Samacheer Class 9 Maths - Mensuration

(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:
\[
\boxed{7.2\text{ m}^2}
\]

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

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:
\[
\boxed{₹26,400}
\]

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

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

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:
\[
\boxed{₹1050}
\]

Use:

• Area of larger figure
• Subtract shaded area

✓ Exact numerical answer requires the figure dimensions.

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:
\[
\boxed{138\text{ cm}^2}
\]

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:
\[
\boxed{354\text{ m}^2}
\]

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:
\[
\boxed{1536\text{ m}^2}
\]

\[
34\text{ m},\ 20\text{ m}
\]

Diagonal:
\[
42\text{ m}
\]

Diagonal divides parallelogram into two congruent triangles.

Triangle sides:
\[
34,\ 20,\ 42
\]

Semi-perimeter:

\[
s=\frac{34+20+42}{2}=48
\]

Area of triangle:

\[
A
=
\sqrt{48(14)(28)(6)}
\]

\[
=
\sqrt{112896}
\]

\[
=336
\]

Area of parallelogram:

\[
2\times336
\]

\[
=672
\]

✓ Area:
\[
\boxed{672\text{ m}^2}
\]

# Surface Area of Cuboid and Cube
# Exercise 7.2

Validated & Corrected Answers

# Important Formulae

Cuboid

Total Surface Area (TSA)

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

Lateral Surface Area (LSA)

\[
LSA=2h(l+b)
\]

Cube

Total Surface Area

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

Lateral Surface Area

\[
LSA=4a^2
\]

• Length = 20 cm
• Breadth = 15 cm
• Height = 8 cm

Find TSA and LSA.

Total Surface Area

\[
TSA
=
2(lb+bh+hl)
\]

\[
=
2[(20\times15)+(15\times8)+(8\times20)]
\]

\[
=
2[300+120+160]
\]

\[
=
2(580)
\]

\[
=1160
\]

✓ TSA:
\[
\boxed{1160\text{ cm}^2}
\]

Lateral Surface Area

\[
LSA
=
2h(l+b)
\]

\[
=
2(8)(20+15)
\]

\[
=
16\times35
\]

\[
=560
\]

✓ LSA:
\[
\boxed{560\text{ cm}^2}
\]

\[
6\text{ m}\times400\text{ cm}\times1.5\text{ m}
\]

Convert:
\[
400\text{ cm}=4\text{ m}
\]

Thus:

• \(l=6\) m
• \(b=4\) m
• \(h=1.5\) m

Total Surface Area

\[
TSA
=
2(lb+bh+hl)
\]

\[
=
2[(6\times4)+(4\times1.5)+(1.5\times6)]
\]

\[
=
2(24+6+9)
\]

\[
=
2(39)
\]

\[
=78
\]

✓ Area:
\[
\boxed{78\text{ m}^2}
\]

Cost of painting

Rate:
\[
₹22/\text{m}^2
\]

\[
\text{Cost}
=
78\times22
\]

\[
=1716
\]

✓ Cost:
\[
\boxed{₹1716}
\]

• Length = 10 m
• Breadth = 9 m
• Height = 8 m

Find cost of whitewashing walls and ceiling.

Area of four walls

\[
LSA
=
2h(l+b)
\]

\[
=
2(8)(10+9)
\]

\[
=
16\times19
\]

\[
=304
\]

Area of ceiling

\[
10\times9=90
\]

Total area

\[
304+90=394
\]

✓ Area to be whitewashed:
\[
\boxed{394\text{ m}^2}
\]

Cost

Rate:
\[
₹8.50/\text{m}^2
\]

\[
394\times8.5
\]

\[
=3349
\]

✓ Cost:
\[
\boxed{₹3349}
\]

(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:
\[
\boxed{225\text{ cm}^2}
\]

Find LSA.

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

\[
a^2=400
\]

Now:

\[
LSA=4a^2
\]

\[
=4(400)
\]

\[
=1600
\]

✓ LSA:
\[
\boxed{1600\text{ cm}^2}
\]

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:
\[
\boxed{₹6084}
\]

Resulting cuboid dimensions:

• Length = \(3\times4=12\) cm
• Breadth = 4 cm
• Height = 4 cm

Total Surface Area

\[
TSA
=
2(lb+bh+hl)
\]

\[
=
2[(12\times4)+(4\times4)+(4\times12)]
\]

\[
=
2(48+16+48)
\]

\[
=
2(112)
\]

\[
=224
\]

✓ TSA:
\[
\boxed{224\text{ cm}^2}
\]

Lateral Surface Area

\[
LSA
=
2h(l+b)
\]

\[
=
2(4)(12+4)
\]

\[
=
8(16)
\]

\[
=128
\]

✓ LSA:
\[
\boxed{128\text{ cm}^2}
\]

# Volume of Cuboid and Cube
# Exercise 7.3

Validated & Corrected Answers

# Important Formulae

Volume of Cuboid

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

Volume of Cube

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

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

\[
6\text{ cm}\times3.5\text{ cm}\times2.5\text{ cm}
\]

Find volume of packet containing 12 boxes.

Volume of one match box

\[
V=6\times3.5\times2.5
\]

\[
=52.5
\]

\[
=52.5\text{ cm}^3
\]

Volume of 12 boxes

\[
12\times52.5
\]

\[
=630
\]

✓ Volume:
\[
\boxed{630\text{ cm}^3}
\]

\[
5:4:3
\]

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

• Length = 20.5 m
• Breadth = 16 m
• Depth = 8 m

Find capacity in litres.

Volume of pond

\[
V=20.5\times16\times8
\]

\[
=328\times8
\]

\[
=2624
\]

\[
=2624\text{ m}^3
\]

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

Capacity:

\[
2624\times1000
\]

\[
=2624000
\]

✓ Capacity:
\[
\boxed{2624000\text{ litres}}
\]

\[
24\text{ cm}\times12\text{ cm}\times8\text{ cm}
\]

Wall dimensions:

• Length = 20 m
• Breadth = 48 cm
• Height = 6 m

Find number of bricks.

Convert to cm:
\[
20\text{ m}=2000\text{ cm}
\]

\[
6\text{ m}=600\text{ cm}
\]

Volume of one brick

\[
24\times12\times8
\]

\[
=2304\text{ cm}^3
\]

Volume of wall

\[
2000\times48\times600
\]

\[
=57600000\text{ cm}^3
\]

Number of bricks

\[
\frac{57600000}{2304}
\]

\[
=25000
\]

✓ Number of bricks:
\[
\boxed{25000}
\]

\[
1440\text{ m}^3
\]

Length:
\[
15\text{ m}
\]

Breadth:
\[
8\text{ m}
\]

Find height.

Using:
\[
V=lbh
\]

\[
1440=15\times8\times h
\]

\[
1440=120h
\]

\[
h=12
\]

✓ Height:
\[
\boxed{12\text{ m}}
\]

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

\[
125000\text{ litres}
\]

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:
\[
\boxed{5\text{ m}}
\]

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:
\[
\boxed{15\text{ cm}}
\]

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

Semi-perimeter:

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

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

\[
=30
\]

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

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

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

\[
3a=30
\]

\[
a=10
\]

Area formula:

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

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

\[
=
25\sqrt3
\]

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

Formula:

\[
LSA=4a^2
\]

\[
=4(12^2)
\]

\[
=4(144)
\]

\[
=576
\]

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

\[
4a^2=600
\]

Then:

\[
a^2=150
\]

Total surface area:

\[
TSA=6a^2
\]

\[
=6(150)
\]

\[
=900
\]

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

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

Formula:

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

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

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

\[
=
2(140)
\]

\[
=280
\]

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

Surface area ratio:

\[
2^2:3^2
\]

\[
4:9
\]

✓ Answer:
\[
\boxed{(2)\ 4:9}
\]

Area of base = 33 cm²

Find height.

Using:

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

\[
660=33h
\]

\[
h=20
\]

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

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

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.

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:
\[
\boxed{(1)\ 1000}
\]