Chapter 10 — Heron's Formula | Class 9 CBSE Maths

Section 10.1 — Foundation

The Problem: Finding Area Without Height

We know area = ½ × base × height. But what if we only know the three sides — and not the height?

Standard Formula (Height Known)

When base and perpendicular height are given:

Area = ½ × base × height

Works perfectly for right triangles and when the altitude is given directly.

The Problem Heron Solved

Triangular park with sides 40 m, 32 m, 24 m.

Without height → can't use ½×b×h directly
Must find height using Pythagoras first
OR use Heron's Formula directly!

💡

Heron's formula uses only the three side lengths — ideal for any scalene triangle.

🏛️

Heron of Alexandria (10 C.E. – 75 C.E.)

Heron was born around 10 AD, possibly in Alexandria in Egypt. He worked in applied mathematics. His works on mathematical and physical subjects are so numerous and varied that he is considered to be an encyclopedic writer in these fields.

His geometrical works deal largely with problems on mensuration, written in three books. Book I deals with the area of squares, rectangles, triangles, trapezoids, regular polygons, circles, surfaces of cylinders, cones, spheres etc. In this book, Heron derived the famous formula for the area of a triangle in terms of its three sides. The formula is also known as Hero's formula.

Review: Area Formulas We Already Know

Shape	Formula	What's needed
Any triangle	½ × b × h	Base b and perpendicular height h
Right triangle	½ × leg₁ × leg₂	Two perpendicular sides (legs)
Equilateral triangle	(√3/4) × a²	Side a (derived from Heron's)
Any triangle (no height)	√s(s−a)(s−b)(s−c)	Only three sides a, b, c

Section 10.1 — The Formula

Heron's Formula — Stated

The formula given by Heron about the area of a triangle, stated exactly as in NCERT:

Heron's Formula (Also known as Hero's Formula)

Area of a triangle = √s(s − a)(s − b)(s − c)

where s = (a + b + c) / 2

a, b, c = the three sides of the triangle
s = semi-perimeter = half the perimeter of the triangle

Every Term Explained

Symbol	Full Name	How to Calculate	Example (a=40, b=24, c=32)
a, b, c	Three sides of the triangle	Read from the problem	40 m, 24 m, 32 m
a+b+c	Perimeter	Sum of all three sides	40+24+32 = 96 m
s	Semi-perimeter	(a+b+c) ÷ 2	96÷2 = 48 m
s−a	s minus first side	s − a	48−40 = 8 m
s−b	s minus second side	s − b	48−24 = 24 m
s−c	s minus third side	s − c	48−32 = 16 m
Area	Area of the triangle	√[s×(s−a)×(s−b)×(s−c)]	√(48×8×24×16) = 384 m²

Before You Apply — Triangle Validity Check

A valid triangle requires: sum of any two sides > third side.
Heron's formula only works when (s−a), (s−b), (s−c) are ALL positive — this is automatically true for any valid triangle.
Also note: Heron's formula is helpful where it is not possible to find the height of the triangle easily.

How to Apply

Step-by-Step Template

Follow this exact sequence every time — earning step marks in CBSE requires showing each step clearly.

The 5-Step Method

Write down the sides: Identify a, b, c from the problem. If the third side is not given but perimeter is, calculate c = Perimeter − a − b.

Calculate semi-perimeter s: s = (a + b + c) / 2
Write: "So, 2s = ___ and s = ___ cm/m"

Find all three differences: Write on separate lines:
s−a = ___ cm/m
s−b = ___ cm/m
s−c = ___ cm/m

Multiply and find product: s × (s−a) × (s−b) × (s−c) = ___
Factor into perfect squares if possible.

Take square root: Area = √[product]. Simplify. Write final answer with square units (cm² or m²).

NCERT Worked Example

Triangular park with sides a = 40 m, b = 32 m, c = 24 m

Step 1 — Given

a = 40 m, b = 24 m, c = 32 m

Step 2 — Semi-perimeter

s = (40 + 24 + 32) / 2 = 96 / 2 = 48 m

Step 3 — Differences

s − a = 48 − 40 = 8 m s − b = 48 − 24 = 24 m s − c = 48 − 32 = 16 m

Step 4 — Product

s(s−a)(s−b)(s−c) = 48 × 8 × 24 × 16 = 147456

Step 5 — Area

Area = √147456 = 384 m²

Area of the triangular park = 384 m²

✅

Verification (NCERT): Since 32² + 24² = 1024 + 576 = 1600 = 40², this is a right triangle! Area = ½ × 32 × 24 = 384 m² — same answer confirms Heron's formula works. ✓

NCERT Verification

Two special triangles verified using Heron's formula

(i) Equilateral: side = 10 cm

s = (10+10+10)/2 = 15 cm s−10 = 5 (for all three) Area = √(15×5×5×5) = √1875 = 25√3 cm²

(ii) Isosceles: base=8, equal sides=5 cm

s = (8+5+5)/2 = 9 cm s−8=1, s−5=4, s−5=4 Area = √(9×1×4×4) = √144 = 12 cm²

Derived Formulas

Special Cases of Heron's Formula

Simplified expressions for common triangle types — save time in MCQs and exams.

Equilateral Triangle (side = a)

All sides equal → a = b = c

s = 3a/2 s−a = s−b = s−c = a/2 Area = √[(3a/2)(a/2)(a/2)(a/2)] = √[3a⁴/16] = (√3/4)a²

Area = (√3/4) × a²

E.g., a = 10 cm → Area = 25√3 cm²

Isosceles Triangle (equal sides=a, base=c)

Two equal sides of length a, base = c

s = a + c/2 s−a = c/2 (for each equal side) s−c = a − c/2 Area = (c/4)√(4a²−c²)

Area = (c/4)√(4a²−c²)

E.g., a=5, c=8 → Area = 2×6 = 12 cm²

Right Triangle — Always Check First!

If c² = a² + b² (Pythagorean triple), use the simpler formula: Area = ½ × a × b (product of the two legs). Heron's formula will give the same answer, but the standard formula is faster.

Common Pythagorean triples to recognise: (3,4,5), (5,12,13), (7,24,25), (8,15,17), (9,40,41), (20,21,29) and their multiples.

Decision Guide — Which Formula to Use?

Situation	Best Formula	Why
Base and height both given	½ × b × h	Simplest — direct substitution
Right triangle (legs given)	½ × leg₁ × leg₂	Legs are perpendicular → direct
Equilateral triangle (side given)	(√3/4)a²	Derived formula — faster
Any triangle, only 3 sides known	√s(s−a)(s−b)(s−c)	Heron's formula — no height needed
Quadrilateral (diagonal given)	Heron's for each triangle	Split into 2 triangles, apply twice
Sides in ratio + perimeter given	Find sides first, then Heron's	Set sides = px, qx, rx

Textbook Examples

All 3 NCERT Examples — Fully Solved

Every example from NCERT Class 9 Chapter 10, with every step explained.

Example 1

Find area of a triangle: two sides 8 cm and 11 cm, perimeter = 32 cm.

Identify what's given: a = 8 cm, b = 11 cm, Perimeter = 32 cm.
Find third side: c = 32 − (8 + 11) = 32 − 19 = 13 cm

Semi-perimeter: 2s = 32 cm → s = 16 cm

s − a = 16 − 8 = 8 cm
s − b = 16 − 11 = 5 cm
s − c = 16 − 13 = 3 cm

Apply Heron's Formula:
Area = √[s(s−a)(s−b)(s−c)] = √[16 × 8 × 5 × 3]
= √1920 = √(64 × 30) = 8√30 cm²

Area of the triangle = 8√30 cm²

💡

Simplifying √1920: Factor 1920 = 64 × 30 = 8² × 30. So √1920 = 8√30. Always factor out perfect squares before leaving the answer.

Example 2

Triangular park ABC: sides 120 m, 80 m, 50 m. Find area for grass. Find cost of fencing at ₹20/m leaving 3 m for gate.

Part (i): Area of the park

Given: sides 120 m, 80 m, 50 m.
2s = 50 + 80 + 120 = 250 m → s = 125 m

s − a = 125 − 120 = 5 m
s − b = 125 − 80 = 45 m
s − c = 125 − 50 = 75 m

Area = √(125 × 5 × 45 × 75) m²
= √2,109,375 = 375√15 m²

Part (ii): Cost of fencing

Perimeter of the park = AB + BC + CA = 250 m

Length of wire needed = 250 − 3 = 247 m (3 m left for gate)

Cost of fencing = ₹20 × 247 = ₹4940

Area to plant grass = 375√15 m² | Cost of fencing = ₹4940

Example 3

Sides of a triangular plot in ratio 3:5:7, perimeter = 300 m. Find its area.

Suppose sides are 3x, 5x and 7x (in metres).
3x + 5x + 7x = 300 (perimeter) → 15x = 300 → x = 20

So the sides of the triangle are:
3 × 20 = 60 m, 5 × 20 = 100 m, 7 × 20 = 140 m

s = (60 + 100 + 140) / 2 = 150 m

s − 60 = 90 m
s − 100 = 50 m
s − 140 = 10 m

Area = √(150 × 90 × 50 × 10) m²
= √6,750,000 = √(2,250,000 × 3) = 1500√3 m²

Area of the triangular plot = 1500√3 m²

✅

Verify: 150×90=13500; 50×10=500; 13500×500=6,750,000; (1500)²×3=6,750,000 ✓

NCERT Exercise 10.1

All 6 Questions — Solved Step by Step

A traffic signal board ('SCHOOL AHEAD') is an equilateral triangle with side 'a'. Find its area using Heron's formula. If perimeter = 180 cm, find area.

Part 1

Equilateral triangle, all sides = a.
s = (a+a+a)/2 = 3a/2
s−a = s−b = s−c = 3a/2 − a = a/2

Area = √[(3a/2)(a/2)(a/2)(a/2)] = √[3a⁴/16] = (√3/4)a²

Part 2

Perimeter = 180 cm → each side a = 180/3 = 60 cm
Area = (√3/4) × 60² = (√3/4) × 3600 = 900√3 cm²

General formula: Area = (√3/4)a² | For perimeter 180 cm: Area = 900√3 cm²

Flyover triangular side walls: sides 122 m, 22 m and 120 m. Ads earn ₹5000/m²/year. Company hired one wall for 3 months. How much rent?

Check for right triangle: 22² + 120² = 484 + 14400 = 14884 = 122²
→ It is a right triangle! (122 m is the hypotenuse)

Area = ½ × 22 × 120 = 1320 m²
(Verify with Heron's: s=132; s−122=10, s−22=110, s−120=12; √(132×10×110×12) = √1742400 = 1320 ✓)

Earnings per year = ₹5000 × 1320 = ₹66,00,000
Rent for 3 months = 66,00,000 × (3/12) = ₹16,50,000

Area = 1320 m² | Rent for 3 months = ₹16,50,000

Slide side wall painted with "KEEP THE PARK GREEN AND CLEAN". Sides = 15 m, 11 m and 6 m. Find area painted in colour.

a=15m, b=11m, c=6m
s = (15+11+6)/2 = 32/2 = 16 m

s−15 = 1 m | s−11 = 5 m | s−6 = 10 m

Area = √(16 × 1 × 5 × 10) = √800 = √(400 × 2) = 20√2 m²

Area painted = 20√2 m² ≈ 28.28 m²

Two sides of a triangle are 18 cm and 10 cm and the perimeter is 42 cm. Find the area.

Third side c = 42 − (18 + 10) = 14 cm
s = 42/2 = 21 cm

s−18 = 3 cm | s−10 = 11 cm | s−14 = 7 cm

Area = √(21 × 3 × 11 × 7) = √4851
4851 = 9 × 539 = 9 × 7 × 77 = 9 × 7 × 7 × 11 = 9 × 49 × 11
√4851 = 3 × 7 × √11 = 21√11 cm²

Area = 21√11 cm² ≈ 69.65 cm²

Sides of a triangle are in ratio 12:17:25 and its perimeter is 540 cm. Find its area.

Let sides be 12x, 17x, 25x.
12x + 17x + 25x = 540 → 54x = 540 → x = 10
Sides: 120 cm, 170 cm, 250 cm

s = 540/2 = 270 cm
s−120 = 150 | s−170 = 100 | s−250 = 20

Area = √(270 × 150 × 100 × 20)
= √(270 × 150 × 2000)
= √81,000,000 = 9000 cm²

Area = 9000 cm²

✅

Check: 9000² = 81,000,000. And 270×150×100×20 = 81,000,000 ✓

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area.

Equal sides = 12 cm each.
Base = 30 − 12 − 12 = 6 cm
s = 30/2 = 15 cm

s−12 = 3 cm (for each equal side) | s−6 = 9 cm

Area = √(15 × 3 × 3 × 9) = √(15 × 81) = 9√15 cm²

Area = 9√15 cm² ≈ 34.86 cm²

Study Strategy

10 Expert Tips for Full Marks

Always Start with s

Write "s = (a+b+c)/2 = ___" as your very first step. CBSE gives step marks — the semi-perimeter calculation earns marks even if a later arithmetic error is made.

Write All Three Differences

Show all three: (s−a), (s−b), (s−c) on separate lines. This earns step marks and helps catch errors. Never skip this even if they look obvious.

Check for Right Triangle First

Before using Heron's, test if a²+b²=c² for the largest side. If yes, use simpler ½×l₁×l₂. The NCERT Q2 (122,22,120) is a right triangle — saves a lot of calculation.

Ratio Problems: Set Sides = kx, ky, kz

Sides in ratio p:q:r with perimeter P → set sides = px, qx, rx. Solve (p+q+r)x = P to find x. This is the only correct method for ratio+perimeter problems (Examples 3, Q5).

Third Side from Perimeter

If two sides (a, b) and perimeter P are given: c = P − a − b. Then s = P/2. Applied in Example 1 (Q1 type). Always do this step BEFORE computing s.

Simplify √ by Factoring

Always factor the product to extract perfect squares. E.g., √1920 = √(64×30) = 8√30. CBSE expects simplified surd form — don't leave it as √1920 in the answer.

Never Forget Square Units

Area is ALWAYS in square units. If sides are in cm → area in cm². If sides are in m → area in m². Missing the unit (²) costs marks in CBSE. Write "cm²" or "m²" always.

Memorise (√3/4)a² for Equilateral

The NCERT Q1 always asks for equilateral triangle formula. Derivation: s=3a/2, s−a=a/2 → (√3/4)a². Also remember: for a=60cm (P=180), Area = 900√3 cm².

Cost Problems: Identify Rate Type

Fencing → Cost = Rate × Perimeter (subtract gate). Painting → Cost = Rate × Area. Advertisement → Rent = Rate per m²/year × Area × (months/12). Identify which before calculating.

Quadrilateral = Two Triangles

Divide quadrilateral along the given diagonal. Apply Heron's to each triangle separately. Total area = sum. This is the key extension of Heron's formula to 4-sided figures.

Quick Reference

Chapter 10 — Formula & Fact Sheet

Formula / Concept	Expression	Notes
Heron's Formula	Area = √s(s−a)(s−b)(s−c)	Core formula — no height needed
Semi-perimeter	s = (a+b+c)/2	Always compute first
Standard area	½ × base × height	Use when height is given
Equilateral triangle	(√3/4) × a²	Derived from Heron's, side = a
Isosceles (equal sides=a, base=c)	(c/4)√(4a²−c²)	Derived from Heron's
Right triangle legs p, q	½ × p × q	Faster than Heron's for right △
Check right triangle	c² = a² + b² (c largest)	Pythagorean condition
Ratio sides with perimeter P	Sides=px,qx,rx; (p+q+r)x=P	Find x, then apply Heron's
Fencing cost	Rate × (Perimeter − gate space)	Perimeter = 2s = a+b+c
Painting / flooring cost	Rate × Area	Area from Heron's
Advertisement rent (n months)	Rate/yr × Area × n/12	Convert months to fraction of year
Quadrilateral with diagonal d	Area = Area(△₁) + Area(△₂)	Split along diagonal

📌

Summary from NCERT: Area of a triangle with sides a, b, c = √s(s−a)(s−b)(s−c) where s = (a+b+c)/2. This is the one formula stated in the NCERT summary for this chapter.

CBSE Pattern Practice

50 Practice Questions

All CBSE question types — MCQ, 1-mark, 2-mark, 3-mark, and 5-mark questions with complete answers.

Section A — MCQ (1 mark each)

In Heron's formula Area = √s(s−a)(s−b)(s−c), the value of 's' is:

MCQ

(a) Perimeter

(b) Semi-perimeter

(d) Largest side

Answer: (b) Semi-perimeter
s = semi-perimeter = (a+b+c)/2 = half the perimeter of the triangle. This is defined explicitly in NCERT.

Area of an equilateral triangle with side 6 cm is:

MCQ

(a) 6√3 cm²

(b) 9√3 cm²

(d) 36 cm²

Answer: (b) 9√3 cm²
(√3/4) × 6² = (√3/4) × 36 = 9√3 cm²

Triangle sides 3 cm, 4 cm, 5 cm. Its area is:

MCQ

(a) 7 cm²

(b) 6 cm²

(d) 5 cm²

Answer: (b) 6 cm²
3²+4²=25=5². Right triangle! Area = ½×3×4 = 6 cm².
Heron's check: s=6; √(6×3×2×1)=√36=6 ✓

Semi-perimeter of a triangle with sides 5 cm, 12 cm, 13 cm is:

MCQ

(a) 15 cm

(b) 30 cm

(d) 12 cm

Answer: (a) 15 cm
s = (5+12+13)/2 = 30/2 = 15 cm

Triangular plot with sides 40 m, 32 m, 24 m (NCERT park). Its area is:

MCQ

(a) 480 m²

(b) 384 m²

(d) 576 m²

Answer: (b) 384 m²
This is the NCERT example. s=48; √(48×8×24×16)=384 m². Also ½×32×24=384 (right triangle) ✓

Sides in ratio 3:5:7, perimeter = 300 m (NCERT Example 3). Area is:

MCQ

(a) 1500 m²

(b) 1500√2 m²

(d) 3000 m²

Answer: (c) 1500√3 m²
x=20; sides 60,100,140; s=150; √(150×90×50×10)=1500√3 m²

Equilateral triangle perimeter = 12 cm. Area =

MCQ

(a) √3 cm²

(b) 4√3 cm²

(d) 6√3 cm²

Answer: (b) 4√3 cm²
Side = 12/3 = 4 cm. Area = (√3/4)×16 = 4√3 cm²

Sides 122 m, 22 m, 120 m (flyover Q2). The triangle is:

MCQ

(a) Equilateral

(b) Isosceles

(d) Obtuse

Answer: (c) Right-angled
22²+120² = 484+14400 = 14884 = 122². Pythagorean condition holds → right triangle.

Area of isosceles triangle with equal sides 5 cm, base 8 cm is: (from NCERT)

MCQ

(a) 10 cm²

(b) 12 cm²

(d) 20 cm²

Answer: (b) 12 cm²
s=9; s−8=1, s−5=4, s−5=4; Area=√(9×1×4×4)=√144=12 cm²

Heron's formula is MOST useful when:

MCQ

(a) Height is given

(b) The triangle is equilateral

(d) The triangle is right-angled

Answer: (c) Only three sides are known
Heron's formula is helpful "where it is not possible to find the height of the triangle easily" — exact NCERT language. It requires only the three sides.

Section B — Very Short Answer (1 mark each)

State Heron's formula for the area of a triangle with sides a, b, c.

1 Mark

Area = √s(s−a)(s−b)(s−c)
where s = (a+b+c)/2 is the semi-perimeter.

Find s for a triangle with sides 7 cm, 8 cm, 9 cm.

1 Mark

s = (7+8+9)/2 = 24/2 = 12 cm

Write the formula for the area of an equilateral triangle with side a.

1 Mark

Area = (√3/4) × a²
Derived by applying Heron's formula to an equilateral triangle where all three sides are equal to a.

An equilateral triangle has side 10 cm. Find its area using Heron's formula.

1 Mark

s=15; s−10=5 (all three). Area = √(15×5×5×5) = √1875 = 25√3 cm²

Find s, (s−a), (s−b), (s−c) for triangle with sides 5, 7, 8 cm.

1 Mark

s = (5+7+8)/2 = 10 cm
s−5 = 5 cm, s−7 = 3 cm, s−8 = 2 cm

Two sides of a triangle are 8 cm and 11 cm. Perimeter = 32 cm. What is the third side?

1 Mark

Third side = 32 − 8 − 11 = 13 cm
(From NCERT Example 1)

Equilateral triangle with perimeter 180 cm. What is each side? What is the area?

1 Mark

Each side = 180/3 = 60 cm
Area = (√3/4)×3600 = 900√3 cm²
(NCERT Q1)

Sides 122 m, 22 m, 120 m. Find area. (NCERT Q2)

1 Mark

Right triangle (22²+120²=122²). Area = ½×22×120 = 1320 m²

Name the book by Heron where the famous triangle area formula was derived.

1 Mark

Book I of his three-book geometrical works on mensuration (referred to in NCERT). The complete work is known as Metrica.

Sides of isosceles triangle: perimeter 30 cm, equal sides 12 cm each. Find the base.

1 Mark

Base = 30 − 12 − 12 = 6 cm
(From NCERT Q6)

Section C — Short Answer (2 marks each)

Find area of triangle with sides 6 cm, 8 cm, 10 cm.

2 Mark

6²+8²=100=10² → Right triangle! Area = ½×6×8 = 24 cm²
Heron's: s=12; 12×6×4×2=576; √576=24 ✓

Find area of triangle sides 13 cm, 14 cm, 15 cm.

2 Mark

s=(13+14+15)/2=21. s−13=8, s−14=7, s−15=6.
Area = √(21×8×7×6) = √7056 = 84 cm²

Triangular wall: sides 15 m, 11 m, 6 m. Find area painted. (NCERT Q3)

2 Mark

s=16; s−15=1, s−11=5, s−6=10.
Area = √(16×1×5×10) = √800 = 20√2 m²
Area = 20√2 m² ≈ 28.28 m²

Triangle: two sides 18 cm, 10 cm, perimeter 42 cm. Find area. (NCERT Q4)

2 Mark

c=42−18−10=14 cm. s=21. s−18=3, s−10=11, s−14=7.
Area = √(21×3×11×7) = √4851 = 21√11 cm²

Find area of equilateral triangle with side 4 cm using Heron's formula.

2 Mark

s=6. s−4=2 (all three).
Area = √(6×2×2×2) = √48 = 4√3 cm²
Check: (√3/4)×16 = 4√3 ✓

Isosceles triangle: perimeter 30 cm, equal sides 12 cm. Find area. (NCERT Q6)

2 Mark

Base=6 cm. s=15. s−12=3, s−6=9.
Area = √(15×3×3×9) = √1215 = 9√15 cm²
Area = 9√15 cm²

Triangular park: perimeter 250 m. Fencing cost at ₹20/m with 3 m gate. Find cost. (NCERT Ex 2)

2 Mark

Wire needed = 250−3 = 247 m
Cost = ₹20 × 247 = ₹4940

Triangle sides 7 cm, 24 cm, 25 cm. Find area (use the most efficient method).

2 Mark

Check: 7²+24²=49+576=625=25². Right triangle!
Area = ½×7×24 = 84 cm²

Flyover wall (NCERT Q2): Ads earn ₹5000/m²/year, area 1320 m². Find rent for 3 months.

2 Mark

Annual earnings = 5000 × 1320 = ₹66,00,000
For 3 months = 66,00,000 × 3/12 = ₹16,50,000

A triangle has area 30 cm² and base 12 cm. Find its height.

2 Mark

Area = ½ × base × height → 30 = ½ × 12 × h → h = 60/12 = 5 cm

Section D — Short Answer II (3 marks each)

Sides in ratio 12:17:25, perimeter 540 cm. Find area. (NCERT Q5)

3 Mark

Step 1: 54x=540 → x=10. Sides: 120, 170, 250 cm.
Step 2: s=270. s−120=150, s−170=100, s−250=20.
Step 3: Area=√(270×150×100×20)=√81,000,000=9000 cm²

Triangular plot with sides 3:5:7, perimeter = 300 m. Find area. (NCERT Example 3)

3 Mark

Step 1: 15x=300 → x=20. Sides: 60m, 100m, 140m.
Step 2: s=150. s−60=90, s−100=50, s−140=10.
Step 3: Area=√(150×90×50×10)=√6,750,000=1500√3 m²

Two sides 8 cm and 11 cm, perimeter 32 cm. Find area. (NCERT Example 1)

3 Mark

Step 1: c=32−8−11=13 cm. s=16 cm.
Step 2: s−8=8, s−11=5, s−13=3.
Step 3: Area=√(16×8×5×3)=√1920=8√30 cm²

Find area of equilateral triangle with side 'a' using Heron's formula. (NCERT Q1)

3 Mark

Step 1: All sides=a. s=3a/2.
Step 2: s−a=a/2 for all three differences.
Step 3: Area=√[(3a/2)(a/2)(a/2)(a/2)]=√[3a⁴/16]=(√3/4)a²
If perimeter=180cm → a=60 cm → Area=900√3 cm²

Triangular park: sides 120 m, 80 m, 50 m. Find area and cost of fencing at ₹20/m, 3 m gate. (NCERT Ex 2)

3 Mark

Area: s=125; s−120=5, s−80=45, s−50=75. Area=√(125×5×45×75)=375√15 m²
Fencing: Wire=250−3=247m. Cost=20×247=₹4940

Find area of triangle with sides 18 cm, 24 cm, 30 cm. Verify with the standard formula.

3 Mark

Check: 18²+24²=324+576=900=30². Right triangle!
Standard: ½×18×24=216 cm²
Heron's: s=36; 36×18×12×6=46656; √46656=216 cm² ✓
Area = 216 cm²

Right triangle perimeter = 60 cm, hypotenuse = 26 cm. Find area.

3 Mark

Legs sum p+q = 60−26 = 34. Also p²+q² = 676.
(p+q)²=1156 → p²+2pq+q²=1156 → 676+2pq=1156 → pq=240.
Area = ½×p×q = ½×240 = 120 cm²

Isosceles triangle: equal sides 17 cm, base 16 cm. Find area.

3 Mark

s=(17+17+16)/2=25. s−17=8, s−17=8, s−16=9.
Area=√(25×8×8×9)=√14400=120 cm²

Rhombus perimeter = 40 cm, one diagonal = 12 cm. Find area.

3 Mark

Each side = 40/4 = 10 cm. Half of d₁ = 6 cm.
Half of d₂ = √(10²−6²) = √64 = 8 cm → d₂ = 16 cm.
Area = ½×d₁×d₂ = ½×12×16 = 96 cm²

Cost of painting a triangular wall at ₹25/m². Sides 15 m, 11 m, 6 m. Find cost.

3 Mark

Area = 20√2 m² (from Q3/NCERT).
Cost = 25 × 20√2 = 500√2 ≈ ₹707.1

Section E — Long Answer (5 marks each)

Triangular park ABC with sides 120 m, 80 m, 50 m. (i) Find area for planting grass (ii) Find perimeter (iii) Cost of fencing at ₹20/m with 3 m gate (iv) Cost of planting grass at ₹5/m².

5 Mark

(i) s=125; Area=√(125×5×45×75)=375√15 m² ≈ 1451.4 m²
(ii) Perimeter = 250 m
(iii) Wire=247 m; Cost=₹20×247=₹4940
(iv) Cost=₹5×375√15=1875√15 ≈ ₹7257

Flyover wall: sides 122 m, 22 m, 120 m. (i) Show it's a right triangle (ii) Find area (iii) Company pays rent for 3 months at ₹5000/m²/year — how much? (NCERT Q2 in full)

5 Mark

(i) 22²+120²=484+14400=14884=122² → Right triangle ✓
(ii) Area=½×22×120=1320 m²
(iii) Annual=5000×1320=₹66,00,000; For 3 months=₹66,00,000×¼=₹16,50,000

Sides in ratio 3:5:7 and perimeter 300 m. (i) Find x and actual sides (ii) Find s (iii) Find s−a, s−b, s−c (iv) Find area (v) Find height on longest side. (NCERT Example 3 extended)

5 Mark

(i) 15x=300 → x=20. Sides: 60m, 100m, 140m.
(ii) s=150m
(iii) s−60=90, s−100=50, s−140=10
(iv) Area=√(150×90×50×10)=1500√3 m²
(v) 1500√3=½×140×h → h=3000√3/140=150√3/7 ≈ 37.1 m

Sides ratio 12:17:25, perimeter 540 cm. (i) Find sides (ii) Find s (iii) Find all differences (iv) Find area (v) Find height on side 170 cm. (NCERT Q5 extended)

5 Mark

(i) x=10. Sides: 120, 170, 250 cm.
(ii) s=270 cm
(iii) s−120=150, s−170=100, s−250=20
(iv) Area=√(270×150×100×20)=9000 cm²
(v) 9000=½×170×h → h=18000/170=1800/17 cm ≈ 105.9 cm

ABCD is a quadrilateral with AB=9m, BC=40m, CD=28m, DA=15m and ∠ABC=90°. Find the area of ABCD.

5 Mark

Step 1: Draw diagonal AC. In △ABC: ∠B=90°, AB=9, BC=40.
AC=√(9²+40²)=√(81+1600)=√1681=41 m. Area(△ABC)=½×9×40=180 m².
Step 2: △ACD: sides 41, 28, 15. s=(41+28+15)/2=42.
s−41=1, s−28=14, s−15=27.
Area=√(42×1×14×27)=√15876=126 m².
Total area = 180+126 = 306 m²

Derive Heron's formula for equilateral triangle. Verify with an equilateral triangle of side 8 cm using both Heron's method and the standard formula for equilateral triangle.

5 Mark

Derivation: All sides=a. s=3a/2. s−a=a/2 for each.
Area=√[(3a/2)(a/2)³]=√[3a⁴/16]=(√3/4)a²
Verify for a=8 cm:
Heron's: s=12; s−8=4; Area=√(12×4×4×4)=√768=8√12=16√3 cm²
Standard: (√3/4)×64=16√3 cm² ✓ Same!

An isosceles triangle has perimeter 30 cm and each equal side = 12 cm. (i) Find base (ii) Find area using Heron's formula (iii) Find height to the base (iv) Verify using ½×base×height.

5 Mark

(i) Base=30−12−12=6 cm
(ii) s=15; s−12=3, s−6=9. Area=√(15×3×3×9)=9√15 cm²
(iii) Area=½×6×h → h=2×9√15/6=3√15 cm
(iv) ½×6×3√15=9√15 cm² ✓ Same!

A farmer has a triangular field with sides 40 m, 32 m, 24 m. She wants to: (i) Find area (ii) Plant wheat on ½ the area at ₹50/m² cost (iii) Plant vegetables on other half at ₹80/m² cost. Find total expense.

5 Mark

(i) s=48; Area=√(48×8×24×16)=384 m²
(ii) Wheat area=192 m². Cost=192×50=₹9,600
(iii) Vegetable area=192 m². Cost=192×80=₹15,360
(iv) Total=9600+15360=₹24,960

A triangular park has sides 120 m, 80 m and 50 m. The park is to be divided into 4 equal triangular plots for 4 families. Find the area of each plot. Also find the cost of leveling each plot at ₹10/m².

5 Mark

s=125; Area=375√15 m² ≈ 1451.4 m²
Each plot = 375√15/4 m² ≈ 362.85 m²
Cost per plot = 10×375√15/4 = 937.5√15 ≈ ₹3628.5

Three triangles are given: (A) sides 40m, 32m, 24m (B) equilateral side 10 cm (C) isosceles 5,5,8 cm. Find area of each using Heron's formula. State in which case a simpler formula can also be used.

5 Mark

(A) 40,32,24: s=48; Area=√(48×8×24×16)=384 m². Also: right triangle (24²+32²=40²), so ½×32×24=384 m² is simpler.
(B) 10,10,10: s=15; Area=√(15×5×5×5)=25√3 cm². Also: (√3/4)×100=25√3 cm² is simpler.
(C) 5,5,8: s=9; s−5=4, s−8=1. Area=√(9×4×4×1)=√144=12 cm². No simpler formula here — Heron's is needed (or use height of isosceles via Pythagoras separately).