| Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
Exercise 14.1
Question 1
Three cubes each of side 5 cm are joined end to end to form a cuboid. Find the surface area of the resulting cuboid.Sol :
Length of each side of cube = 5cm
According to question,
Three cubes are joined end to end to form a cuboid
So, the length of the resulting cuboid = 5 + 5 + 5 = 15cm
But, the breadth and height will remain the same
So, Breadth = 5cm
and height = 5cm
Surface Area of the Cuboid
= 2(Length × Breadth + Breadth × Height + Height × Length)
= 2(lb + bh + hl)
= 2(15×5 + 5×5 + 5×15)
= 2(75 + 25 + 75)
= 2(175)
= 350cm2
Hence, the surface area of cuboid is 350cm2.
Question 2
Cardboard boxes of two different sizes are made. The bigger has dimensions 20cm, 15cm and 5cm and the smaller dimensions 16cm, 12cm and 4cm. 5% of the total surface area is required extra for all overlaps. If the cost of the cardboard is Rs. 20 for one square metre, find the cost of the cardboard for supplying 200 boxes of each kind.Sol :
For bigger box:
Length of the bigger box = 20cm
Breadth of the bigger box = 15cm
Height of the bigger box = 5cm
So, Total Surface Area of bigger box = 2(lb + bh + hl)
= 2(20×15 + 15×5 + 5×20)
= 2(300 + 75 + 100)
= 2(475)
= 950cm2
For Smaller box:
Length of the smaller box = 16cm
Breadth of the smaller box = 12cm
Height of the smaller box = 4cm
So, Total Surface Area of bigger box = 2(lb + bh + hl)
= 2(16×12 + 12×4 + 4×16)
= 2(192 + 48 + 64)
= 2(304)
= 608cm2
Total surface area of 200 boxes of each type = 200(950 +608)
= 200 × 1558
= 311600cm2
Extra Area Required = 5% of (950 + 608)
$=\frac{5}{100} \times 1558 \times 200$
= 15580cm2
So, Total cardboard Required = 311600 + 15580
= 327180cm2
$=\frac{327180}{1000}=327.18 \mathrm{m}^{2}$
Cost of Cardboard for 1m2 = Rs 20
Cost of Cardboard for 327.18 m2 = 20 × 327.18
= Rs 654.36
Hence, the cost of the cardboard for supplying 200 boxes of each kind is Rs 654.36.
Question 3
The length of cold storage is double its breadth. Its height is 3 metres. The area of its four walls (including doors) is 108 m2. Find its volume.Sol :
Given: Area = 108m2, Height = 3m
and Length of a cold storage is double its breadth
So, Let breadth of a cold storage = x
∴ length of the cold storage = 2x
Area of four walls = 108m2
∴ 2(l + b)× h = 108
⇒ 2 (x + 2x)×3 = 108
⇒ 6(3x) = 108
$\Rightarrow \mathrm{x}=\frac{108}{18}$
⇒ x = 6
∴ Breadth of a cold storage = 6m
and Length of a cold storage = 2×6 = 12m
Hence, Volume of the cold storage = l × b × h
= 12 × 6 × 3
= 216m3
Question 4
Find:(i) the lateral surface,
(ii) the whole surface, and
(iii) the volume of a right circular cylinder whose height is 13.5 cm and radius of the base 7cm
Sol :
Given: Radius of base, r = 7cm
Height , h = 13.5cm
(i) Lateral Surface Area of right circular cylinder = 2πrh
$=2 \times \frac{22}{7} \times 7 \times 13.5$
= 2×22×13.5
= 594cm2
(ii) Total Surface Area of cylinder = 2πrh + 2πr2
= 2πr(h + r)
$=2 \times \frac{22}{7} \times 7(13.5+7)$
= 2×22×20.5
= 902cm2
(iii) Volume of cylinder = πr2h
$=\frac{22}{7} \times 7 \times 7 \times 13.5$
= 22×7×13.5
= 2079cm3
Question 5
The radius and height of a right circular cone are in the ratio of 5:12. If its volume is 314 cm3, find its slant height.
[Take π =3.14]Sol :
Given: Volume of a right circular cone = 314cm3
and r : h = 5 : 12
Let r = 5x and h = 12x
The volume of a right circular cone$=\frac{1}{3} \pi r^{2} h$
$\Rightarrow 314=\frac{1}{3} \times 3.14 \times(5 \mathrm{x})^{2} \times 12 \mathrm{x}$
$\Rightarrow \frac{314 \times 3}{3.14}=25 \mathrm{x}^{2} \times 12 \mathrm{x}$
$\Rightarrow \frac{314 \times 3 \times 100}{314 \times 25 \times 12}=\mathrm{x}^{3}$
⇒ x3 = 1Hence, x = 1
∴ r = 5x = 5×1 = 5cm
and h = 12x = 12×1 = 12cm
Now, we have to find the slant height, l
Slant height, l = √(r2 + h2)
= √{(5)2 + (12)2}
= √25+144
= √169
= ±13
= 13cm
[taking positive root, because slant height can’t be negative]
∴ Slant Height = 13cm
Question 6
A cylinder, whose height is two-thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.Sol :
Let height of a cylinder = h
and diameter = d
According to question,
$\mathrm{h}=\frac{2}{3} \mathrm{d}$
$\Rightarrow \mathrm{h}=\frac{2}{3} \times 2 \mathrm{r}$
$\Rightarrow \mathrm{h}=\frac{4}{3} \mathrm{r}$
Volume of Cylinder = Volume of Sphere (Given)
⇒ r3 = 43
⇒ r = 4cm
Hence, Radius of base of the cylinder is 4cm.
$\pi r^{2} h=\frac{4}{3} \pi R^{3}$
$\Rightarrow r^{2} \times \frac{4}{3} r=\frac{4}{3} R^{3}$
⇒ r3 = R3⇒ r3 = 43
⇒ r = 4cm
Hence, Radius of base of the cylinder is 4cm.
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
No comments:
Post a Comment