KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

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KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

Try These (Page 170)

Question (a).
Match the following figures with their respective areas in the box.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 170 Q1
Solution:
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 170 Q1.1
Area of parallelogram ABCD
= Base × height
= 14 × 7
= 98 cm2
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 170 Q1.2
Area of semicircle = \(\frac{1}{2} \pi r^{2}\)
= \(\frac{1}{2} \times \frac{22}{7} \times \frac{14}{2} \times \frac{14}{2}\)
= 77 cm2
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 170 Q1.3
Area of triangle = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 14 × 7
= 49 cm2
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 170 Q1.4
Area of rectangle = l × b
= 14 × 7
= 98 cm2
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 170 Q1.5
Area of square = (side)2
= 7 × 7
= 49 cm2

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

Question (b).
Write the perimeter of each shape.
Solution:
Perimeter of parallelogram = 2(14 + b) cm

Perimeter of semicircle = \(\frac{2 \pi r}{2}\)
= πr
= \(\frac{22}{7}\) × 7
= 22 cm

Perimeter of triangle = (14 + 11 + 9)
= (14 + 20)
= 34 cm

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

Perimeter of square = 4 × side
= 4 × 7
= 28 cm

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

Try These (Page 172)

Question 1.
Nazma’s sister also has a trapezium-shaped plot. Divide it into three parts as shown in the figure below. Show that the area of trapezium WXYZ = \(h \frac{(a+b)}{2}\)
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 172 Q1
Solution:
Draw ZM ⊥ XW and YN ⊥ XW
Now, trapezium XYZW is divided into three parts. Two triangles and one rectangle.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 172 Q1.1
area of ΔZWM = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × c × h
area of ΔYNX = \(\frac{1}{2}\) × d × h
area of rectangle NYZM = b × h
area of trapezium XYZW = area of ΔZWM + area of ΔYNX + area of rectangle NYZM
= \(\frac{1}{2}\) × c × h + \(\frac{1}{2}\) × d × h + b × h
= \(h\left[\frac{1}{2} c+\frac{1}{2} d+b\right]\)
= \(h\left[\frac{c+d+2 b}{2}\right]\)
= \(h\left[\frac{c+d+b+b}{2}\right]\)
= \(h \frac{(a+b)}{2}\)

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

Question 2.
If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separately 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
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 172 Q2
area of ∆ZWM = \(\frac{1}{2}\) × c × h
= \(\frac{1}{2}\) × 6 × 10
= 30 cm2
area of ∆YNX = \(\frac{1}{2}\) × d × h
= \(\frac{1}{2}\) × 4 × 10
= 20 cm2
area of rectangle = b × h
= 12 × 10
= 120 cm2
Area of trapezium XYZW = 30 + 20 + 120 = 170 cm2
area of trapezium XYZW = \(\frac{h(a+b)}{2}\)
= \(\frac{h[c+d+b+b]}{2}\)
= \(\frac{10[6+4+12+12]}{2}\)
= 5(34)
= 170 cm2

Try These (Page 173)

Question 1.
Find the area of the following trapeziums.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 173 Q1
Solution:
(i) Area of trapezium = \(\frac{1}{2}\) (Sum of parallel side) × h
= \(\frac{1}{2}\) (9 + 7) × 3
= \(\frac{1}{2}\) × 16 × 3
= 24 cm2

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

(iii) Area of trapezium = \(\frac{1}{2}\) (Sum of parallel side) × h
= \(\frac{1}{2}\) (10 + 5) × 6
= \(\frac{1}{2}\) × 15 × 6
= 15 × 3
= 45 cm2

Try These (Page 174)

Question 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? (Fig. 11.12).
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 174 Q1
Solution:
Area of parallelogram = Base × Altitute
Area of parallelogram = Area of two triangles
= \(\frac{1}{2}\) × base × height + \(\frac{1}{2}\) × base × height
= 2 × \(\frac{1}{2}\) × b × h
= b × h
∴ b × h = b × h
Yes, this agrees with the formula that we already know. It means an area of a parallelogram is also equal to the sum of the area of two triangles.

Try These (Page 175)

Question 1.
Find the area of these quadrilaterals.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 175 Q1
Solution:
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 175 Q1.1
(i) Area of quadrilateral ABCD = ar ΔABC + ar ΔACD
= \(\frac{1}{2}\) × AC × BM + \(\frac{1}{2}\) × AC × CN
= \(\frac{1}{2}\) × 6 × 3 + \(\frac{1}{2}\) × 6 × 5
= 3 × 3 + 3 × 5
= 9 + 15
= 24 cm2

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

(ii) Area of trapezium = \(\frac{1}{2}\) × d1 × d2
= \(\frac{1}{2}\) × 7 × 6
= 21 cm2

(iii) Area of parallelogram = 2 area of Δ
= 2 × \(\frac{1}{2}\) × b × h
= 2 × \(\frac{1}{2}\) × 8 × 2
= 16 cm2

Try These (Page 176)

Question (i).
Divide the following polygons into parts (triangles and trapezium) to find out their area.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 176 Q1
FI is a diagonal of polygon EFGHI; NQ is a diagonal of polygon MNOPQR
Solution:
Polygon EFGHI is divided into two parts. ΔEFG and trapezium EIHG.
Polygon MNOPQ is divided into two triangles OPN and MRQ and trapezium PQMN.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 176 Q1.1

Question (ii).
Polygon ABCDE is divided into parts as shown below. 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.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 176 Q2
Area of Polygon ABCDE = area of ΔAFB + ……..
Area of ΔAFB = \(\frac{1}{2}\) × AF × BF = \(\frac{1}{2}\) × 3 × 2 = …….
Area of trapezium FBCH = FH × \(\frac{(B F+C H)}{2}\) = 3 × \(\frac{(2+3)}{2}\) [FH = AH – AF]
Area of ΔCHD = \(\frac{1}{2}\) × HD × CH = …….., Area of ΔADE = \(\frac{1}{2}\) × AD × GE = …. So, the area of polygon ABCDE = …….
Solution:
Area of polygon ABCDE = area of ΔAFB + area of trapezium FBCH + area of ΔCHD + area of ΔADE
= \(\frac{1}{2}\) × AF × BF + \(\frac{F H \times(B F+C H)}{2}\) + \(\frac{1}{2}\) × HD × CH + \(\frac{1}{2}\) × AD × GE
= \(\frac{1}{2}\) × 3 × 2 + \(\frac{(A H-A F) \times[2+3]}{2}\) + \(\frac{1}{2}\) × (AD – AH) × 3 + \(\frac{1}{2}\) × 8 × 2.5
= 3 + \(\frac{(6-3)(5)}{2}+\frac{1}{2}\) × (8 – 6) × 3 + 10
= 3 + \(\frac{15}{2}+\frac{1}{2}\) × 2 × 3 + 10
= 13 + \(\frac{15}{2}+\frac{6}{2}\)
= \(\frac{26+15+6}{2}\)
= \(\frac{47}{2}\)
= 23.5 cm2

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

Question (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.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 176 Q3
Solution:
Area of polygon MNOPQR = area of ∆AMN + area of trapezium ANOC + area of ∆OCP + area of ∆PDQ + area of trap. BDQR + area of ∆MBR
= \(\frac{1}{2}\) × AM × AN + \(\frac{1}{2}\) (AN + OC) × AC + \(\frac{1}{2}\) × CP × OC + \(\frac{1}{2}\) × DP × DQ + \(\frac{1}{2}\) (BR + DQ) × BD + \(\frac{1}{2}\) × MB × BR
= \(\frac{1}{2}\) × 2 × 2.5 + \(\frac{1}{2}\) (2.5 + 3) × (MC – AM) + \(\frac{1}{2}\) × (MP – MC) × 3 + \(\frac{1}{2}\) × (MP – MD) × 2 + \(\frac{1}{2}\) (2.5 + 2) × (MD – MB) + \(\frac{1}{2}\) × 4 × 2.5
= 2.5 + \(\frac{1}{2}\) × 5.5(6 – 2) + \(\frac{1}{2}\) (9 – 6) × 3 + \(\frac{1}{2}\) (9 – 7) × 2 + \(\frac{1}{2}\) × 4.5(7 – 4) + 5
= 2.5 + \(\frac{1}{2}\) × 5.5 × 4 + \(\frac{1}{2}\) × 3 × 3 + \(\frac{1}{2}\) × 2 × 2 + \(\frac{1}{2}\) × 4.5 × 3 + 5
= 2.5 + 11 + \(\frac{9}{2}\) + 2 + \(\frac{13.5}{2}\) + 5
= 15.5 + \(\frac{9}{2}+\frac{13.5}{2}\) + 5
= \(\frac{31+9+13.5+10}{2}\)
= \(\frac{63.5}{2}\)
= 31.75 cm2

Try These (Page 181)

Question 1.
Find the total surface area of the following cuboids:
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 181 Q1
Solution:
(i) Length (l) of cuboid = 6 cm
Breadth (b) of cuboid = 4 cm
Height (h) of cuboid = 2 cm
Total surface area of cuboid = 2(lb + bh + lh)
= 2(6 × 4 + 4 × 2 + 2 × 6]
= 2[24 + 8 + 12]
= 2 × 44
= 88 cm2

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

(ii) Length (l) of cuboid = 4 cm
Breadth (b) of cuboid = 4 cm
Height (h) of cuboid = 10 cm
Total surface area of cuboid = 2 (lb + bh + lh)
= 2(4 × 4 + 4 × 10 + 10 × 4)
= 2[16 + 40 + 40]
= 2 × 96
= 192 cm2

Try These (Page 182)

Question 1.
Find the surface area of cube A and lateral surface area of cube B.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 182 Q1
Solution:
(A) Side of cube = 10 cm
Surface area of cube = 6(side)2
= 6(10)2
= 6 × 100
= 600 cm2

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

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

Try These (Page 184)

Question 1.
Find the total surface area of the following cylinders.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 184 Q1
Solution:
(i) Radius of cylinder = 14 cm
Height of cylinder = 8 cm
Total surface area of cylinder = CSA of cylinder + area of bar
= 2πrh + 2πr2
= 2πr(h + r)
= 2 × \(\frac {22}{7}\) × 14[8 + 14]
= 2 × 22 × 2[8 + 14]
= 88 × 22
= 1936 cm2

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

(ii) Height of cylinder = 2 m
Diameter of cylinder = 2 m
Total surface area of cylinder = 2πrh + 2πr2
= 2πr(h + r)
= 2 × \(\frac {22}{7}\) × 1(2 + 1)
= \(\frac{44}{7}\) × 3
= \(\frac{132}{7}\)
= 18.86 cm2 (approx.)

Try These (Page 188)

Question 1.
Find the volume of the following cuboids.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 188 Q1
Solution:
(i) Length of cuboid (l) = 8 cm
Breadth of cuboid (b) = 3 cm
Height of cuboid (h) = 2 cm
Volume of cuboid = l × b × h
= 8 × 3 × 2
= 48 cm3

(ii) Area of base = 24 m2
Height of cuboid = 3 cm = \(\frac{3}{100}\) m
Volume of cuboid = Area of base × height
= 24 × \(\frac{3}{100}\)
= \(\frac{72}{100}\)
= 0.72 m3

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

Try These (Page 189)

Question 1.
Find the volume of the following cubes:
(a) with a side 4 cm
(b) with a side 1.5 m
Solution:
(a) Side of cube = 4 cm
Volume of cube = (Side)3
= (4)3
= 64 cm3

(b) Side of cube = 1.5 m
Volume of cube = (Side)3
= (1.5)3
= 1.5 × 1.5 × 1.5
= 3.3750 m3

Try These (Page 189)

Question 1.
Find the volume of the following cylinders.
KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions Page 189 Q1
Solution:
(a) Radius of cylinder = 7 cm
Height of cylinder = 10 cm
Volume of cylinder = πr2h
= \(\frac{22}{7} \times(7)^{2} \times 10\)
= \(\frac{22}{7}\) × 7 × 7 × 10
= 154 × 10
= 1540 cm3

KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

(b) Area of base = 250 m2
Height of cylinder = 2 m
Volume of cylinder = Area of base × height
= 250 × 2
= 500 m3