What is Heron's formula for the area of a triangle?

Area = √[s(s − a)(s − b)(s − c)], where a, b and c are the sides and s = (a + b + c)/2 is the semi-perimeter.

What is the area of a sector of a circle?

Area of a sector = (θ/360°) × πr², where θ is the angle of the sector and r the radius.

What is the curved surface area and volume of a cylinder?

Curved surface area = 2πrh; volume = πr²h, where r is the base radius and h the height.

What is the total surface area of a cone?

Total surface area = πr(l + r), where r is the base radius and l the slant height.

What is the volume of a sphere?

Volume of a sphere = (4/3)πr³, where r is the radius.

Samacheer Class 9 Maths - Mensuration

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Your Progress — Unit 7: Mensuration0% complete

Ex 7.1Application of Heron’s Formula26 questions

Validated & Corrected Answers

# Heron’s Formula

For a triangle with sides:
\[
a,\ b,\ c
\]

Semi-perimeter:

\[
s=\frac{a+b+c}{2}
\]

Area:

:contentReference[oaicite:0]{index=0}

Q.1 Using Heron’s formula, find the area ▾

✓ Solution

(i) Sides = 10 cm, 24 cm, 26 cm

Semi-perimeter:

\[
s=\frac{10+24+26}{2}
\]

\[
=\frac{60}{2}
\]

\[
=30
\]

Area:

\[
A
=
\sqrt{30(30-10)(30-24)(30-26)}
\]

\[
=
\sqrt{30\times20\times6\times4}
\]

\[
=
\sqrt{14400}
\]

\[
=120
\]

✓ Area:
\[
\boxed{120\text{ cm}^2}
\]

(ii) Sides = 1.8 m, 8 m, 8.2 m

Semi-perimeter:

\[
s
=
\frac{1.8+8+8.2}{2}
\]

\[
=\frac{18}{2}
\]

\[
=9
\]

Area:

\[
A
=
\sqrt{9(9-1.8)(9-8)(9-8.2)}
\]

\[
=
\sqrt{9\times7.2\times1\times0.8}
\]

\[
=
\sqrt{51.84}
\]

\[
=7.2
\]

✓ Area:
\[
\boxed{7.2\text{ m}^2}
\]

Q.2 Triangular ground with sides ▾

✓ Solution

\[
22\text{ m},\ 120\text{ m},\ 122\text{ m}
\]

Find area and levelling cost at ₹20/m².

Semi-perimeter:

\[
s
=
\frac{22+120+122}{2}
\]

\[
=\frac{264}{2}
\]

\[
=132
\]

Area:

\[
A
=
\sqrt{132(132-22)(132-120)(132-122)}
\]

\[
=
\sqrt{132\times110\times12\times10}
\]

\[
=
\sqrt{1742400}
\]

\[
=1320
\]

✓ Area:
\[
\boxed{1320\text{ m}^2}
\]

Cost of levelling

\[
\text{Cost}
=
1320\times20
\]

\[
=26400
\]

✓ Cost:
\[
\boxed{₹26,400}
\]

Q.3 Perimeter of triangular plot = 600 m ▾

✓ Solution

Sides are in ratio:
\[
5:12:13
\]

Let sides be:
\[
5x,\ 12x,\ 13x
\]

Then:

\[
5x+12x+13x=600
\]

\[
30x=600
\]

\[
x=20
\]

Sides:
\[
100,\ 240,\ 260
\]

Semi-perimeter:

\[
s=\frac{600}{2}=300
\]

Area:

\[
A
=
\sqrt{300(300-100)(300-240)(300-260)}
\]

\[
=
\sqrt{300\times200\times60\times40}
\]

\[
=
\sqrt{144000000}
\]

\[
=12000
\]

✓ Area:
\[
\boxed{12000\text{ m}^2}
\]

Q.4 Find area of equilateral triangle whose perimeter is 180 cm ▾

✓ Solution

Side:

\[
a=\frac{180}{3}=60\text{ cm}
\]

Area formula:

:contentReference[oaicite:1]{index=1}

\[
A
=
\frac{\sqrt3}{4}(60)^2
\]

\[
=
\frac{\sqrt3}{4}\times3600
\]

\[
=900\sqrt3
\]

✓ Area:
\[
\boxed{900\sqrt3\text{ cm}^2}
\]

Q.5 Advertisement board is an isosceles triangle ▾

✓ Solution

Perimeter:
\[
36\text{ m}
\]

Equal sides:
\[
13\text{ m},\ 13\text{ m}
\]

Third side:

\[
36-26=10\text{ m}
\]

Semi-perimeter:

\[
s=\frac{36}{2}=18
\]

Area:

\[
A
=
\sqrt{18(18-13)(18-13)(18-10)}
\]

\[
=
\sqrt{18\times5\times5\times8}
\]

\[
=
\sqrt{3600}
\]

\[
=60
\]

✓ Area:
\[
\boxed{60\text{ m}^2}
\]

Painting cost

\[
60\times17.50
\]

\[
=1050
\]

✓ Cost:
\[
\boxed{₹1050}
\]

Q.6 Find the area of unshaded region ▾

✓ Solution

Use:

Area of larger figure
Subtract shaded area

✓ Exact numerical answer requires the figure dimensions.

Q.7 Quadrilateral ABCD ▾

✓ Solution

Given:

\(AB=13\) cm
\(BC=12\) cm
\(CD=9\) cm
\(DA=14\) cm
Diagonal \(BD=15\) cm

Split quadrilateral into two triangles.

Area of ΔABD

Sides:
\[
13,\ 14,\ 15
\]

Semi-perimeter:

\[
s_1=\frac{13+14+15}{2}=21
\]

Area:

\[
A_1
=
\sqrt{21(8)(7)(6)}
\]

\[
=
\sqrt{7056}
\]

\[
=84
\]

Area of ΔBCD

Sides:
\[
12,\ 9,\ 15
\]

Semi-perimeter:

\[
s_2=\frac{12+9+15}{2}=18
\]

Area:

\[
A_2
=
\sqrt{18(6)(9)(3)}
\]

\[
=
\sqrt{2916}
\]

\[
=54
\]

Total area

\[
84+54=138
\]

✓ Area:
\[
\boxed{138\text{ cm}^2}
\]

Q.8 Park in shape of quadrilateral ▾

✓ Solution

Sides:
\[
15,\ 20,\ 26,\ 17
\]

Angle between first two sides is \(90^\circ\).

First triangle

\[
A_1
=
\frac12\times15\times20
\]

\[
=150
\]

Diagonal:

\[
\sqrt{15^2+20^2}
=
25
\]

Second triangle

Sides:
\[
25,\ 26,\ 17
\]

Semi-perimeter:

\[
s=\frac{25+26+17}{2}=34
\]

Area:

\[
A_2
=
\sqrt{34(9)(8)(17)}
\]

\[
=
\sqrt{41616}
\]

\[
=204
\]

Total area

\[
150+204=354
\]

✓ Area:
\[
\boxed{354\text{ m}^2}
\]

Q.9 Rhombus with perimeter 160 m and one diagonal 48 m ▾

✓ Solution

Side:

\[
a=\frac{160}{4}=40
\]

Half diagonal:

\[
24
\]

Let other half diagonal be \(x\).

Using Pythagoras theorem:

::contentReference[oaicite:2]{index=2}

\[
24^2+x^2=40^2
\]

\[
576+x^2=1600
\]

\[
x^2=1024
\]

\[
x=32
\]

Other diagonal:

\[
64
\]

Area of rhombus:

\[
A=\frac12 d_1d_2
\]

\[
=\frac12(48)(64)
\]

\[
=1536
\]

✓ Area:
\[
\boxed{1536\text{ m}^2}
\]

Q.10 A parallelogram has adjacent sides 34 m and 20 m and a diagonal of length 42 m. Find the area of the parallelogram. ▾

The diagonal divides the parallelogram into two congruent triangles with sides 34 m, 20 m and 42 m.

Semi-perimeter s = (34 + 20 + 42)/2 = 48

Area of one triangle = √[48(48−34)(48−20)(48−42)] = √[48·14·28·6] = 336 m²

Parallelogram area = 2 × 336 = 672 m²

Answer: 672 m²

Q.1 Cuboid dimensions: Length = 20 cm, Breadth = 15 cm, Height = 8 cm. Find the Total Surface Area (TSA) and Lateral Surface Area (LSA). ▾

TSA = 2(lb + bh + hl) = 2[(20×15) + (15×8) + (8×20)] = 2(300 + 120 + 160) = 2(580) = 1160 cm²

LSA = 2h(l + b) = 2×8×(20 + 15) = 16×35 = 560 cm²

Answers: TSA = 1160 cm², LSA = 560 cm²

Q.2 A cuboidal box measures 6 m × 400 cm × 1.5 m. Find its Total Surface Area (in m²). Also, find the cost of painting the box at the rate of ₹22 per m². ▾

Convert 400 cm = 4 m. So l = 6 m, b = 4 m, h = 1.5 m.

TSA = 2(lb + bh + hl) = 2[(6×4) + (4×1.5) + (1.5×6)] = 2(24 + 6 + 9) = 2(39) = 78 m²

Cost = 78 × 22 = ₹1716

Answers: TSA = 78 m²; Cost = ₹1716

Q.3 A hall measures 10 m (length) × 9 m (breadth) and has height 8 m. Find the cost of whitewashing its four walls and ceiling at the rate of ₹8.50 per m². ▾

Area of four walls (LSA) = 2h(l + b) = 2×8×(10 + 9) = 16×19 = 304 m²

Area of ceiling = l × b = 10 × 9 = 90 m²

Total area = 304 + 90 = 394 m²

Cost = 394 × 8.5 = ₹3349

Answers: Area = 394 m²; Cost = ₹3349

Q.4 Find TSA and LSA of cube ▾

✓ Solution

(i) Side = 8 m

TSA

\[
6(8^2)
=
6(64)
=
384
\]

✓ TSA:
\[
\boxed{384\text{ m}^2}
\]

LSA

\[
4(8^2)
=
4(64)
=
256
\]

✓ LSA:
\[
\boxed{256\text{ m}^2}
\]

(ii) Side = 21 cm

TSA

\[
6(21^2)
=
6(441)
=
2646
\]

✓ TSA:
\[
\boxed{2646\text{ cm}^2}
\]

LSA

\[
4(21^2)
=
4(441)
=
1764
\]

✓ LSA:
\[
\boxed{1764\text{ cm}^2}
\]

(iii) Side = 7.5 cm

TSA

\[
6(7.5^2)
=
6(56.25)
=
337.5
\]

✓ TSA:
\[
\boxed{337.5\text{ cm}^2}
\]

LSA

\[
4(7.5^2)
=
4(56.25)
=
225
\]

✓ LSA:
\[
\boxed{225\text{ cm}^2}
\]

Q.5 Total surface area of cube = 2400 cm² ▾

✓ Solution

Find LSA.

Using:
\[
6a^2=2400
\]

\[
a^2=400
\]

Now:

\[
LSA=4a^2
\]

\[
=4(400)
\]

\[
=1600
\]

✓ LSA:
\[
\boxed{1600\text{ cm}^2}
\]

Q.6 Cubical container of side 6.5 m ▾

✓ Solution

Find area to be painted and painting cost.

Total Surface Area

\[
TSA
=
6(6.5)^2
\]

\[
=
6(42.25)
\]

\[
=253.5
\]

✓ Area:
\[
\boxed{253.5\text{ m}^2}
\]

Painting cost

Rate:
\[
₹24/\text{m}^2
\]

\[
253.5\times24
\]

\[
=6084
\]

✓ Cost:
\[
\boxed{₹6084}
\]

Q.7 Three identical cubes of side 4 cm are joined end to end to form a cuboid. Find the Total Surface Area (TSA) and Lateral Surface Area (LSA) of the resulting cuboid. ▾

Resulting cuboid dimensions: length = 3×4 = 12 cm, breadth = 4 cm, height = 4 cm.

TSA = 2(lb + bh + hl) = 2[(12×4) + (4×4) + (4×12)] = 2(48 + 16 + 48) = 2(112) = 224 cm²

LSA = 2h(l + b) = 2×4×(12 + 4) = 8×16 = 128 cm²

Answers: TSA = 224 cm²; LSA = 128 cm²

Q.1 Find the volume of the cuboid ▾

✓ Solution

(i)

Dimensions:

Length = 12 cm
Breadth = 8 cm
Height = 6 cm

\[
V=l\times b\times h
\]

\[
=12\times8\times6
\]

\[
=576
\]

✓ Volume:
\[
\boxed{576\text{ cm}^3}
\]

(ii)

Dimensions:

Length = 60 m
Breadth = 25 m
Height = 1.5 m

\[
V=60\times25\times1.5
\]

\[
=1500\times1.5
\]

\[
=2250
\]

✓ Volume:
\[
\boxed{2250\text{ m}^3}
\]

Q.2 A match box measures 6 cm × 3.5 cm × 2.5 cm. Find the volume of a packet containing 12 such boxes. ▾

Volume of one box = 6 × 3.5 × 2.5 = 52.5 cm³

Volume of 12 boxes = 12 × 52.5 = 630 cm³

Answer: 630 cm³

Q.3 Dimensions of chocolate box are in ratio ▾

✓ Solution

\[
5:4:3
\]

Volume:
\[
7500\text{ cm}^3
\]

Let dimensions be:
\[
5x,\ 4x,\ 3x
\]

Then:

\[
(5x)(4x)(3x)=7500
\]

\[
60x^3=7500
\]

\[
x^3=125
\]

\[
x=5
\]

Thus dimensions:

\[
5x=25
\]

\[
4x=20
\]

\[
3x=15
\]

✓ Dimensions:
\[
\boxed{25\text{ cm},\ 20\text{ cm},\ 15\text{ cm}}
\]

Q.4 A rectangular pond is 20.5 m long, 16 m wide and 8 m deep. Find its capacity in litres. ▾

Volume = 20.5 × 16 × 8 = 2624 m³

1 m³ = 1000 litres, so capacity = 2624 × 1000 = 2,624,000 litres

Answer: 2,624,000 litres

Q.5 Bricks measure 24 cm × 12 cm × 8 cm. A wall is 20 m long, 48 cm thick and 6 m high. Find the number of bricks required (ignore mortar). ▾

Convert lengths to cm: 20 m = 2000 cm, height 6 m = 600 cm. Wall thickness = 48 cm.

Volume of one brick = 24 × 12 × 8 = 2304 cm³

Volume of wall = 2000 × 48 × 600 = 57,600,000 cm³

Number of bricks = 57,600,000 / 2304 = 25,000

Answer: 25,000 bricks

Q.6 The volume of a container is 1440 m³. If its length is 15 m and breadth is 8 m, find its height. ▾

V = l × b × h ⇒ 1440 = 15 × 8 × h = 120 h

h = 1440 / 120 = 12 m

Answer: 12 m

Q.7 Find volume of cube ▾

✓ Solution

(i) Side = 5 cm

\[
V=5^3
\]

\[
=125
\]

✓ Volume:
\[
\boxed{125\text{ cm}^3}
\]

(ii) Side = 3.5 m

\[
V=(3.5)^3
\]

\[
=42.875
\]

✓ Volume:
\[
\boxed{42.875\text{ m}^3}
\]

(iii) Side = 21 cm

\[
V=21^3
\]

\[
=9261
\]

✓ Volume:
\[
\boxed{9261\text{ cm}^3}
\]

Q.8 Cubical milk tank holds ▾

✓ Solution

\[
125000\text{ litres}
\]

Find side length.

Convert litres to cubic metres:

\[
125000\text{ litres}=125\text{ m}^3
\]

Let side be \(a\).

\[
a^3=125
\]

\[
a=5
\]

✓ Side length:
\[
\boxed{5\text{ m}}
\]

Q.9 Metallic cube side = 15 cm ▾

✓ Solution

Melted into cuboid.

Cuboid dimensions:

Length = 25 cm
Height = 9 cm
Breadth = ?

Volume of cube

\[
15^3
\]

\[
=3375
\]

Volume of cuboid

\[
25\times b\times9
\]

Since volumes are equal:

\[
25\times b\times9=3375
\]

\[
225b=3375
\]

\[
b=15
\]

✓ Breadth:
\[
\boxed{15\text{ cm}}
\]

Ex 7.4Multiple Choice Questions10 questions

Mensuration – Validated & Corrected Answers

Q.1 Semi-perimeter of triangle with sides ▾

✓ Solution

\[
15\text{ cm},\ 20\text{ cm},\ 25\text{ cm}
\]

Semi-perimeter:

\[
s=\frac{15+20+25}{2}
\]

\[
=\frac{60}{2}
\]

\[
=30
\]

✓ Answer:
\[
\boxed{(3)\ 30\text{ cm}}
\]

Q.2 Area of triangle with sides ▾

✓ Solution

\[
3\text{ cm},\ 4\text{ cm},\ 5\text{ cm}
\]

Using Heron’s formula:

:contentReference[oaicite:0]{index=0}

Semi-perimeter:

\[
s=\frac{3+4+5}{2}=6
\]

Area:

\[
A
=
\sqrt{6(6-3)(6-4)(6-5)}
\]

\[
=
\sqrt{6\times3\times2\times1}
\]

\[
=
\sqrt{36}
\]

\[
=6
\]

✓ Answer:
\[
\boxed{(2)\ 6\text{ cm}^2}
\]

Q.3 Perimeter of equilateral triangle = 30 cm ▾

✓ Solution

\[
3a=30
\]

\[
a=10
\]

Area formula:

:contentReference[oaicite:1]{index=1}

\[
A
=
\frac{\sqrt3}{4}(10)^2
\]

\[
=
25\sqrt3
\]

✓ Answer:
\[
\boxed{(4)\ 25\sqrt3\text{ cm}^2}
\]

Q.4 Lateral surface area of cube of side 12 cm ▾

✓ Solution

Formula:

\[
LSA=4a^2
\]

\[
=4(12^2)
\]

\[
=4(144)
\]

\[
=576
\]

✓ Answer:
\[
\boxed{(3)\ 576\text{ cm}^2}
\]

Q.5 If lateral surface area of cube is 600 cm² ▾

✓ Solution

\[
4a^2=600
\]

Then:

\[
a^2=150
\]

Total surface area:

\[
TSA=6a^2
\]

\[
=6(150)
\]

\[
=900
\]

✓ Answer:
\[
\boxed{(3)\ 900\text{ cm}^2}
\]

Q.6 Total surface area of cuboid ▾

✓ Solution

\[
10\text{ cm}\times6\text{ cm}\times5\text{ cm}
\]

Formula:

:contentReference[oaicite:2]{index=2}

\[
=
2[(10\times6)+(6\times5)+(5\times10)]
\]

\[
=
2(60+30+50)
\]

\[
=
2(140)
\]

\[
=280
\]

✓ Answer:
\[
\boxed{(1)\ 280\text{ cm}^2}
\]

Q.7 Ratio of sides of two cubes = 2:3 ▾

✓ Solution

Surface area ratio:

\[
2^2:3^2
\]

\[
4:9
\]

✓ Answer:
\[
\boxed{(2)\ 4:9}
\]

Q.8 Volume of cuboid = 660 cm³ ▾

✓ Solution

Area of base = 33 cm²

Find height.

Using:

\[
V=\text{Base area}\times h
\]

\[
660=33h
\]

\[
h=20
\]

✓ Answer:
\[
\boxed{(3)\ 20\text{ cm}}
\]

Q.9 Capacity of water tank ▾

✓ Solution

\[
10\text{ m}\times5\text{ m}\times1.5\text{ m}
\]

Volume:

\[
V=10\times5\times1.5
\]

\[
=75\text{ m}^3
\]

Since:
\[
1\text{ m}^3=1000\text{ litres}
\]

Capacity:

\[
75\times1000
\]

\[
=75000
\]

✓ Correct Answer:
\[
\boxed{(4)\ 75000\text{ litres}}
\]

❗ Correction:
The originally marked answer \((3)\ 7500\text{ litres}\) is incorrect.

Q.10 Number of bricks required ▾

✓ Solution

Brick dimensions:
\[
50\text{ cm}\times30\text{ cm}\times20\text{ cm}
\]

Wall dimensions:
\[
5\text{ m}\times3\text{ m}\times2\text{ m}
\]

Convert wall dimensions to cm:

\[
500\text{ cm}\times300\text{ cm}\times200\text{ cm}
\]

Volume of wall

\[
500\times300\times200
\]

\[
=30000000
\text{ cm}^3
\]

Volume of one brick

\[
50\times30\times20
\]

\[
=30000
\text{ cm}^3
\]

Number of bricks

\[
\frac{30000000}{30000}
\]

\[
=1000
\]

✓ Answer:
\[
\boxed{(1)\ 1000}
\]

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Samacheer Class 9 Maths - Mensuration

Mensuration — key concepts & quick answers

Validated & Corrected Answers

(i) Sides = 10 cm, 24 cm, 26 cm

(ii) Sides = 1.8 m, 8 m, 8.2 m

Cost of levelling

Painting cost

Area of ΔABD

Area of ΔBCD

Total area

First triangle

Second triangle

Total area

(i) Side = 8 m

TSA

LSA

(ii) Side = 21 cm

TSA

LSA

(iii) Side = 7.5 cm

TSA

LSA

Total Surface Area

Painting cost

(i)

(ii)

(i) Side = 5 cm

(ii) Side = 3.5 m

(iii) Side = 21 cm

Volume of cube

Volume of cuboid

Mensuration – Validated & Corrected Answers

Volume of wall

Volume of one brick

Number of bricks

Revise Mensuration with confidence.