 – 18 = 3\text{ cm}$$
$$s – b = 21 – 10 = 11\text{ cm}$$
$$s – c = 21 – 14 = 7\text{ cm}$$
Step 4: Compute Area Using Heron’s Formula
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)}$$
$$\Delta = \sqrt{21 \times 3 \times 11 \times 7}$$
Factorise $21 = 7 \times 3$:
$$\Delta = \sqrt{(7 \times 3) \times 3 \times 11 \times 7} = \sqrt{7^2 \times 3^2 \times 11} = 7 \times 3 \times \sqrt{11} = 21\sqrt{11}\text{ cm}^2$$
Substitute $\sqrt{11} \approx 3.317$:
$$\Delta = 21 \times 3.317 \approx 69.65\text{ cm}^2$$
Final Answer:
The area of the triangle is $21\sqrt{11}\text{ cm}^2$ (or approximately $69.65\text{ cm}^2$).
Question 5. [Sides in Given Ratio with Perimeter]
Sides of a triangle are in the ratio of $12 : 17 : 25$ and its perimeter is $540\text{ cm}$. Find its area.
Answer:
Step 1: Determine Actual Side Lengths
- Let common ratio multiplier be $x$.
- $a = 12x$, $b = 17x$, $c = 25x$.
$$\text{Perimeter} = 12x + 17x + 25x = 540\text{ cm}$$
$$54x = 540 \implies x = 10\text{ cm}$$
The actual side lengths are:
- $a = 12 \times 10 = 120\text{ cm}$
- $b = 17 \times 10 = 170\text{ cm}$
- $c = 25 \times 10 = 250\text{ cm}$
Step 2: Calculate Semi-Perimeter ($s$)
$$s = \frac{540}{2} = 270\text{ cm}$$
Step 3: Calculate Differences
$$s – a = 270 – 120 = 150\text{ cm}$$
$$s – b = 270 – 170 = 100\text{ cm}$$
$$s – c = 270 – 250 = 20\text{ cm}$$
Step 4: Compute Area Using Heron’s Formula
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)}$$
$$\Delta = \sqrt{270 \times 150 \times 100 \times 20}$$
Break terms into clean factors:
- $270 = 9 \times 3 \times 10$
- $150 = 3 \times 5 \times 10$
- $100 = 10^2$
- $20 = 2 \times 10 = 5 \times 4$
$$\Delta = \sqrt{(9 \times 30) \times (5 \times 30) \times 100 \times (4 \times 5)}$$
$$\Delta = \sqrt{9 \times 4 \times 100 \times 30^2 \times 5^2} = 3 \times 2 \times 10 \times 30 \times 5 = 60 \times 150 = 9000\text{ cm}^2$$
Final Answer:
The area of the triangle is $9000\text{ cm}^2$.
Question 6. [Isosceles Triangle Area Calculation]
An isosceles triangle has perimeter $30\text{ cm}$ and each of the equal sides is $12\text{ cm}$. Find the area of the triangle.
Answer:
Step 1: Find the Unequal Third Side ($c$)
- Equal sides: $a = 12\text{ cm}$, $b = 12\text{ cm}$.
- Perimeter $P = a + b + c = 30\text{ cm}$.
$$12 + 12 + c = 30 \implies 24 + c = 30 \implies c = 30 – 24 = 6\text{ cm}$$
Step 2: Calculate Semi-Perimeter ($s$)
$$s = \frac{P}{2} = \frac{30}{2} = 15\text{ cm}$$
Step 3: Calculate Differences
$$s – a = 15 – 12 = 3\text{ cm}$$
$$s – b = 15 – 12 = 3\text{ cm}$$
$$s – c = 15 – 6 = 9\text{ cm}$$
Step 4: Compute Area Using Heron’s Formula
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)}$$
$$\Delta = \sqrt{15 \times 3 \times 3 \times 9} = \sqrt{15 \times 3^2 \times 3^2} = 3 \times 3 \times \sqrt{15} = 9\sqrt{15}\text{ cm}^2$$
Substitute $\sqrt{15} \approx 3.873$:
$$\Delta = 9 \times 3.873 \approx 34.86\text{ cm}^2$$
Final Answer:
The area of the isosceles triangle is $9\sqrt{15}\text{ cm}^2$ (or approximately $34.86\text{ cm}^2$).
Question 7. [Area of a Trapezium via Triangular Decomposition]
A field is in the shape of a trapezium whose parallel sides are $25\text{ m}$ and $10\text{ m}$. The non-parallel sides are $14\text{ m}$ and $13\text{ m}$. Find the area of the field.
Answer: D C A B 10 m 25 m 14 m 13 m E
Step 1: Geometric Setup and Decomposition
Let $ABCD$ be the trapezium with $AB \parallel CD$, $AB = 25\text{ m}$, $CD = 10\text{ m}$, $AD = 14\text{ m}$, and $BC = 13\text{ m}$.
Draw line segment $CE \parallel DA$ with point $E$ lying on $AB$.
- Quadrilateral $AECD$ forms a parallelogram because $AE \parallel CD$ and $CE \parallel DA$.
- Therefore:
$$AE = CD = 10\text{ m}$$
$$CE = AD = 14\text{ m}$$ - Base of remaining triangle $\triangle CEB$:
$$EB = AB – AE = 25\text{ m} – 10\text{ m} = 15\text{ m}$$
Step 2: Area of $\triangle CEB$ Using Heron’s Formula
The sides of $\triangle CEB$ are $a = 15\text{ m}$, $b = 14\text{ m}$, $c = 13\text{ m}$.
$$s = \frac{15 + 14 + 13}{2} = \frac{42}{2} = 21\text{ m}$$
$$s – a = 21 – 15 = 6\text{ m}$$
$$s – b = 21 – 14 = 7\text{ m}$$
$$s – c = 21 – 13 = 8\text{ m}$$
$$\text{Area}(\triangle CEB) = \sqrt{21 \times 6 \times 7 \times 8}$$
$$\text{Area}(\triangle CEB) = \sqrt{(7 \times 3) \times (3 \times 2) \times 7 \times (2 \times 4)}$$
$$\text{Area}(\triangle CEB) = \sqrt{7^2 \times 3^2 \times 2^2 \times 2^2} = 7 \times 3 \times 2 \times 2 = 84\text{ m}^2$$
Step 3: Determine Vertical Height ($h$) of Trapezium
$$\text{Area}(\triangle CEB) = \frac{1}{2} \times \text{base} \times h = 84$$
$$\frac{1}{2} \times 15 \times h = 84 \implies h = \frac{84 \times 2}{15} = \frac{168}{15} = 11.2\text{ m}$$
Step 4: Compute Area of Trapezium $ABCD$
$$\text{Area} = \frac{1}{2}(AB + CD) \times h = \frac{1}{2}(25 + 10) \times 11.2$$
$$\text{Area} = \frac{1}{2} \times 35 \times 11.2 = 35 \times 5.6 = 196\text{ m}^2$$
(Alternatively: $\text{Area of Parallelogram } AECD = \text{base} \times h = 10 \times 11.2 = 112\text{ m}^2$. Total area $= 112 + 84 = 196\text{ m}^2$).
Final Answer:
The total area of the field is $196\text{ m}^2$.
Question 8. [Rhombus Grazing Field for 18 Cows]
A rhombus-shaped field has green grass for $18\text{ cows}$ to graze. If each side of the rhombus is $30\text{ m}$ and its longer diagonal is $48\text{ m}$, how much area of grass field will each cow be getting?
Answer:
Step 1: Divide Rhombus into Two Congruent Triangles
A rhombus has four equal sides ($30\text{ m}$). The diagonal ($48\text{ m}$) divides the rhombus into two congruent isosceles triangles, each having side lengths $a = 30\text{ m}$, $b = 30\text{ m}$, and $c = 48\text{ m}$.
Step 2: Area of One Triangle via Heron’s Formula
$$s = \frac{30 + 30 + 48}{2} = \frac{108}{2} = 54\text{ m}$$
$$s – a = 54 – 30 = 24\text{ m}$$
$$s – b = 54 – 30 = 24\text{ m}$$
$$s – c = 54 – 48 = 6\text{ m}$$
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{54 \times 24 \times 24 \times 6}$$
$$\Delta = \sqrt{(9 \times 6) \times 24^2 \times 6} = \sqrt{9 \times 6^2 \times 24^2} = 3 \times 6 \times 24 = 432\text{ m}^2$$
Step 3: Total Area of Rhombus Field
$$\text{Total Area} = 2 \times \text{Area}(\triangle) = 2 \times 432 = 864\text{ m}^2$$
Step 4: Calculate Area per Cow
$$\text{Area per Cow} = \frac{\text{Total Area}}{\text{Number of Cows}} = \frac{864\text{ m}^2}{18} = 48\text{ m}^2$$
Final Answer:
Each cow gets $48\text{ m}^2$ of grazing area.
Question 9. [Two-Colour Striped Umbrella]
An umbrella is made by stitching $10\text{ triangular pieces}$ of cloth of two different colours, each piece measuring $20\text{ cm}$, $50\text{ cm}$, and $50\text{ cm}$. How much cloth of each colour is required for the umbrella?
Answer:
Step 1: Identify Dimensions of One Triangular Piece
Sides of one triangle: $a = 20\text{ cm}$, $b = 50\text{ cm}$, $c = 50\text{ cm}$.
Total pieces = $10$ (meaning $5$ pieces of each colour).
Step 2: Compute Area of One Piece Using Heron’s Formula
$$s = \frac{20 + 50 + 50}{2} = \frac{120}{2} = 60\text{ cm}$$
$$s – a = 60 – 20 = 40\text{ cm}$$
$$s – b = 60 – 50 = 10\text{ cm}$$
$$s – c = 60 – 50 = 10\text{ cm}$$
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{60 \times 40 \times 10 \times 10}$$
$$\Delta = \sqrt{(6 \times 10) \times (4 \times 10) \times 10^2} = \sqrt{24 \times 10^4} = 100\sqrt{24} = 100 \times 2\sqrt{6} = 200\sqrt{6}\text{ cm}^2$$
Step 3: Compute Cloth Required for Each Colour (5 pieces each)
$$\text{Cloth of Each Colour} = 5 \times 200\sqrt{6} = 1000\sqrt{6}\text{ cm}^2$$
Substitute $\sqrt{6} \approx 2.449$:
$$\text{Area} \approx 1000 \times 2.449 = 2449\text{ cm}^2$$
Final Answer:
The cloth required for each colour is $1000\sqrt{6}\text{ cm}^2$ (or approximately $2449.5\text{ cm}^2$).
Question 10. [Three-Shaded Kite Area]
A kite in the shape of a square with a diagonal $32\text{ cm}$ and an isosceles triangle of base $8\text{ cm}$ and sides $6\text{ cm}$ each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?
Answer:
Step 1: Compute Area of Square (Shades I and II)
- A square has perpendicular diagonals of equal length: $d_1 = d_2 = 32\text{ cm}$.
$$\text{Area of Square} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 32 \times 32 = 512\text{ cm}^2$$ - The horizontal diagonal bisects the square into two equal triangles: Shade I (top) and Shade II (bottom).
$$\text{Area of Shade I} = \frac{512}{2} = 256\text{ cm}^2$$
$$\text{Area of Shade II} = \frac{512}{2} = 256\text{ cm}^2$$
Step 2: Compute Area of Isosceles Triangle Tail (Shade III)
Sides of tail: $a = 6\text{ cm}$, $b = 6\text{ cm}$, $c = 8\text{ cm}$.
$$s = \frac{6 + 6 + 8}{2} = \frac{20}{2} = 10\text{ cm}$$
$$s – a = 10 – 6 = 4\text{ cm}$$
$$s – b = 10 – 6 = 4\text{ cm}$$
$$s – c = 10 – 8 = 2\text{ cm}$$
$$\Delta_{\text{III}} = \sqrt{10 \times 4 \times 4 \times 2} = \sqrt{16 \times 20} = 4\sqrt{20} = 4 \times 2\sqrt{5} = 8\sqrt{5}\text{ cm}^2$$
Substitute $\sqrt{5} \approx 2.236$:
$$\Delta_{\text{III}} = 8 \times 2.236 = 17.888 \approx 17.89\text{ cm}^2$$
Final Answer:
- Paper of Shade I = $256\text{ cm}^2$
- Paper of Shade II = $256\text{ cm}^2$
- Paper of Shade III = $8\sqrt{5}\text{ cm}^2$ (or $17.89\text{ cm}^2$)
Question 11. [Floral Floor Design Polishing Cost]
A floral design on a floor is made up of $16\text{ tiles}$ which are triangular, the sides of the triangle being $9\text{ cm}$, $28\text{ cm}$ and $35\text{ cm}$. Find the cost of polishing the tiles at the rate of $50\text{p per cm}^2$.
Answer:
Step 1: Area of One Triangular Tile Using Heron’s Formula
Sides: $a = 9\text{ cm}$, $b = 28\text{ cm}$, $c = 35\text{ cm}$.
$$s = \frac{9 + 28 + 35}{2} = \frac{72}{2} = 36\text{ cm}$$
$$s – a = 36 – 9 = 27\text{ cm}$$
$$s – b = 36 – 28 = 8\text{ cm}$$
$$s – c = 36 – 35 = 1\text{ cm}$$
$$\Delta = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{36 \times 27 \times 8 \times 1}$$
$$\Delta = \sqrt{36 \times (9 \times 3) \times (4 \times 2) \times 1} = \sqrt{36 \times 9 \times 4 \times 6}$$
$$\Delta = 6 \times 3 \times 2 \times \sqrt{6} = 36\sqrt{6}\text{ cm}^2$$
Substitute $\sqrt{6} \approx 2.45$:
$$\Delta \approx 36 \times 2.45 = 88.2\text{ cm}^2$$
Step 2: Compute Total Area of 16 Tiles
$$\text{Total Area} = 16 \times 36\sqrt{6} = 576\sqrt{6}\text{ cm}^2 \approx 16 \times 88.2 = 1411.2\text{ cm}^2$$
Step 3: Calculate Polishing Cost at $50\text{ paise}$ (₹$0.50$) per $\text{cm}^2$
$$\text{Cost} = \text{Total Area} \times \text{Rate} = 1411.2 \times 0.50 = ₹705.60$$
Final Answer:
The total cost of polishing the tiles is ₹$705.60$.
[👉 Also Read: Class 9 Math Chapter 7 Triangles NCERT Solutions]
Master High-Yield Board FAQs (Rank Math Schema Ready)
What is Heron’s formula and when is it preferred over the standard triangle formula?
Heron’s formula calculates the area of a triangle given only its three side lengths: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$, where $s = \frac{a+b+c}{2}$. It is preferred when all three side lengths of a triangle are known but the perpendicular height (altitude) is unknown.
What is the semi-perimeter of a triangle?
The semi-perimeter, denoted by $s$, represents half of the total boundary perimeter of a triangle. It is calculated by adding the lengths of all three sides and dividing the sum by two: $s = \frac{a + b + c}{2}$.
Can Heron’s formula be used for right-angled triangles?
Yes. Heron’s formula applies to any triangle, including right-angled triangles. However, for right-angled triangles, the standard formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ is faster because the two legs forming the right angle serve directly as base and height.
How do you derive the area of an equilateral triangle from Heron’s formula?
For an equilateral triangle with equal sides $a$, the semi-perimeter is $s = \frac{3a}{2}$. Subtracting each side yields $(s – a) = \frac{a}{2}$. Substituting into Heron’s formula gives $\sqrt{\left(\frac{3a}{2}\right)\left(\frac{a}{2}\right)^3} = \sqrt{\frac{3a^4}{16}} = \frac{\sqrt{3}}{4}a^2$.
How do you find the area of an irregular quadrilateral using Heron’s formula?
An irregular quadrilateral is divided into two triangles by drawing one of its diagonals. If the lengths of the four sides and the chosen diagonal are known, Heron’s formula is applied to each triangle separately, and the two areas are added together to give the quadrilateral’s total area.
How do you find the altitude of a triangle after calculating its area with Heron’s formula?
Once the area $\Delta$ is determined using Heron’s formula, equate it to the standard area formula: $\Delta = \frac{1}{2} \times \text{base} \times h$. Rearranging this equation allows you to calculate the vertical altitude corresponding to that base: $h = \frac{2\Delta}{\text{base}}$.
What happens if $(s – a)$ is zero or negative in Heron’s formula?
If $(s – a) \le 0$, the given side lengths violate the Triangle Inequality Theorem, which requires the sum of any two sides of a triangle to be strictly greater than the third side. A negative or zero term under the square root indicates that the sides cannot form a valid closed triangle.
How do you find the area of a trapezium using Heron’s formula?
Draw a line from one upper vertex parallel to the opposite non-parallel leg to split the trapezium into a parallelogram and a triangle. Use Heron’s formula on the triangle to find its area, compute its vertical height from $h = \frac{2 \times \text{Area}}{\text{base}}$, and use $h$ in the trapezium area formula $\frac{1}{2}(\text{sum of parallel sides}) \times h$.
Why must units be squared for area but linear for perimeter?
Perimeter measures boundary distance in one dimension, so it is expressed in linear units like metres ($\text{m}$). Area measures two-dimensional enclosed surface space—the product of two linear dimensions—so it is always expressed in squared units like $\text{m}^2$ or $\text{cm}^2$.
How do you solve for the sides of a triangle when given their ratio and perimeter?
Let the sides be $ax, bx,$ and $cx$, where $x$ is a common multiplier. Set their sum equal to the given perimeter ($ax + bx + cx = P$) and solve for $x$. Multiply each ratio part by $x$ to find the actual side lengths, then proceed with Heron’s formula.
What is the area of a rhombus if its diagonals are known?
The area of a rhombus whose diagonals are $d_1$ and $d_2$ is $\text{Area} = \frac{1}{2} \times d_1 \times d_2$. Because the diagonals of a rhombus bisect each other perpendicularly, they divide the rhombus into four congruent right-angled triangles.
Can Heron’s formula be used if side lengths are given in different units?
No. All side lengths must be converted to the same unit (all metres, all centimetres, etc.) before calculating the semi-perimeter $s$ and applying Heron’s formula. Mixing units results in an incorrect area.
How do you find the area of an isosceles triangle using Heron’s formula?
Identify the two equal sides ($a$ and $a$) and the base ($b$). Compute the semi-perimeter $s = \frac{2a + b}{2}$, calculate $(s – a)$ and $(s – b)$, and substitute them into $\Delta = \sqrt{s(s – a)^2(s – b)} = (s – a)\sqrt{s(s – b)}$.
What is the most efficient way to simplify the square root in Heron’s formula?
Instead of multiplying large numbers under the radical into a multi-digit product, break each term ($s, s-a, s-b, s-c$) into its prime factors. Pair matching factors to pull them outside the square root directly, keeping calculations straightforward and reducing errors.
What are the key presentation steps to score 100% on Heron’s formula questions in CBSE exams?
To secure full marks: (1) state the given side lengths with uniform units, (2) write the semi-perimeter formula and compute $s$ explicitly, (3) show the differences $(s-a)$, $(s-b)$, and $(s-c)$ on separate lines, (4) write Heron’s formula algebraically before substituting values, and (5) conclude with the final numerical answer and its square units clearly highlighted.