PSEB 8th Class Maths Solutions Chapter 11 Mensuration InText Questions

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 11 Mensuration InText Questions

From Textbook : [Textbook Page No. 170]

Can you write an expression for the perimeter of each of the above shapes?
Solution:
1. Rectangle → a × b
2. Square → a × a
3. Triangle → \(\frac {1}{2}\) × b × h
4. Parallelogram → b × h
5. Circle → πb²

Yes, perimeter of above shapes are as follows:
1. Perimeter of rectangle = 2 (length + breadth)
2. Perimeter of square = 4 × (side)
3. Perimeter of triangle = sum of lengths of three sides
4. Perimeter of parallelogram = 2 × (sum of adjacent sides)
5. Perimeter (circumference) of circle = 2πr

Try These : [Textbook Page No. 170]

(a) Match the following figures with their respective areas in the box.

Solution:

Area of square
= l × l = 7 × 7

(b) Write the perimeter of each shape.
Solution:
Perimeter of given shapes:
1. Perimeter of this figure can’t be found as breadth is not given [7 cm is the height, not breadth].

2. Perimeter of semicircle = πr + 2r
= \(\frac {22}{7}\) × 7 + 2 × 7
= 22 + 14
= 36 cm

3. Perimeter of triangle = sum of lengths of three sides
= 14 + 11 + 19 = 34cm

4. Perimeter of rectangle = 2(l + b)
= 2 (14 + 7)
= 2 × 21
= 42cm

5. Perimeter of square = 4l
= 4 × 7 = 28cm

Try These : [Textbook Page No. 172]

1. Nazma’s sister also has a trapezium-shaped plot. Divide it into three parts as shown in figure. Show that the area of trapezium WXYZ = \(\frac {a + b}{2}\).

Solution:
Area of ∆ PWZ = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × c × h
= \(\frac {1}{2}\)ch
Area of rectangle PQYZ = length × breadth
= b × h = bh
Area of ∆ QXY = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × d × h
= \(\frac {1}{2}\)dh
Now, Area of trapezium WXYZ
Area of ∆ PWZ + Area of rectangle PQYZ + Area of ∆ QXY
= \(\frac {1}{2}\) ch + bh + \(\frac {1}{2}\)dh
= \(\frac {1}{2}\) ch + \(\frac {1}{2}\)dh + bh
= \(\frac {1}{2}\) (c + d)h + bh
= \(\frac {1}{2}\)(a – b)h + bh (∵ c + d = a – b)
= [\(\frac {1}{2}\)(a – b) + b]h
= [\(\frac {a – b}{2}\) + b] h
= \(\frac {a-b+2b}{2}\) h
= \(\frac {h (a + b)}{2}\)
Thus, area of trapezium WXYZ = \(\frac{h(a+b)}{2}\)

2. If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression \(\frac{h(a+b)}{2}\).
Solution:
h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm.
Area of ∆ PWZ = \(\frac {1}{2}\) × c × h
= (\(\frac {1}{2}\) × 6 × 10)cm²
= 30 cm²
Area of ∆ QXY = \(\frac {1}{2}\) × d × h
= (\(\frac {1}{2}\) × 4 × 10)cm²
= 20 cm²
Area of rectangle PQYZ
= length × breadth
= 12 × 10
= 120 cm²
∴ Area of trapezium WXYZ = Area of ∆ PWZ + Area of ∆ QXY + Area of rectangle PQYZ
= (30 + 20 + 120) cm²
= 170 cm²
Now, Area of trapezium WXYZ
= \(\frac{h(a+b)}{2}\)
= \(\frac{10(22+12)}{2}\)
= \(\frac{10(34)}{2}\)
= 170cm²
∵ a = c + b + d
= 6 + 12 + 4
= 22 cm
Thus, area of trapezium verified.

Try These : [Textbook Page No. 173]

1. Find the area of the following trapeziums:

Solution:
(i) Area of given trapezium = \(\frac {1}{2}\) × (9 + 7) × 3
= \(\frac {1}{2}\) × 16 × 3 cm² = 24 cm²
(ii) Area of given trapezium = \(\frac {1}{2}\) × (10 + 5) × 6
= \(\frac {1}{2}\) × 15 × 6 cm² = 45 cm²

Try These : [Textbook Page No. 174]

1. We know that parallelogram is also a quadrilateral.

Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already?
Solution:
Let XYZW be a given quadrilateral, which is a parallelogram.

Join the diagonal WY of the parallelogram XYZW. It divides the parallelogram into two triangles ∆WXY and ∆YZW.
Then, area of parallelogram WXYZ
= Area of ∆ WXY + Area of ∆ YZW
= (\(\frac {1}{2}\) × xy × h) + (\(\frac {1}{2}\) × wz × h)
= (\(\frac {1}{2}\) × b × h) + (\(\frac {1}{2}\) × b × h)
= \(\frac {1}{2}\) bh + \(\frac {1}{2}\) bh
= bh sq units
Area of parallelogram = base × height
= b × h
= bh sq units
As we have studied that, a parallelogram is a special case of a trapezium, where parallel sides are equal.
∴ Area of trapezium XYZW
= \(\frac {1}{2}\) × (b + b) × h
= (\(\frac {1}{2}\) × 2b × h) sq. units
= bh sq units
Thus, the above relation prooves the formula which already we know.

Think, Discuss and Write: [Textbook Page No. 175]

1. A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?
Solution:
No, by dividing trapezium by diagonal two congruent triangles cannot be obtained.

Let us understand from given figures. Here, by drawing diagonals of a quadrilateral ABCD (which is a trapezium) congruent triangles are not obtained.

Try These : [Textbook Page No. 175]

1. Find the area of these quadrilaterals:

Solution:
(i) h₁ = 3 cm, h₂ = 5 cm
d = length of diagonal AC = 6 cm
∴ Area of quadrilateral ABCD
= \(\frac {1}{2}\)d(h₁ + h₂)
= \(\frac {1}{2}\) × 6 × (3 + 5)
= 3 × 8 cm²
= 24 cm²

(ii) d₁ = 7 cm, d₂ = 6 cm
∴ Area of rhombus ABCD
= \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 7 × 6
= 7 × 3 cm²
= 21 cm²

(iii) [Note : Here, given figure is a parallelogram. It’s diagonals divides it into two congruent triangles. From this figure, we can see that base of the triangle is 8 cm and height is 2 cm.]
∴ Area of parallelogram
= 2 (area of ∆ ADC)
= 2 × (\(\frac {1}{2}\) × b × h)
= 2 × (\(\frac {1}{2}\) × 8 × 2) cm²
= 2 × 8
= 16 cm²

Try These : [Textbook Page No. 176]

(i) Divide the following polygons into parts (triangles and trapezium) to find out its area.

Solution:
(a)

Let’s draw perpendicular on diagonal \(\overline{\mathrm{FI}}\).
Here, GA ⊥ FI, EB ⊥ FI and HC ⊥ FI are drawn.
Area of polygon EFGHI = Area of ∆ GFA + Area of the trapezium ACHG + Area of ∆ HCI + Area of ∆ BIE + Area of ∆ FBE
= [\(\frac {1}{2}\) × FA × GA] + [\(\frac {1}{2}\) (AG + CH) × AC] + [\(\frac {1}{2}\) × CI × HC] + [\(\frac {1}{2}\) × BI × BE] + [\(\frac {1}{2}\) × FB × BE]
[Note : We can also use Area of ∆ EFI in place of Area of ∆ BIE + Area of ∆ FBE]

(b)

Let’s draw perpendicular on diagonal \(\overline{\mathrm{NQ}}\).
\(\overline{\mathrm{OE}}\) ⊥ \(\overline{\mathrm{NQ}}\), \(\overline{\mathrm{MF}}\) ⊥ \(\overline{\mathrm{NQ}}\), \(\overline{\mathrm{PG}}\) ⊥ \(\overline{\mathrm{NQ}}\) and \(\overline{\mathrm{RH}}\) ⊥ \(\overline{\mathrm{NQ}}\) are drawn.
Area of polygon MNOPQR = Area of ∆ NEO + Area of trapezium EGPO + Area of ∆ GQP + Area of ∆ HQR + Area of trapezium MRHF + Area of ∆ NFM
= [\(\frac {22}{7}\) × NE × OE] + [\(\frac {22}{7}\) × (OE + PG) × EG] + [\(\frac {22}{7}\) × GQ × PG] + [\(\frac {22}{7}\) × HQ × HR] + [\(\frac {22}{7}\) × (FM + HR) × FH] + [\(\frac {22}{7}\) × NF × FM]

(ii) Polygon ABODE is divided into parts as shown in figure.

Find its area if AD = 8 cm, AH = 6 cm,
AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.
Area of polygon ABODE = area of ∆ AFB + ….
Area of ∆ AFB = \(\frac {1}{2}\) × AF × BF
= \(\frac {1}{2}\) × 3 × 2 = ………..
Area of trapezium FBCH
= FH × \(\frac{(BF + CH)}{2}\)
= 3 × \(\frac{(2+3)}{2}\) [FH = AH – AF]
Area of ∆ CHD = \(\frac {1}{2}\) × HD × CH= …………;
Area of ∆ ADE = \(\frac {1}{2}\) × AD × GE = ………….
So the area of polygon ABCDE = …………
Solution:

Here, AD = 8 cm,
AH = 6 cm,
HD = 2 cm,
AG = 4 cm,
GD = 4 cm,
AF = 3 cm and
GF = 1 cm
Area of polygon ABCDE = Area of ∆ AFB + Area of trapezium FBCH + Area of ∆ CHD + Area of ∆ ADE
Now,
Area of ∆ AFB = \(\frac {1}{2}\) × AF × BF
= \(\frac {1}{2}\) × 3 × 2 = 3 cm2 … (i)
Area of trapezium FBCH
= \(\frac {1}{2}\) × (BF × CH) × FH
= \(\frac {1}{2}\) × (2 + 3) × 3 [∵ FH = AH – AF]
= \(\frac {1}{2}\) × 5 × 3 = \(\frac {15}{2}\)
= 7.5 cm² … (ii)
Area of ∆ CHD = \(\frac {1}{2}\) × HD × CH
= \(\frac {1}{2}\) × (AD – AH) × CH
[∵ HD = AD – AH]
= \(\frac {1}{2}\) × (8 – 6) × 3
= \(\frac {1}{2}\) × 2 × 3
= 3 cm² … (iii)
Area of ∆ ADE = \(\frac {1}{2}\) × AD × GE
= \(\frac {1}{2}\) × 8 × 2.5
= 4 × 2.5
= 10 cm² ………. (iv )
∴ Area of polygon ABCDE = Area of [(i) + (ii) + (iii) + (iv)]
= (3 + 7.5 + 3 + 10) cm²
= 23.5 cm²

(iii) Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm NA, OC, QD and RB are perpendiculars to diagonal MP.

Solution:

Here,
MP = 9 cm,
MD = 7 cm
∴ DP = MP – MD
= (9 – 7) cm
= 2 cm
MC = 6 cm ∴ CP = 3 cm
MB = 4 cm ∴ BP = 5 cm
MA = 2 cm ∴ AC = MC – MA
= (6 – 2) cm = 4 cm
MB + BD = MD
∴ BD = (7 – 4) cm
= 3 cm
Area of ∆ MAN = \(\frac {1}{2}\) × MA × AN
= \(\frac {1}{2}\) × 2 × 2.5
= 2.5 cm² ……. (i)
Area of trapezium ACON
= \(\frac {1}{2}\) × (AN + OC) × AC
= \(\frac {1}{2}\) × (2.5 + 3) × 4
= \(\frac {1}{2}\) × 4 × 5.5
= 2 × 5.5
= 11 cm² … (ii)
Area of ∆ CPO = \(\frac {1}{2}\) × CP × CO
= \(\frac {1}{2}\) × 3 × 3
= \(\frac {9}{2}\) = 4.5 cm² … (iii)
Area of ∆ MBR = \(\frac {1}{2}\) × MB × BR
= \(\frac {1}{2}\) × 4 × 2.5
= 2 × 2.5
= 5 cm² … (iv )
Area of trapezium BRQD
= \(\frac {1}{2}\) × (BR + DQ) × BD
= \(\frac {1}{2}\) × (2.5 + 2) × 3
= \(\frac {1}{2}\) × 4.5 × 3
= 6.75 cm² ………. (v)
Area of ∆ DQP = \(\frac {1}{2}\) × DP × DQ
= \(\frac {1}{2}\) × 2 × 2
= 2 cm² … ( vi )
∴ Area of polygon MNOPQR = Area of [(i) + (ii) + (iii) + (iv) + (v) + (vi)]
= (2.5 + 11 + 4.5 + 5 + 6.75 + 2) cm²
= 31.75 cm²

Think, Discuss and Write : [Textbook Page No. 180]

1. Why is it incorrect to call the solid shown here a cylinder?
Solution:

It is incorrect to call given solid as a cylinder because, a cylinder has two identical circular faces, parallel to each other. The radii of both the faces are the same.

Try These :[Textbook Page No. 181]

1. Find the total surface area of the following cuboids:

Solution:
(i) Here, length (i) = 6 cm,
breadth (b) = 4 cm and
height (h) = 2 cm.
Total surface area of a cuboid = 2 (lb + bh + hl)
= 2(6 × 4 + 4 × 2 + 2 × 6)
= 2(24 + 8 + 12)
= 2 (44)
= 88 cm²

(ii) Here, length (l) = 4 cm,
breadth (b) = 4 cm and
height (h) = 10 cm.
Total surface area of a cuboid = 2 (lb + bh + hl)
= 2 (4 × 4 + 4 × 10 + 10 × 4)
= 2 (16 + 40 + 40)
= 2(96) = 192 cm²

Think, Discuss and Write :[Textbook Page No. 181]

1. Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base?
Solution:
Yes, the total surface area of cuboid = lateral surface area + 2 × area of base [Note: Area of base and area of top of a cuboid, both are same.]

2. If we interchange the lengths of the base and the height of a cuboid to get another cuboid, will its lateral surface area change?

Solution:
(a) Lateral surface area of cuboid (i)
= 2(l + b) × h
(b) Lateral surface area of cuboid (ii)
= 2 (h + b) × l
Thus, it is clear that by interchanging the lengths of the base and the height of a cuboid, its lateral surface area will change.

Try These : [Textbook Page No. 182]

1. Find the surface area of cube A and lateral surface area of cube B.

Solution:
For cube A:
Side = 10 cm
Surface area of cube A = 6 × (side)²
= 6 × (10)²
= 6 × 100
= 600 cm²

For cube B:
Side = 8 cm
Lateral surface area of a cube B
= 4 × (side)²
= 4 × (8)²
= 4 × 64
= 256 cm²

Think, Discuss and Write : [Textbook Page No. 183]

Question (i)
Two cubes each with side b are joined to form a cuboid. What is the surface area of this cuboid? Is it 12b²? Is the surface area of cuboid formed by joining three such cubes, 18b²? Why?

Solution:
(a) By joining two cubes end-to-end with side b, a cuboid shape formed.
For cuboid:
length = b + b = 2b, breadth = b and height = b
Total surface area of this cuboid = 2 (lb + bh + lh)
= 2 [(2b × b) + (b × b) + (2b × b)]
= 2 (2b² + b² + 2b²)
= 2 (5b²)
= 10b²
So surface area of the cuboid formed by joining two cubes is not equal to 12b²), it is 10b²).

(b) By joining three cubes end-to-end with side b, a cuboid shape forms.
For cuboid:
length = b + b + b = 3b, breadth = b and height = b
Total surface area of this cuboid = 2 (lb + bh + lh)
= 2 [(3b × b) + (b × b) + (3b × b)]
= 2 (3b² + b² + 3b²)
= 2(7b²) = 14b²
So the surface area of cuboid formed by joining three such cubes is not equal to 18b², it is 14b².

(ii) How will you arrange 12 cubes of equal length to form a cuboid of smallest surface area?
Solution:
Let us arrange 12 cubes of equal length say b to form a cuboid in different situations.
(a)

Here, length = 12b, breadth = b and height = b
Total surface area of a cuboid = 2 (lb + bh + lh)
= 2 [(12b × b) + (b × b) + (12b × b)]
= 2 (12b² + b² + 12b²)
= 2 (25b²) – 50b²

(b)

Here, length = 3b, breadth = 2b and height = 2b
Total surface area of a cuboid
= 2 (lb+ bh+ lh)
= 2 [(3b × 2b) + (2b × 2b) + (2b × 3b)]
= 2 (6b² + 4b² + 6b²)
= 2 (16b²) = 32b²

(c)

Here, length = 6b, breadth = 2b and height = b
Total surface area of a cuboid
= 2 (lb + bh + lh)
= 2 [(6b × 2b) + (2b × b) + (b × 6b)]
= 2 (12b² + 2b² + 6b²)
= 2 (20b²)
= 40b²
Hence, from above results we can conclude that, if we arrange 12 cubes of equal length according to situation (b), we get smallest surface area.

(iii) After the surface area of a cube is painted, the cube is cut into 64 smaller cubes of same dimensions. How many have no face painted? 1 face painted? 2 faces painted? 3 faces painted?
Solution:
1. 8 cubes, which have no face painted. (∵ Middle 4 × 2)
2. 24 cubes, which have 1 face painted. (∵ on each surface 4 × 6)
3. 24 cubes, which have 2 faces painted. (∵ on each surface 4 × 6)
4. 8 cubes, which have 3 faces painted. (∵ on each surface 2 × 4)

Try These : [Textbook Page No. 184]

1. Find total surface area of the following cylinders:

Solution:
(i) Radius of cylinder r = 14 cm
Height of cylinder h = 8 cm
Total surface area of a cylinder = 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 14(14 + 8)
= 2 × 22 × 2 × 22
= 44 × 44
= 1936 cm²

(ii) Radius of cylinder r = \(\frac{\text { diameter }}{2}=\frac{2}{2}\) = 14 cm
Height of cylinder h = 2 cm
Total surface area of a cylinder = 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 1(1 + 2)
= 2 × \(\frac {22}{7}\) × 1 × 3
= \(\frac {132}{7}\)
= 18\(\frac {6}{7}\) m²

Think, Discuss and Write: [Textbook Page No. 184]

1. Note that lateral surface area of a cylinder is the circumference of base × height of cylinder. Can we write lateral surface area of a cuboid as perimeter of base × height of cuboid?
Solution:

Let l be the height, b be the breadth and h be the height of cuboid.
∴ Lateral surface area of a cuboid = Area of 4 walls of the cuboid
= (l × h) + (l × h) + (b × h) + (b × h)
= 2 lh + 2 bh
= 2(1 + b) × h
= Perimeter of base × height
Yes, we can write lateral surface area of a cuboid as perimeter of base × height of cuboid.

Try These : [Textbook Page No. 188]

1. Find the volume of the following cuboids:

Solution:
(i) For given cuboid:
length (l) = 8 cm, breadth (b) = 3 cm s and height (h) = 2 cm
Volume of a cuboid
= Area of base × height
= (l × b) × h
= (8 × 3) × 2
= 24 × 2 = 48 cm³
OR
Volume of cuboid = l × b × h
= (8 × 3 × 2) cm³
= 48 cm³

(ii) Area of base of cuboid = 24 m³
height (h) = 3 cm = \(\frac {3}{100}\) m
Volume of cuboid
= Area of base × height
= 24 × \(\frac {3}{100}\) = 0.72 m³

Try These: [Textbook Page No. 189]

1. Find the volume of the following cubes
(a) With a side 4 cm
Solution:
For given cube : side = 4 cm
∴ Volume of the cube = (side)³
= (4)³ = 4 × 4 × 4
= 64 cm³

(b) With a side 1.5m
Solution:
For given cube : side = 1.5 m
∴ Volume of the cube = (side)³
= (1.5)³ = 1.5 × 1.5 × 1.5
= \(\frac{15}{10} \times \frac{15}{10} \times \frac{15}{10}=\frac{3375}{1000}\)
= 3.375 m³

Think, Discuss and Write: [Textbook Page No. 189]

1. A company sells biscuits. For packing purpose they are using cuboidal boxes:
box A → 3 cm × 8 cm × 20 cm,
box B → 4 cm × 12 cm × 10 cm. What size of the box will be economical for the company? Why? Can you suggest any other size (dimensions) which has the same volume but is more economical than these?
Solution:
For box A:
Given dimension = 3 cm × 8 cm × 20 cm
length = 20 cm, breadth = 8 cm and height = 3 cm
∴ Volume of the box A = l × b × h
= 20 × 8 × 3
= 480 cm³
Total surface area of the box A
= 2 (lb + bh + lh)
= 2 [(20 × 8) + (8 × 3) + (20 × 3)]
= 2 (160 + 24 + 60 )
= 2 (244)
= 488 cm³

For box B:
Given dimension = 4 cm × 12 cm × 10 cm
length = 12 cm, breadth = 10 cm and height = 4 cm
∴ Volume of the box B = l × b × h
= 12 × 10 × 4
= 480 cm³
Total surface area of the box B
= 2 (lb + bh + lh)
= 2 [(12 × 10) + (10 × 4) + (12 × 4)]
= 2(120 + 40 + 48 )
= 2(208)
= 416 cm³
From above results, we can conclude that both boxes have same volume, but surface area of box B is less than that of box A.
∴ Box B is more economical than box A.
Now, let another box of size be 8 cm × 6 cm × 10 cm.
∴ length = 8 cm, breadth = 6 cm and height =10 cm
Volume = l × b × h
= 8 × 6 × 10 = 480 cm³
It’s surface area = 2 (lb + bh + Ih)
= 2 [(8 × 6) + (6 × 10) + (8 × 10)]
= 2 (48 + 60 + 80)
= 2 (188) = 376 cm²
Surface area of this box is less than that of box B.
∴ This box is more economical for company.

Try These: [Textbook Page No. 189]

1. Find the volume of the following cylinders:

Solution:
(i) For given cylinder :
radius (r) = 7 cm, height (h) = 10 cm
Volume of the cylinder = πr²h
= \(\frac {22}{7}\) × 7² × 10
= \(\frac {22}{7}\) × 7 × 7 × 10
= 22 × 70
= 1540cm³

(ii) For given cylinder:
base area 250 m², height (h) = 2m
Volume of the cylinder
= base area × height
= 250 × 2
= 500 m³