Lesson 6 Prism
CDC Class 9 Maths Solution
Cross-section area of prism
Lateral surface area of prism
Total surface area of prism
Volume of prism
Exercise 6.1
1. What is the volume of prism whose cross-sectional area is 35 cm² and height is 10 cm?
Solution:
Here,
Cross sectional area of the prism (A) = 35 cm²
Height of the prism (h) = 10 cm
We know that,
Volume of the prism (V) = A × h
= 35 × 10
= 350 cm³
∴ Volume of the prism is 350 cm³.
2. Find the lateral surface area of the prism if the perimeter of the base is 36 cm and height of 8 cm.
Solution:
Here,
Perimeter of the base (P) = 36 cm
Height of the prism (h) = 8 cm
We know that,
Lateral surface area of the prism = P × h
= 36 × 8
= 288 cm²
∴ Lateral surface area of the prism is 288 cm².
3. Find the total surface area of a prism if cross-sectional area and lateral surface area are 40 cm² and 175 cm² respectively.
Solution:
Here,
Cross sectional area of the prism (A) = 40 cm²
Lateral surface area of the prism (LSA) = 175 cm²
We know that,
Total surface area of the prism (TSA) = 2 × A + LSA
= 2 × 40 + 175
= 80 + 175
= 255 cm²
∴ Total surface area of the prism is 255 cm².
4. Find the cross-sectional area, lateral surface area, total surface area, and volume of the following prisms.
(a)
Solution:
Here,
In given prism,
length (l) = 10 cm, breadth (b) = 6 cm and height (h) = 8 cm
We know that,
Cross sectional area of prism = base area of prism
= l × b = 10 × 6 = 60 cm²
Again, perimeter of the base (P) = 2(l + b) = 2(10 + 6) cm = 32 cm
Lateral surface area of prism = P × h = 32 cm × 8 cm = 256 cm²
Total surface area of the prism (TSA) = 2 × A + LSA
= 2 × 60 + 256
= 120 + 256
= 376 cm²
Volume of the prism (V) = A × h
= 60 × 8
= 480 cm³
∴ Cross sectional area of the prism is 60 cm², lateral surface area of the prism is 256 cm², total surface area of the prism is 376 cm² and volume of the prism is 480 cm³.
(b)
Solution:
Here,
In given prism,
length (l) = 5 cm, breadth (b) = 5 cm and height (h) = 5 cm
We know that,
Cross sectional area of prism = base area of prism
= l × b = 5 × 5 = 25 cm²
Again, perimeter of the base (P) = 2(l + b) = 2(5 + 5) cm = 20 cm
Lateral surface area of prism = P × h = 20 cm × 5 cm = 100 cm²
Total surface area of the prism (TSA) = 2 × A + LSA
= 2 × 25 + 100
= 50 + 100
= 150 cm²
Volume of the prism (V) = A × h
= 25 × 5
= 125 cm³
∴ Cross sectional area of the prism is 25 cm², lateral surface area of the prism is 100 cm², total surface area of the prism is 150 cm² and volume of the prism is 125 cm³.
(c)
Solution:
In the given figure,
Cross sectional area (A) = (9 x 4) + (4 x 3) = 36 + 12 = 48 cm²
Lateral surface area (LSA) = (9 + 4 + 3 + 4 + 3 + 4 + 3 + 4) x 6 = 34 x 6 = 204 cm²
Total surface area (TSA) = 2 x A + LSA = 2 x 48 + 204 = (96 + 204) cm² = 300 cm²
Volume (V) = A x h = 48 x 6 = 288 cm³
(c)
Solution:
In the given figure,
Cross sectional area (A) = (10 x 3) + (4 x 3) = 36 + 12 = 48 cm²
Lateral surface area (LSA) = (9 + 4 + 3 + 4 + 3 + 4 + 3 + 4) x 6 = 34 x 6 = 204 cm²
Total surface area (TSA) = 2 x A + LSA = 2 x 48 + 186 = (96 + 204) cm² = 300 cm²
Volume (V) = A x h = 48 x 6 = 288 cm³
(d)
Solution:
In the given figure,
Cross sectional area (A) = (10 x 3) + (4 x 4) + (4 x 4)= 30 + 16 + 16= 62 cm²
Lateral surface area (LSA) = (4 + 4 + 3 + 3 + 3 + 4 + 4 + 4 + 3 + 3 + 3 + 4) = 42 x 8 = 336 cm²
Total surface area (TSA) = 2 x A + LSA = 2 x 62 + 336 = (124 + 336) cm² = 460 cm²
Volume (V) = A x h = 62 x 8 = 496 cm³
(e)
Solution:
Here,
(i) The area of cross section = area of rectangle + area of triangle
The area of rectangle = l × b = 4 × 6 = 24 cm²
Semi-perimeter of Δ = (a + b + c) / 2 = (3 m + 3 m + 4 m) / 2 = 5 m
Area of Δ = √[s(s - a)(s - b)(s - c)] = √[5(5 - 3)(5 - 3)(5 - 4)] = √[20] = 4.47 m²
∴ The area of cross section (A) = area of rectangle + area of Δ
= 24 m² + 4.47 m² = 28.47 m²
(ii) Perimeter of the cross section (P) = 4 m + 6 m + 3 m + 3 m + 6 m = 22 m
Length of prism (h) = 10 m
∴ Lateral surface area (L.S.A.) = P × h = 22 m × 10 m = 220 m²
(iii) Total surface area of prism = L.S.A. + 2A = 220 m² + 2 × 28.47 m² = 276.94 m²
(iv) Volume of the prism (V) = A × h = 28.47 m² × 10 m = 284.7 m³
(f)
Solution:
Here,
(i) The area of cross section = (30 x 10) + (20 x 10) + (10 x 10) = 600 cm²
(ii) Perimeter of the cross section (P) = 30 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm = 120 cm
Length of prism (h) = 30 cm
∴ Lateral surface area (L.S.A.) = P × h = 120 cm × 30 cm = 3600 cm²
(iii) Total surface area of prism = L.S.A. + 2A = 3600 cm² + 2 × 300 cm² = 4200 cm²
(iv) Volume of the prism (V) = A × h = 600 cm² × 30 cm = 18000 cm³
∴ The cross sectional area of the prism is 300 cm², lateral surface area of the prism is 3600 cm², total surface area of the prism is 4200 cm² and volume of the prism is 18000 cm³.
5. If there is a cuboidal prism having 6 cm length, 5 cm breadth, and 8 cm height:
(a) Find the area of cross section?
Solution:
Here,
The cross section of the prism = area of rectangle with length 6 cm and breadth 5 cm
∴ The area of cross section (A) = l × b = 6 cm × 5 cm = 30 cm²
(b) Find the lateral surface area?
Solution:
Here,
Lateral surface area (L.S.A.) = 2h (l + b) = 2 x 8(6 + 5) = 16 x 11 = 176 cm²
(c) Find the volume?
Solution:
Here,
Volume of the prism (V) = A × h = 30 cm² × 8 cm = 240 cm³
6. A rectangular tank is 2 m long, 1.5 m broad, and 1 m high. Find the capacity of the tank in liters.
Solution:
Here,
Volume of the tank = l × b × h = 2 m × 1.5 m × 1 m = 3 m³
We know that,
1 m³ = 1000 liters
∴ Capacity of the tank = 3 m³ × 1000 = 3000 liters
7. Divide a timber cube of 20 cm long into 8 equal parts. Find the length of each piece.
Solution:
Here,
To divide a timber cube with a side length of 20 cm into 8 equal parts, we split it into 8 smaller cubes (2 × 2 × 2).
Original side length = 20 cm
New side length of each smaller cube = 20 cm / 2 = 10 cm
We know that,
Original volume = 20 cm × 20 cm × 20 cm = 8000 cm³
Volume of each smaller cube = 10 cm × 10 cm × 10 cm = 1000 cm³
Total volume of 8 smaller cubes = 8 × 1000 cm³ = 8000 cm³ (matches the original volume)
∴ Length of each piece = 10 cm
8. Find the cross-sectional area, lateral surface area, total surface area, and volume of the following prism.
Solution:
a)
The trianguar base is a right angled triangle.
Here, AB = 3 cm and BC = 4 cm
By using Pythagoras theorem, CA = √[AB² + BC²] = √[3² + 4²] = √[25] = 5 cm
Now, perimeter of the triangular base (P) = AB + BC + CA = 3 cm + 4 cm + 5 cm = 12 cm
(i) Area of triangular base (A) = (1/2) × b × p = (1/2) × 4 cm × 3 cm = 6 cm²
(ii) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l)
= 12 cm × 12 cm
= 144 cm²
(iii) Total surface area of the prism = Lateral surface area + 2 × Area of triangular base (A)
= 144 cm² + 2 × 6 cm²
= 156 cm²
(iv) Volume of the prism (V) = Area of triangular base (A) × length (l)
= 6 cm² × 12 cm
= 72 cm³
b)
The triangular base is an equilateral triangle.
Here, each side (a) = 2√[6] cm
Area of triangular base (A) = (√[3] / 4) × a² = (√3 / 4) × (6)² = (√3 / 4) × 36 = 9 √3 cm²
Now, perimeter of the triangular base (P) = 3 × 6 = 18 cm
(i) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l) = 18 x 15 = 270 cm²
(ii) Total surface area of the prism = Lateral surface area + 2 × Area of triangular base (A) = 270 + 2 x 9 √3 = 270 + 18 √3 cm²
(iii) Volume of the prism (V) = Area of triangular base (A) × length (l) = 9 √3 x 15 = 135 √3 cm³
(c)
The given prism is isosceles triangle base prism.
Area of the triangular base (A) = (b / 4) × √[4a² - b²]
= (8 / 4) × √[4 × (6)² - (8)²]
= 2 × √80 = 8 √5 cm²
Perimeter of the base (P) = a + a + b = (6 + 8 + 6) = 20 cm
Lateral surface area of the prism (LSA) = P × h = 20 × 13 = 260 cm²
Total surface area of the prism (TSA) = 2 × A + LSA = 2 x 8 √5 + 260 = 16 √5 + 260 cm² = 276 √5 cm²
Volume of the prism (V) = A × h = 8 √5 x 13 cm³ = 104 √5 cm³
9. The volume of the right triangular prism is 864 m³. If the length of side to make right angle of right triangles are 8 m and 9 m respectively find the height of prism.
Solution:
Here,
The triangular base is a right angled triangle.
where, legs of the triangle = 8 m and 9 m
Area of triangular base (A) = (1/2) × base × height
= (1/2) × 8 m × 9 m
= 36 m²
Volume of the prism (V) = A × h
Given volume = 864 m³
or, 864 m³ = 36 m² × h
or, h = 864 m³ / 36 m²
or, h = 24 m
∴ The height of the prism is 24 m.
9. The volume of the right triangular prism is 864 m³. If the length of side to make right angle of right triangles are 8 m and 9 m respectively find the height of prism.
Solution:
Here,
The triangular base is a right angled triangle.
where, sides of the triangle = 8 m and 9 m
Area of triangular base (A) = (1/2) × base × height
= (1/2) × 8 m × 9 m
= 36 m²
Volume of the prism (V) = A × h
Given volume = 864 m³
864 m³ = 36 m² × h
or, h = 864 m³ / 36 m² = 24 m
∴ The height of the prism is 24 m.
10. The height of the right triangle prism is 30 cm. If the length of the base and perpendicular of the right triangle is 4 cm and 3 cm respectively. Find the areas of the rectangular faces.
Solution:
Here, the base of the prism is a right triangle with base 4 cm and perpendicular 3 cm.
By using Pythagoras theorem, hypotenuse = √[4² + 3²]
= √[16 + 9]
= √[25] = 5 cm
∴ The sides of the triangular base are 4 cm, 3 cm, and 5 cm.
The length of the prism (h) = 30 cm
The prism has three rectangular faces, each with one dimension as the length of the prism and the other as a side of the triangular base.
Area of the first rectangular face = 4 × 30
= 120 cm²
Area of the second rectangular face = 3 × 30
= 90 cm²
Area of the third rectangular face = 5 × 30
= 150 cm²
Hence, the areas of the rectangular faces are 120 cm², 90 cm², and 150 cm².
11. The total surface area of the triangular prism is 660 cm² and the base of the prism is a right triangle. The length of the base and hypotenuse of the triangle are 12 cm and 13 cm respectively. Find the height/length of the prism.
Solution:
Here, the base of the prism is a right triangle with base 12 cm and hypotenuse 13 cm.
By using Pythagoras theorem, perpendicular = √[13² - 12²]
= √[169 - 144]
= √[25] = 5 cm
∴ The sides of the triangular base are 12 cm, 5 cm, and 13 cm.
Area of the triangular base (A) = ½ × base × perpendicular = ½ × 12 × 5 = 30 cm²
Perimeter of the triangular base (P) = 12 + 5 + 13 = 30 cm
The total surface area of the prism = 2 × (area of the triangular base) + (perimeter of the base) × (height of the prism)
or,
660 cm² = 2 × 30 + 30 × h
or, 660 = 60 + 30h
or, 30h = 600
or, h = 20
Hence, the height of the prism is 20 cm.
12. The volume of the triangular prism is 480 cm³. If the length of the prism is l cm and the sides of the base are 4 cm, 13 cm and 15 cm:
(a) Find the length (l) of the prism.
(b) Find the areas of the rectangular faces.
(c) Find the total surface area of the prism.
Solution:
Here, the base of the prism is a triangle with sides 4 cm, 13 cm, and 15 cm.
Semi-perimeter (s) = (4 + 13 + 15) / 2 = 32 / 2 = 16 cm
Area of the triangular base (A) = √[s(s - a)(s - b)(s - c)] = √[16(16 - 4)(16 - 13)(16 - 15)]
= √[16 × 12 × 3 × 1] = √[576] = 24 cm²
(a)
The volume of the prism (V) = area of the triangular base (A) × length (l)
480 cm³ = 24 cm² × l
or l = 480 / 24 = 20 cm
Hence, the length of the prism is 20 cm.
(b)
The prism has three rectangular faces, each with one dimension as the length of the prism and the other as a side of the triangular base.
Length of the prism (l) = 20 cm
Area of the first rectangular face = 4 × 20
= 80 cm²
Area of the second rectangular face = 13 × 20
= 260 cm²
Area of the third rectangular face = 15 × 20
= 300 cm²
Hence, the areas of the rectangular faces are 80 cm², 260 cm², and 300 cm².
(c)
Perimeter of the triangular base (P) = 4 + 13 + 15 = 32 cm
The total surface area of the prism = 2 × (area of the triangular base) + (perimeter of the base) × (length of the prism)
= 2 × 24 + 32 × 20
= 48 + 640
= 688 cm²
Hence, the total surface area of the prism is 688 cm².
13. The lateral surface area of the triangular prism is 2160 cm². The perimeter of the base is 54 cm and the cross-sectional area is 126 cm². Now find:
(a) Height of the prism
(b) Volume of the prism
Solution:
Here, the base of the prism is a triangle with perimeter 54 cm and area 126 cm².
(a)
The lateral surface area of the prism (LSA) = (perimeter of the base) × (height of the prism)
or, 2160 cm² = 54 cm × h
or, h = 2160 / 54
or, h = 40 cm
Hence, the height of the prism is 40 cm.
(b)
The volume of the prism (V) = (area of the triangular base) × (height of the prism)
Area of the triangular base = 126 cm²
Height of the prism (h) = 40 cm
Volume = 126 × 40
= 5040 cm³
Hence, the volume of the prism is 5040 cm³.
Vedanta Excel in Mathematics (Book 9) Solution
Exercise 6.1
1. (a) What is a prism? Define with examples.
Solution:
A prism is a three-dimensional shape with two parallel, congruent bases connected by lateral faces. The bases are polygons, and the lateral faces are parallelograms (rectangles in a right prism).
Example 1: A triangular prism has triangular bases, e.g., a prism with a triangular base of sides 3 cm, 4 cm, and 5 cm, and height 10 cm.
Example 2: A rectangular prism (cuboid) has rectangular bases, e.g., a box with dimensions 5 cm × 4 cm × 3 cm.
(b) Define a cross section of a prism. What is the shape of the cross section of a cuboid?
Solution:
A cross section of a prism is the shape obtained by cutting the prism with a plane parallel to its base. It is congruent to the base.
For a cuboid, the base is a rectangle, so the cross section is a rectangle.
(c) Define the lateral surface of a prism. What are the lateral surfaces of a triangular prism? If the area of a base is 24 cm², what is its cross sectional area?
Solution:
The lateral surface of a prism is the sum of the areas of its lateral faces (excluding the bases). For a triangular prism, the lateral surfaces are the three rectangular faces connecting the triangular bases.
For a triangular prism with base sides 3 cm, 4 cm, and 5 cm, and height 10 cm, the lateral faces are:
- 3 cm × 10 cm = 30 cm²
- 4 cm × 10 cm = 40 cm²
- 5 cm × 10 cm = 50 cm²
If the area of a base is 24 cm², the cross-sectional area is also 24 cm².
(d) If the area of a base of a prism is 24 cm², what is its cross sectional area?
Solution:
The cross-sectional area of a prism is the same as the area of its base. Hence, the cross-sectional area is 24 cm².
2. (a) The area of a base of a prism is 24 cm², its cross height is 36 cm² and height is 15 cm. Find its volume.
Solution:
Here, the area of the base of the prism is 24 cm², and the height is 15 cm.
The volume of the prism (V) = (area of the base) × (height)
= 24 cm² × 15 cm
= 360 cm³
Hence, the volume of the prism is 360 cm³.
(b) The perimeter of cross section of a prism of length 10 cm is 48 cm. Find its lateral surface area.
Solution:
Here, the perimeter of the cross section (base) is 48 cm, and the length of the prism is 10 cm.
The lateral surface area of the prism (LSA) = (perimeter of the base) × (length)
= 48 cm × 10 cm
= 480 cm²
Hence, the lateral surface area of the prism is 480 cm².
(c) The lateral surface area and the cross sectional area of a prism are 175 cm² and 40 cm² respectively. Find total surface area of the prism.
Solution:
Here, the lateral surface area of the prism is 175 cm², and the cross-sectional area (area of the base) is 40 cm².
The total surface area of the prism (TSA) = 2 × (area of the base) + (lateral surface area)
or, = 2 × 40 cm² + 175 cm²
= 80 cm² + 175 cm²
= 255 cm²
Hence, the total surface area of the prism is 255 cm².
(d) The perimeter of cross section and the area of cross section of a prism are 36 cm and 54 cm² respectively. If the length of the prism is 15 cm, find its total surface area.
Solution:
Here, the perimeter of the cross section (base) is 36 cm, the area of the cross section (base) is 54 cm², and the length of the prism is 15 cm.
The total surface area of the prism (TSA) = 2 × (area of the base) + (perimeter of the base) × (length)
= 2 × 54 cm² + 36 cm × 15 cm
= 108 cm² + 540 cm²
= 648 cm²
Hence, the total surface area of the prism is 648 cm².
3. (a) The volume of a cube is 125 cm³. Find its total surface area.
Solution:
Here, the volume of the cube is 125 cm³.
Volume of a cube = (side)³
or, 125 = (side)³
or, side = 5 cm
The total surface area of a cube = 6 × (side)²
= 6 × 5²
= 6 × 25
= 150 cm²
Hence, the total surface area of the cube is 150 cm².
(b) The total surface area of a cube is 96 cm². Find its volume.
Solution:
Here, the total surface area of the cube is 96 cm².
Total surface area of a cube = 6 × (side)²
or,
96 = 6 × (side)²
or,(side)² = 96 / 6
or, (side)² = 16
or, side = 4 cm
The volume of a cube = (side)³
= 4³
= 64 cm³
Hence, the volume of the cube is 64 cm³.
(c) The length and the breadth of a cuboid are 15 cm and 8 cm respectively. If the volume of the cuboid is 720 cm³, find its height and calculate the total surface area.
Solution:
Here, the length of the cuboid is 15 cm, the breadth is 8 cm, and the volume is 720 cm³.
The volume of the cuboid (V) = length × breadth × height
or,
720 cm³ = 15 cm × 8 cm × h
or, 720 = 120 × h
or, h = 720 / 120
or, h = 6 cm
The total surface area of the cuboid (TSA) = 2(length × breadth + breadth × height + height × length)
= 2(15 × 8 + 8 × 6 + 6 × 15)
= 2(120 + 48 + 90)
= 2 × 258
= 516 cm²
Hence, the height of the cuboid is 6 cm, and the total surface area is 516 cm².
(d) The area of the rectangular base of a metallic block is 192 cm² and its volume is 768 cm³. Find its thickness.
Solution:
Here, the area of the rectangular base of the metallic block is 192 cm², and its volume is 768 cm³.
Area of the base = length × breadth = 192 cm²
The volume of the block (V) = length × breadth × height
or, 768 cm³ = 192 cm² × h
or,h = 768 / 192
or, h = 4 cm
Hence, the thickness of the metallic block is 4 cm.
(e) A rectangular metallic block is 40 cm long, 24 cm broad and 10 cm high. How many pieces of rectangular slices each of 8 mm thick can be cast lengthwise from the block?
Solution:
Here, the dimensions of the metallic block are: length = 40 cm, breadth = 24 cm, and height = 10 cm.
The thickness of each slice is 8 mm = 8 / 10 = 0.8 cm.
The slices are cut lengthwise, so the thickness is along the length of the block.
Number of slices = total length / thickness of each slice
or,
= 40 cm / 0.8 cm
= 50
Hence, 50 pieces of rectangular slices can be cast from the block.
4. Calculate the cross sectional area, the lateral surface area, the total surface area and volume of each of the following prisms.
(a)
Solution:
Cross Sectional area (A)= (16 x 4) + (8 x 4) = 96 96 cm²
Lateral surface area (LSA) = (16 + 4 + 8 + 4 + 8 + 8) x 6 = 48 x 6 = 2888 cm²
Total surface area (TSA) = 2 x 96 + 288 = 480 cm²
Volume = A x h = 96 x 6 = 576 cm³
(b)
Solution:
Cross Sectional area (A)= (10 x 3) + (2 x 2) + (2 x 2) = 38 cm²
Lateral surface area (LSA) = (10 + 5 + 2 + 2 + 6 + 2 + 2 + 5) x 5 = 34 x 5 = 170 cm²
Total surface area (TSA) = 2 x 38 + 170 = 246 cm²
Volume = A x h = 38 x 5 = 190 cm³
(c)
Solution:
Cross Sectional area (A)= (3 + 2 + 3)x 3 + 2 x 2 + 2 x 2 = 48 cm²
Lateral surface area (LSA) = (9 + 4 + 3 + 4 + 3 + 4 + 3 + 4) x 6 = 34 x 6 = 204 cm²
Total surface area (TSA) = 2 x 48 + 204 = 300 cm²
Volume = A x h = 48 x 6 = 288 cm³
(d)
Solution:
Cross Sectional area (A)= (8 x 3) + 4 + 4) = 32 cm²
Lateral surface area (LSA) = (3 + 3 + 2 + 2 + 2 + 3 + 3 + 3 + 2 + 2 + 2 + 3) x 5 = 30 x 5 = 150 cm²
Total surface area (TSA) = 2 x 32 + 150 = 64 + 150 = 214 cm²
Volume = A x h = 32 x 5 = 160 cm³
(e)
Solution:
Cross Sectional area (A)= (15 x 3) + (4 x 3) + (4 x 3) = 69 cm²
Lateral surface area (LSA) = (15 + 6 + 4 + 3 + 9 + 3 + 4 + 6) x 5 = 48 x 5 = 240 cm²
Total surface area (TSA) = 2 x 69 + 240 = 138 + 240 = 278 cm²
Volume = A x h = 69 x 5 = 345 cm³
(f)
Solution:
Here,
(i) The area of cross section = (30 x 10) + (20 x 10) + (10 x 10) = 600 cm²
(ii) Perimeter of the cross section (P) = 30 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm + 10 cm = 120 cm
Length of prism (h) = 30 cm
∴ Lateral surface area (L.S.A.) = P × h = 120 cm × 30 cm = 3600 cm²
(iii) Total surface area of prism = L.S.A. + 2A = 3600 cm² + 2 × 300 cm² = 4200 cm²
(iv) Volume of the prism (V) = A × h = 600 cm² × 30 cm = 18000 cm³
5. (a) A cubic water tank is filled in 1296 seconds at the rate of 1 litre per 6 seconds.
(i) Calculate the internal volume and the length of the side of the tank.
(ii) Calculate the total internal surface area of the tank.
Solution:
(i)
Here, the tank is filled in 1296 seconds at the rate of 1 litre per 6 seconds.
Number of 6-second intervals = 1296 / 6
= 216
Total volume = 216 litres
Volume in cm³ = 216 × 1000 = 216000 cm³
The tank is cubic, so volume = (side)³
or,
(side)³ = 216000
or,
side = 60 cm (since 60³ = 216000)
Side in metres = 60 / 100 = 0.6 m
Hence, the internal volume is 216 litres, and the length of the side is 60 cm.
(ii)
The total internal surface area of a cube = 6 × (side)²
= 6 × 60²
= 6 × 3600
= 21600 cm²
Surface area in m² = 21600 / 10000 = 2.16 m²
Hence, the total internal surface area is 21600 cm² or 2.16 m².
(b) A lidless rectangular water tank made of zinc plates is 2 m long, 1.5 m broad and 1 m high.
(i) How many square metres of zinc plates are used in the tank?
(ii) How many litres of water does it hold when it is full?
(iii) Find the cost of zinc plates at Rs 1200 per sq. m.
Solution:
(i)
Here, the tank is 2 m long, 1.5 m broad, and 1 m high, and it is lidless.
Surface area of the tank (base + 4 lateral faces) = (length × breadth) + 2(length × height + breadth × height)
= (2 × 1.5) + 2(2 × 1 + 1.5 × 1)
= 3 + 2(2 + 1.5)
= 3 + 2 × 3.5
= 3 + 7
= 10 m²
Hence, 10 square metres of zinc plates are used.
(ii)
Volume of the tank = length × breadth × height
= 2 × 1.5 × 1
= 3 m³
Volume in litres = 3 × 1000 = 3000 litres
Hence, the tank holds 3000 litres of water when full.
(iii)
Cost of zinc plates = surface area × cost per sq. m
= 10 × 1200
= 12000 Rs
Hence, the cost of the zinc plates is Rs 12000.
(c) A rectangular carton is 80 cm × 60 cm × 40 cm.
(i) How many packets of soaps can each 10 cm × 5 cm × 4 cm be kept inside the carton?
(ii) By how many centimetres should the height of the carton be increased to keep 1200 packets of soaps?
Solution:
(i)
Here, the carton is 80 cm long, 60 cm broad, and 40 cm high, and each soap packet is 10 cm × 5 cm × 4 cm.
Number of packets along the length = 80 / 10 = 8
Number of packets along the breadth = 60 / 5 = 12
Number of packets along the height = 40 / 4 = 10
Total number of packets = 8 × 12 × 10
= 960
Hence, 960 packets of soaps can be kept inside the carton.
(ii)
Number of packets along the length = 80 / 10 = 8
Number of packets along the breadth = 60 / 4 = 15
Number of packets along the height = 40 / 5 = 8
Total packets with this orientation = 8 × 15 × 8 = 960
Packets per layer (length × breadth) = 8 × 15 = 120 packets
Current capacity = 960 packets, required capacity = 1200 packets.
Additional packets needed = 1200 - 960 = 240 packets
Additional layers needed = 240 / 120 = 2 layers
Height per layer = 5 cm
Additional height = 2 × 5 = 10 cm
Hence, the height of the carton should be increased by 10 cm.
6. (a) A cubic wooden block of length 12 cm is cut into 8 equal cubical pieces. Find the length of the edge of each piece.
Solution:
Here, the cubic wooden block has a side length of 12 cm.
Volume of the block = (side)³ = 12³
= 1728 cm³
The block is cut into 8 equal cubical pieces.
Volume of each smaller cube = 1728 / 8
= 216 cm³
Side length of each smaller cube = √[216]³
= 6 cm (since 6³ = 216)
Hence, the length of the edge of each piece is 6 cm.
(b) 8 metallic cubical blocks of equal size are melted and joined together to form a bigger cubical block. If each smaller block is 10 cm thick, find the thickness of the bigger block.
Solution:
Here, each smaller cubical block has a side length (thickness) of 10 cm.
Volume of one smaller cube = 10³ = 1000 cm³
Total volume of 8 smaller cubes = 8 × 1000
= 8000 cm³
The bigger block is a cube, so its volume = (side)³
(side)³ = 8000
Side length (thickness) of the bigger cube = √[8000]³
= 20 cm (since 20³ = 8000)
Hence, the thickness of the bigger block is 20 cm.
(c) A rectangular metallic block is 50 cm × 20 cm × 8 cm. If it is melted and re-formed into a cubical block, find the length of the edge of the cube.
Solution:
Here, the rectangular block has dimensions 50 cm × 20 cm × 8 cm.
Volume of the rectangular block = 50 × 20 × 8
= 8000 cm³
The volume of the cubical block is the same, so (side)³ = 8000
Side length of the cube = √[8000]³
= 20 cm (since 20³ = 8000)
Hence, the length of the edge of the cube is 20 cm.
(d) Vegetable ghee is stored in a rectangular vessel of internal dimensions 60 cm × 10 cm × 45 cm. It is transferred into the identical cubical vessels. If the internal length of each cubical vessel is 10 cm, how many vessels are required to empty the rectangular vessel?
Solution:
Here, the rectangular vessel has internal dimensions 60 cm × 10 cm × 45 cm.
Volume of the rectangular vessel = length × breadth × height
= 60 × 10 × 45
= 27000 cm³
Each cubical vessel has an internal side length of 10 cm.
Volume of one cubical vessel = (side)³ = 10³
= 1000 cm³
Number of cubical vessels required = (volume of rectangular vessel) / (volume of one cubical vessel)
= 27000 / 1000
= 27
Hence, 27 cubical vessels are required to empty the rectangular vessel.
7. (a) The adjoining figure is a rectangular glass vessel of length 40 cm, breadth 30 cm, and height 20 cm. If it contains some water up to the height of 12 cm, how many litres of water are needed to fill the vessel completely? (1 l = 1000 cm³)
Solution:
Here, the rectangular vessel has dimensions 40 cm × 30 cm × 20 cm, and the water level is at a height of 12 cm.
Base area of the vessel = length × breadth = 40 × 30
= 1200 cm²
Current volume of water = base area × height of water
= 1200 × 12
= 14400 cm³
Total volume of the vessel = length × breadth × height
= 40 × 30 × 20
= 24000 cm³
Additional volume needed = total volume - current volume
= 24000 - 14400
= 9600 cm³
Additional volume in litres = 9600 / 1000
= 9.6 litres
Hence, 9.6 litres of water are needed to fill the vessel completely.
(b) In the given figure, a cubical vessel of length 20 cm is completely filled with water. If the water is poured into a rectangular vessel of length 32 cm, breadth 25 cm, and height 15 cm, find the height of the water level in the rectangular vessel. How much more water is required to fill the rectangular vessel completely? (1 l = 1000 cm³)
Solution:
Here, the cubical vessel has a side length of 20 cm and is completely filled with water.
Volume of the cubical vessel = (side)³ = 20³
= 8000 cm³
The water is poured into a rectangular vessel of dimensions 32 cm × 25 cm × 15 cm.
Base area of the rectangular vessel = length × breadth = 32 × 25
= 800 cm²
Let the height of the water level be h cm.
Volume of water = base area × h
8000 = 800 × h
h = 8000 / 800
= 10 cm
Total volume of the rectangular vessel = length × breadth × height
= 32 × 25 × 15
= 12000 cm³
Additional volume needed = total volume - current volume
= 12000 - 8000
= 4000 cm³
Additional volume in litres = 4000 / 1000
= 4 litres
Hence, the height of the water level in the rectangular vessel is 10 cm, and 4 litres of water are required to fill the rectangular vessel completely.
Surface area and volume of triangular prisms
EXERCISE 6.2
General section
1. a) From the given triangular prism, write the formulae to find its
(i) cross sectional area = (1/2)xy
(ii) lateral surface area = (x + y + z) × l
(iii) total surface area = xy + (x + y + z) × l
(iv) volume = (1/2)xy × l
b) If p, q, r are the three sides of a triangle of a triangular base prism of length h, write the formulae to find its
(i) cross sectional area = (1/4) × √[(p + q + r)(p + q - r)(p + r - q)(q + r - p)]
(ii) lateral surface area = (p + q + r) × h
(iii) total surface area = (1/2) × √[(p + q + r)(p + q - r)(p + r - q)(q + r - p)] + (p + q + r) × h
(iv) volume = (1/4) × √[(p + q + r)(p + q - r)(p + r - q)(q + r - p)] × h
c) The base of a prism of length h cm is an equilateral triangle with each side a cm, write the formulae to find its
(i) cross sectional area = (√[3] / 4) a² cm²
(ii) lateral surface area = 3ah cm²
(iii) total surface area = (√[3] / 2) a² + 3ah cm²
(iv) volume = (√[3] / 4) a² h cm³
EXERCISE 6.2
General section
2. a) A prism has a right-angled triangular base with perpendicular 8 cm and base 6 cm. Find its cross sectional area.
Solution:
Here,
The cross-sectional area of the prism is the area of its right-angled triangular base.
Area of the triangle = (1/2) × base × height
= (1/2) × 6 cm × 8 cm
= (1/2) × 48 cm²
= 24 cm²
b) A right prism base is a triangle whose sides are 3 cm, 25 cm, and 26 cm. Find the area of its cross section.
Solution:
Here,
The cross-sectional area of the prism is the area of its triangular base with sides 3 cm, 25 cm, and 26 cm.
Semi-perimeter s = (3 + 25 + 26) / 2
= 54 / 2
= 27 cm
Area of the triangle = √[s(s - 3)(s - 25)(s - 26)]
= √[27(27 - 3)(27 - 25)(27 - 26)]
= √[27 × 24 × 2 × 1]
= √[1296]
= 36 cm²
c) If the perimeter of the triangular base of a prism is 18 cm and its length is 15 cm, find its lateral surface area.
Solution:
Here,
Perimeter of the triangular base = 18 cm
Length of the prism = 15 cm
We know that,
Lateral surface area = (perimeter of the base) × (length)
= 18 cm × 15 cm
= 270 cm²
d) The perimeter of the triangular base of a prism is 14.5 cm and its lateral surface area is 290 cm². Find the length of the prism.
Solution:
Here,
Perimeter of the triangular base = 14.5 cm
Lateral surface area = 290 cm²
We know that,
Lateral surface area = (perimeter of the base) × (length)
290 cm² = 14.5 cm × length
or, length = 290 cm² / 14.5 cm = 20 cm
e) The area of the triangular base of a prism is 30.5 cm² and its lateral surface area is 305 cm². Find the total surface area of the prism.
Solution:
Here,
Area of the triangular base = 30.5 cm²
Lateral surface area = 305 cm²
We know that,
Total surface area = (lateral surface area) + 2 × (area of the base)
= 305 cm² + 2 × 30.5 cm²
= 305 cm² + 61 cm²
= 366 cm²
f) The perimeter of the triangular base of a prism is 24 cm and the area of its cross section is 24 cm². If the prism is 10 cm long, find its total surface area.
Solution:
Here,
Perimeter of the triangular base = 24 cm
Area of the cross-section (area of the base) = 24 cm²
Length of the prism = 10 cm
We know that,
Lateral surface area = (perimeter of the base) × (length)
= 24 cm × 10 cm
= 240 cm²
Now, total surface area = (lateral surface area) + 2 × (area of the base)
= 240 cm² + 2 × 24 cm²
= 240 cm² + 48 cm²
= 288 cm²
3. a) If the area of the triangular base of a prism 25 cm long is 16.4 cm², find the volume of the prism.
Solution:
Here,
Area of the triangular base = 16.4 cm²
Length of the prism = 25 cm
We know that,
Volume of the prism = (area of the base) × (length)
= 16.4 cm² × 25 cm
= 410 cm³
b) A triangular based prism is 30 cm long. If the length of the sides of its triangular base are 3 cm, 4 cm and 5 cm, find its volume.
Solution:
Here,
The base is a triangle with sides 3 cm, 4 cm, and 5 cm.
We know that,
Area of the triangle = (1/2) × base × height
= (1/2) × 3 cm × 4 cm
= (1/2) × 12 cm²
= 6 cm²
Next, length of the prism = 30 cm
Then, volume of the prism = (area of the base) × (length)
= 6 cm² × 30 cm
= 180 cm³
c) In a prism, if the volume is 400 cm³ and area of its triangular base is 50 cm², find the length of the prism.
Solution:
Here,
Volume of the prism = 400 cm³
Area of the triangular base = 50 cm²
We know that,
Volume of the prism = (area of the base) × (length)
or, 400 cm³ = 50 cm² × length
or, Length = 400 cm³ / 50 cm²
= 8 cm
Solution:
Here,
In triangle ABC, AB (perpendicular) = 12 cm, BC (base) = 5 cm, and angle ABC = 90°.
We need to find AC (hypotenuse).
Using the Pythagorean theorem: (AC)² = (AB)² + (BC)²
= (12)² + (5)²
= 144 + 25
= 169
AC = √[169]
= 13 cm
4. a) (i) Find the volume of the prism shown in the diagram.
Solution:
Here,
The base is a right-angled triangle with perpendicular AB = 12 cm and base BC = 5 cm.
Area of the triangular base = (1/2) × base × height
= (1/2) × 5 cm × 12 cm
= (1/2) × 60 cm²
= 30 cm²
Length of the prism = 25 cm
Volume of the prism = (area of the base) × (length)
= 30 cm² × 25 cm
= 750 cm³
(ii)
Solution:
Here,
The base is an equilateral triangle with each side = 6 cm.
Area of the equilateral triangle = (√[3] / 4) × (side)²
= (√[3] / 4) × (6)²
= (√[3] / 4) × 36 cm²
= 9 √[3] cm²
Length of the prism = 10 √[3] cm
Volume of the prism = (area of the base) × (length)
= (9 √[3]) cm² × (10 √[3]) cm
= 9 × 10 × (√[3] × √[3]) cm³
= 90 × 3 cm³
= 270 cm³
iii)
Solution:
Here,
The base of the cross-section is a right-angled triangle with base QR = 10 cm and height PQ = 8 cm.
Area of the triangular cross-section (A) = (1/2) × base × height
= (1/2) × 10 cm × 8 cm
= (1/2) × 80 cm²
= 40 cm²
Length of the prism (l) = 30 cm
Volume = (area of the cross-section) × (length)
= 40 cm² × 30 cm
= 1200 cm³
(iv)
The given prism is isosceles triangle base prism.
Area of the triangular base (A) = (b / 4) × √[4a² - b²]
= (10 / 4) × √[4 × (13)² - (10)²]
= (10 / 4) × √[4 × (13)² - (10)²]
= (10 / 4) × √[4 × 169 - 100]
= (10 / 4) × √[676 - 100]
= (10 / 4) × √576
= (10 / 4) × 24
= 2.5 × 24
= 60
Now, volume of the prism (V) = A × h = 60 x 18 cm³ = 1080 cm³
v)
Solution:
Here,
The base of the cross-section is a triangle with sides 15 cm, 17 cm, and 8 cm.
Perimeter of the triangular base (P) = 15 cm + 17 cm + 8 cm = 40 cm
Semi-perimeter (s) = 40 cm / 2 = 20 cm
Area of the triangular cross-section (A) = √s(s - a)(s - b)(s - c)
= √20(20 - 15)(20 - 17)(20 - 8)
= √20 × 5 × 3 × 12
= √3600
= 60 cm²
Length of the prism (l) = 20 cm
Volume = (area of the cross-section) × (length)
= 60 cm² × 20 cm
= 1200 cm³
vi)
Solution:
a)
The trianguar base is a right angled triangle.
Here, p = 12 cm and b = 16 cm
By using Pythagoras theorem, h = √[p² + b²] = √[12² + 16²] = √[400] = 20 cm
Now, area of triangular base (A) = (1/2) × b × p = (1/2) × 16 cm × 12 cm = 96 cm²
Volume of the prism (V) = Area of triangular base (A) × length (l)
= 96 cm² × 40 cm
= 3840 cm³
b) Calculate the lateral surface area and total surface area of the following prisms.
i)
Solution:
Here,
The base of the prism is a triangle with sides 5 cm, 4 cm, and 7 cm.
Perimeter of the triangular base = 5 cm + 4 cm + 7 cm = 16 cm
Semi-perimeter (s) = 16 cm / 2 = 8 cm
Area of the triangular base = √s(s - a)(s - b)(s - c)
= √8(8 - 5)(8 - 4)(8 - 7)
= √8 × 3 × 4 × 1
= √96
= 9.8 cm² (approximately)
Height of the prism (h) = 8 cm
Lateral surface area (LSA) = Perimeter of the base × Height
= 16 cm × 8 cm
= 128 cm²
Total surface area (TSA) = LSA + 2 × Area of the base
= 128 cm² + 2 × 9.8 cm²
= 128 cm² + 19.6 cm²
= 147.6 cm²
ii)
Solution:
Here,
The base of the prism is a triangle with sides 3.5 cm, 3.4 cm, and 6 cm.
Perimeter of the triangular base = 3.5 cm + 3.4 cm + 6 cm = 12.9 cm
Semi-perimeter (s) = 12.9 cm / 2 = 6.45 cm
Area of the triangular base = √s(s - a)(s - b)(s - c)
= √6.45(6.45 - 3.5)(6.45 - 3.4)(6.45 - 6)
= √6.45 × 2.95 × 3.05 × 0.45
= √26.11524375
= 5.11 cm² (approximately)
Height of the prism (h) = 8 cm
Lateral surface area (LSA) = Perimeter of the base × Height
= 12.9 cm × 8 cm
= 103.2 cm²
Total surface area (TSA) = LSA + 2 × Area of the base
= 103.2 cm² + 2 × 5.11 cm²
= 103.2 cm² + 10.22 cm²
= 113.42 cm²
iii)
Solution:
a)
Here, p = 8 cm and b = 6 cm
By using Pythagoras theorem, h = √[p² + b²] = √[8² + 6²] = √[100] = 10 cm
Now, perimeter of the triangular base (P) = 8 cm + 6 cm + 10 cm = 24 cm
(i) Area of triangular base (A) = (1/2) × b × p = (1/2) × 6 cm × 8 cm = 24 cm²
(ii) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l)
= 24 cm × 15 cm
= 360 cm²
(iii) Total surface area of the prism = Lateral surface area + 2 × Area of triangular base (A)
= 360 cm² + 2 × 24 cm²
= 360 cm² + 48 cm² = 408 cm²
iv)
Solution:
a)
Here, p = 12 cm and b = 5 cm
By using Pythagoras theorem, h = √[p² + b²] = √[12² + 5²] = √[169] = 13 cm
Now, perimeter of the triangular base (P) = 12 cm + 5 cm + 13 cm = 30 cm
(i) Area of triangular base (A) = (1/2) × 5 × 12 = 30 cm²
(ii) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l)
= 30 cm × 30 cm
= 900 cm²
(iii) Total surface area of the prism = Lateral surface area + 2 × Area of triangular base (A)
= 900 cm² + 2 × 30 cm²
= 900 cm² + 60 cm² = 960 cm²
(v)
The triangular base is an equilateral triangle.
Here, each side (a) = 10 cm
Area of triangular base (A) = (√[3] / 4) × a² = (√3 / 4) × (10)² = (√3 / 4) × 100 = 25 √3 cm²
Now, perimeter of the triangular base (P) = 3 × 10 = 30 cm
(i) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l) = 30 x 15 = 450 cm²
(ii) Total surface area of the prism = Lateral surface area + 2 × Area of triangular base (A) = 450 + 2 x 25 √3 = 450 + 25 √3 cm²
(vi)
The given prism is isosceles triangle base prism.
Area of the triangular base (A) = (b / 4) × √[4a² - b²]
= (6 / 4) × √[4 × (6)² - (7)²]
= (6 / 4) × √108 = 162 cm²
Perimeter of the base (P) = a + a + b = (6 + 6 + 7) = 19 cm
Lateral surface area of the prism (LSA) = P × h = 19 × 20 = 380 cm²
Total surface area of the prism (TSA) = 2 × A + LSA = 2 x 8 √5 + 260 = 16 √5 + 260 cm² = 276 √5 cm²
5. a)The given diagram is the solid prism of triangular base. If the volume of the prism is 48 cm³, find its height.
Solution:
Here,
The base of the prism is a right-angled triangle with sides 3 cm, 4 cm, and 5 cm.
Area of the triangular base = (1/2) × base × height
= (1/2) × 3 cm × 4 cm
= (1/2) × 12 cm²
= 6 cm²
Volume of the prism = Area of the base × Height
48 cm³ = 6 cm² × height
or, height = 48 cm³ / 6 cm²
or, height = 8 cm
5. b)If the volume of the prism given aside is 432 cm³, find its height.
Solution:
Here,
Here, p = 8 cm and h = 10 cm
By using Pythagoras theorem, b = √[h² - p²] = √[10² - 8²] = √[36] = 6 cm
Now, perimeter of the triangular base (P) = 8 cm + 6 cm + 10 cm = 24 cm
Area of triangular base (A) = (1/2) × b × p = (1/2) × 8 cm × 6 cm = 24 cm²
Volume of the prism = Area of the base × Height
or, 432 = 24 × Height
or, height = 432/24 = 18 cm
5. c)The volume of a prism having its base a right-angled triangle is 6,300 cm³. If the lengths of the sides of the right-angled triangle containing the right angle are 20 cm and 21 cm, find the height of the prism.
Solution:
Here,
The base of the prism is a right-angled triangle with legs 20 cm and 21 cm.
Area of the triangular base = (1/2) × base × height
= (1/2) × 20 cm × 21 cm
= (1/2) × 420 cm²
= 210 cm²
Volume of the prism = Area of the base × Height
6300 cm³ = 210 cm² × Height
Height = 6300 cm³ / 210 cm²
= 30 cm
6. a) The area of rectangular surfaces of a triangular prism having base sides 9 cm, 10 cm and 17 cm is 864 cm². Calculate the height of the prism.
Solution:
Here,
The base is a triangle with sides 9 cm, 10 cm, and 17 cm.
Perimeter of the triangular base = 9 cm + 10 cm + 17 cm
= 36 cm
Area of the rectangular surfaces (LSA) = 864 cm²
LSA = (perimeter of the base) × (height of the prism)
or, 864 cm² = 36 cm × height
or, height = 864 cm² / 36 cm
or, height = 24 cm
6. b) A prism with an equilateral triangular base is 18 cm long. If the area of its rectangular surfaces is 567 cm², find the length of each side of the triangular base.
Solution:
Here,
The base is an equilateral triangle; let each side be s cm.
Perimeter of the triangular base = s + s + s = 3s cm
Length of the prism = 18 cm
Area of the rectangular surfaces (LSA) = 567 cm²
LSA = (perimeter of the base) × (length of the prism)
or, 567 cm² = (3s) cm × 18 cm
or, 567 cm² = 54s cm²
or, s = 567 cm² / 54 cm
or, s = 21/2 cm = 10.5 cm
Creative section-B7. a) The perimeter and area of the cross section of a triangular prism are 44 cm and 66 cm² respectively. If the area of lateral surface area of the prism is 1,100 cm²,
(i) find its volume
Solution:
Here,
Perimeter of the triangular base = 44 cm
We know that,
Area of the cross-section (base) = 66 cm²
Lateral Surface Area (LSA) = 1,100 cm²
Now, LSA = (perimeter of the base) × (length of the prism)
1,100 cm² = 44 cm × length
Length = 1,100 cm² / 44 cm
= 25 cm
Next, Volume = (area of the base) × (length)
= 66 cm² × 25 cm
= 1,650 cm³
(ii) find its total surface area
Solution:
Here,
Total Surface Area (TSA) = LSA + 2 × (area of the base)
= 1,100 cm² + 2 × 66 cm²
= 1,100 cm² + 132 cm²
= 1,232 cm²
b) The area of cross section of a triangular prism is 126 cm² and its volume is 5,040 cm³. If the perimeter of its base is 54 cm,
(i) find its lateral surface area
Solution:
Here,
Area of the cross-section (base) = 126 cm²
Volume = 5,040 cm³
Perimeter of the base = 54 cm
Volume = (area of the base) × (length)
or, 5,040 cm³ = 126 cm² × length
or, length = 5,040 cm³ / 126 cm²
or, length = 40 cm
Lateral Surface Area (LSA) = (perimeter of the base) × (length)
= 54 cm × 40 cm
= 2,160 cm²
(ii) find its total surface area
Solution:
Here,
Total Surface Area (TSA) = LSA + 2 × (area of the base)
= 2,160 cm² + 2 × 126 cm²
= 2,160 cm² + 252 cm²
= 2,412 cm²
c) The perimeter and area of the cross section of a triangular prism are 24 cm and 24 cm². If its total surface area is 480 cm², find
(i) the lateral surface area
Solution:
Here,
Perimeter of the cross-section (base) = 24 cm
Area of the cross-section (base) = 24 cm²
Total Surface Area (TSA) = 480 cm²
TSA = LSA + 2 × (area of the base)
or, 480 cm² = LSA + 2 × 24 cm²
or, 480 cm² = LSA + 48 cm²
or, LSA = 480 cm² - 48 cm² = 432 cm²
(ii) the volume
Solution:
Here,
LSA = 432 cm²
LSA = (perimeter of the base) × (length)
or, 432 cm² = 24 cm × length
or, length = 432 cm² / 24 cm
or, length = 18 cm
Volume = (area of the base) × (length)
= 24 cm² × 18 cm
= 432 cm³
8. A trap for insects is in the shape of a triangular prism which is shown alongside.
The base is an equilateral triangle with each side = 10 cm, and the length of the prism = 6√[3] cm.
(i) Find the total surface area of the trap.
Solution:
Here,
The base is an equilateral triangle with each side = 10 cm.
Area of the triangular base (A) = √[3] / 4 × (side)²
= √[3] / 4 × (10)²
= √[3] / 4 × 100 cm²
= 25 √[3] cm²
Perimeter of the triangular base = 3 × 10 cm = 30 cm
Length of the prism = 6 √[3] cm
Lateral Surface Area (LSA) = (perimeter of the base) × (length)
= 30 cm × 6 √[3] cm
= 180 √[3] cm²
Total Surface Area (TSA) = LSA + 2 × (area of the base)
= 180 √[3] cm² + 2 × 25 √[3] cm²
= 180 √[3] cm² + 50 √[3] cm²
= 230 √[3] cm²
≈ 230 × 1.732 cm² ≈ 398.36 cm²
(ii) Find the volume of the trap.
Solution:
Here,
Volume = Area of the triangular base (A) × length
= (25 √[3] cm²) × (6 √[3] cm)
= 25 × 6 × (√[3] × √[3]) cm³
= 150 × 3 cm³
= 450 cm³
8. b) Two campers made a tent shown alongside in the shape of triangular prism.
The given prism is an isosceles triangle base prism with base = 6 ft, other two sides = 5 ft each, and length of the prism = 8 ft.
(i) Find the amount of fabric required for making it, including the floor.
Solution:
Here,
Area of the triangular base (A) = (b / 4) × √[4a² - b²]
= (6 / 4) × √[4 × (5)² - (6)²]
= (6 / 4) × √[4 × 25 - 36]
= (6 / 4) × √[100 - 36]
= (6 / 4) × √[64]
= (6 / 4) × 8 = 12 ft²
Perimeter of the base (P) = a + a + b = (5 + 5 + 6) = 16 ft
Lateral surface area of the prism (LSA) = P × length = 16 × 8 = 128 ft²
Area of the rectangular floor = 6 ft × 8 ft = 48 ft²
Area of each slanted rectangular side (equal side × length) = 5 ft × 8 ft = 40 ft²
Total area of two slanted sides = 2 × 40 ft² = 80 ft²
Area of two triangular ends = 2 × 12 ft² = 24 ft²
Total fabric required (including the floor) = Floor + Two slanted sides + Two triangular ends
= 48 ft² + 80 ft² + 24 ft²
= 152 ft²
(ii) How much air is occupied in the tent?
Solution:
Here,
Volume of the prism (V) = A × length
= 12 ft² × 8 ft
= 96 ft³
9. a) Mrs. Ghale manages the accommodation for 16 guests in her daughter’s birthday ceremony. For this purpose, she plans to make a tent in the shape of triangular prism of length 12 m. If each person has the space of 6 square meters on the floor and 15 cubic meters of air to breathe in average inside the tent, what is the height of the tent? Find it.
Solution:
Here,
Total floor area required for 16 guests = 16 × 6 m²
= 96 m²
The floor is the rectangular base of the prism with length = 12 m and width = w m.
Area of the floor = length × width
96 m² = 12 m × w
or, w = 96 m² / 12 m = 8 m
Total volume required for 16 guests = 16 × 15 m³
= 240 m³
The base of the prism is a triangle with base = 8 m and height = h m.
Area of the triangular base (A) = (1/2) × base × height
= (1/2) × 8 m × h
= 4h m²
Volume of the prism = Area of the triangular base × length
or, 240 m³ = (4h m²) × 12 m
or, 240 m³ = 48h m³
or, h = 240 m³ / 48 m² = 5 m
Therefore, the height of the tent is 5 m.
9. b) In the marriage ceremony of Suresh's son, there was accommodation for 150 people. For this purpose, he made a tent in the shape of triangular prism in such a way that each person had the space of 4 square meters on the floor and 20 cubic meters of air to breathe in average inside the tent. If the tent was 30 m long, what was the height of the tent?
Solution:
Here,
Floor area per person = 4 m²
Total people = 150
Total floor area = 150 × 4 = 600 m²
Now, finding width of the tent,
Floor area = Length × Width
or, 600 = 30 × width
or, width = 600 / 30 = 20 m
Next, calculating Cross-Sectional Area (Triangular Base),
Base of triangle (b) = Width of tent = 20 m
Area of triangle (A) = (1/2) × b × h = (1/2) × 20 × h = 10h m²
Volume per person = 20 m³
Total volume = 150 × 20 = 3000 m³
Volume of prism = Cross-sectional area × Length
or, 3000 = 10h × 30
or, 3000 = 300h
or, h = 3000 / 300 = 10 m
Therefore, the height of the tent's triangular cross-section is 10 meters.
10 a) Find the volume of dog house given alongside.
The cross-section consists of a rectangle ABCD with length = 4 m, height = 5 m, and triangle EDC with sides 3 m, 4 m, 3 m, and length of the prism = 6 m.
Solution:
Here
(i) The area of cross section = area of rectangle ABCD + area of △EDC
The area of rectangle ABCD = l × b = 4 m × 5 m = 20 m²
Semi-perimeter of △ = (a + b + c) / 2 = (3 m + 4 m + 3 m) / 2 = 5 m
Area of △ = √[s(s-a)(s-b)(s-c)] = √[5 (5-3) (5-4) (5-3)] = √[20] ≈ 4.47 m²
∴ The area of cross section (A) = area of rectangle + area of △
= 20 m² + 4.47 m² = 24.47 m²
Volume of the prism (V) = A×h = 24.47 m² × 6 m = 146.82 m³
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