Mathematics Solutions Solutions for Class 8 Math Chapter 19 Miscellaneous Exercise 2 are provided here with simple step-by-step explanations. These solutions for Miscellaneous Exercise 2 are extremely popular among Class 8 students for Math Miscellaneous Exercise 2 Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Page No 119:

#### Question 1:

Questions and their alternative answers are given. Choose the correct alternative answer.

(1) Find the circumference of a circle whose area is 1386 cm^{2}.

(A) 132 cm^{2}

(B) 132 cm

(C) 42 cm

(D) 21 cm^{2 }

(2) The side of a cube is 4 m . If it is doubled, how many times will be the volume of the new cube, as compared with the original cube?

(A) Two times

(B) Three times

(C) Four times

(D) Eight times

#### Answer:

(1) Area = 1386 cm2

$\mathrm{\pi}{r}^{2}=1386\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}{r}^{2}=1386\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=441\phantom{\rule{0ex}{0ex}}\Rightarrow r=21\mathrm{cm}$

Circumference = $2\mathrm{\pi}r=2\times \frac{22}{7}\times 21=132\mathrm{cm}$

Hence, the correct answer is option B.

(2) Side of a cube = 4 m

Volume = ${a}^{3}={4}^{3}=64{\mathrm{m}}^{3}$

New side = $2\times \mathrm{original}\mathrm{side}=2\times 4=8$ m

Volume of new cube = ${\left(8\right)}^{3}=512{\mathrm{m}}^{3}$

New volume = 512 m^{3 }= $8\times 64=8\times \mathrm{old}\mathrm{volume}$

Thus, the new volume is 8 times the old volume.

Hence, the correct answer is option D.

#### Page No 119:

#### Question 2:

Pranalee was practising for a 100 m running race. She ran 100 m distance 20 times. The time required, in seconds, for each attempt was as follows.

18, 17, 17, 16, 15, 16, 15, 14, 16, 15, 15, 17, 15, 16, 15, 17, 16, 15, 14, 15.

Find the mean of the times taken for running.

#### Answer:

Mean = $\frac{\mathrm{Sum}\mathrm{of}\mathrm{observations}}{\mathrm{Number}\mathrm{of}\mathrm{observations}}$

$\mathrm{Mean}=\frac{18+17+17+16+15+16+15+14+16+15+15+17+15+16+15+17+16+15+14+15}{20}$

$\Rightarrow \mathrm{Mean}=\frac{314}{20}=15.7$ sec

#### Page No 119:

#### Question 3:

∆ DEF and ∆ LMN are congruent in the correspondence EDF ↔ LMN. Write the pairs of congruent sides and congruent angles in the correspondence.

#### Answer:

∆DEF ≅ ∆LMN

$\Rightarrow \angle \mathrm{DEF}=\angle \mathrm{LMN},\angle \mathrm{EFD}=\angle \mathrm{MNL},\angle \mathrm{FDE}=\angle \mathrm{NLM}\phantom{\rule{0ex}{0ex}}\mathrm{And}\mathrm{DE}=\mathrm{LM},\mathrm{EF}=\mathrm{MN},\mathrm{DF}=\mathrm{LN}\phantom{\rule{0ex}{0ex}}$

#### Page No 119:

#### Question 4:

The cost of a machine is ₹2,50,000. It depreciates at the rate of 4% per annum. Find the cost of the machine after three years.

#### Answer:

Cost of a machine = ₹2,50,000

Rate of depreciation = 4%

Time period = 3 years

$A=P{\left(1-\frac{r}{100}\right)}^{n}\phantom{\rule{0ex}{0ex}}\Rightarrow A=250000{\left(1-\frac{4}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow A=250000{\left(\frac{96}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow A=2,21,184$

Amount after three years = Rs 2,21,184

#### Page No 119:

#### Question 5:

In ☐ ABCD side AB ∥ side DC, seg AE ⊥ seg DC. If *l* (AB) = 9 cm, *l* (AE) = 10 cm, A(☐ ABCD) = 115 cm^{2}, find *l *(DC).

#### Answer:

AB = 9 cm

AE = 10 cm

side AB ∥ side DC

So, ABCD is a trapezium

Area of trapezium = $\frac{1}{2}\times \mathrm{AE}\left(\mathrm{AB}+\mathrm{DC}\right)=\frac{1}{2}\times 10\left(9+\mathrm{DC}\right)=115$

$\Rightarrow 9+\mathrm{DC}=\frac{115\times 2}{10}=23\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DC}=14\mathrm{cm}$

#### Page No 119:

#### Question 6:

The diameter and height of a cylindrical tank is 1.75 m and 3.2 m respectively. How much is the capacity of tank in litre? (π = $\frac{22}{7}$)

#### Answer:

Diameter = 1.75 m

Radius = 0.875 m

Height = 3.2 m

Capacity = $\mathrm{\pi}{r}^{2}h=\frac{22}{7}\times {\left(0.875\right)}^{2}\times 3.2=7.7$ m^{3 }

$1{\mathrm{m}}^{3}=1000\mathrm{L}\phantom{\rule{0ex}{0ex}}7.7{\mathrm{m}}^{3}=7.7\times 1000=7700\mathrm{L}$

#### Page No 119:

#### Question 7:

The length of a chord of a circle of 16.8 cm, radius is 9.1 cm. Find its distance from the centre.

#### Answer:

Length of the chord = 16.8 cm

Radius = 9.1 cm

Let the distance from the centre of the chord be *d.*

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

So, AC = CB = $\frac{16.8}{2}=8.4$ cm

In $\u25b3$AOC,

Applying the Pythagoras theorem we have

${\mathrm{OC}}^{2}+{\mathrm{AC}}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{2}+8.{4}^{2}=9.{1}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{2}=82.81-70.56\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{2}=12.25\phantom{\rule{0ex}{0ex}}\Rightarrow d=3.5\mathrm{cm}$

Thus, the distance of chord from the centre is 3.5 cm.

#### Page No 119:

#### Question 8:

The following tables shows the number of male and female workers, under employment gurantee scheme, in villages A, B, C and D.

Village | A | B | C | D |

No. of females | 150 | 240 | 90 | 140 |

No. of males | 225 | 160 | 210 | 110 |

(2) Show the information by a percentage bar diagram.

#### Answer:

(1) Sub-divided bar diagram

(2) Percentage bar diagram

#### Page No 120:

#### Question 9:

Solve the following equations.

(1) 17(*x*+4) + 8(*x*+6) = 11(*x*+5) +15(*x*+3)

(2) $\frac{3y}{2}$+$\frac{y+4}{4}$ = 5−$\frac{y-2}{4}$

(3) 5(1−2*x*) = 9(1−*x*)

#### Answer:

(1) 17(*x*+4) + 8(*x*+6) = 11(*x*+5) +15(*x*+3)

$\Rightarrow 17x+68+8x+48=11x+55+15x+45\phantom{\rule{0ex}{0ex}}\Rightarrow 25x+116=26x+100\phantom{\rule{0ex}{0ex}}\Rightarrow 26x-25x=116-100\phantom{\rule{0ex}{0ex}}\Rightarrow x=16$

(2) $\frac{3y}{2}$+$\frac{y+4}{4}$ = 5−$\frac{y-2}{4}$

$3y\times 2+y+4=20-y-2\phantom{\rule{0ex}{0ex}}\Rightarrow 6y+y+4=18-y\phantom{\rule{0ex}{0ex}}\Rightarrow 8y=14\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{14}{8}=\frac{7}{4}$

(3) 5(1−2*x*) = 9(1−*x*)

$\Rightarrow 5-10x=9-9x\phantom{\rule{0ex}{0ex}}\Rightarrow -10x+9x=9-5\phantom{\rule{0ex}{0ex}}\Rightarrow -x=4\phantom{\rule{0ex}{0ex}}\Rightarrow x=-4\phantom{\rule{0ex}{0ex}}$

#### Page No 120:

#### Question 10:

Complete the activity according to the given steps.

(1) Draw rhombus ABCD. Draw diagonal AC.

(2) Show the congruent parts in the figure by identical marks.

(3) State by which test and in which correspondence ∆ ADC and ∆ ABC are congruent.

(4) Give reason to show ∠DCA ≌ ∠BCA, and ∠DAC ≌ ∠BAC

(5) State which property of a rhombus is revealed from the above steps.

#### Answer:

(1)

(2)

(3) In ∆ ADC and ∆ ABC

AC = AC (Common)

AD = AB (ABCD is a rhombus so all sides are equal)

DC = BC (ABCD is a rhombus so all sides are equal)

So, by SSS congruency ∆ ADC is congruent to ∆ ABC

(4) Since ∆ ADC is congruent to ∆ ABC so,

∠DCA ≌ ∠BCA and ∠DAC ≌ ∠BAC by CPCT (corresponding parts of congruent triangles)

(5) Since ∠DCA ≌ ∠BCA and ∠DAC ≌ ∠BAC so, AC acts as the angle bisector.

Thus, we can say that diagonals of a rhombus act as angle bisectors.

#### Page No 120:

#### Question 11:

The shape of a farm is a quadrilateral. Measurements taken of the farm, by naming its corners as P, Q, R, S in order are as follows. *l*(PQ) = 170 m, *l*(QR) = 250 m, *l*(RS) = 100 m, *l*(PS) = 240 m, *l*(PR) = 260 m.

Find the area of the field in hectare (1 hectare = 10,000 sq.m)

#### Answer:

In $\u2206$PSR,

$s=\frac{100+260+240}{2}=\frac{600}{2}=300\phantom{\rule{0ex}{0ex}}\mathrm{Ar}\left(\u25b3\mathrm{PSR}\right)=\sqrt{300\left(300-100\right)\left(300-260\right)\left(300-240\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{300\times 200\times 40\times 600}\phantom{\rule{0ex}{0ex}}=\sqrt{144000000}\phantom{\rule{0ex}{0ex}}=12000\mathrm{sq}\mathrm{m}$

In $\u2206$PQR,

$s=\frac{260+250+170}{2}=340\phantom{\rule{0ex}{0ex}}\mathrm{Ar}\left(\u25b3\mathrm{PQR}\right)=\sqrt{340\left(340-260\right)\left(340-250\right)\left(340-170\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{340\times 80\times 90\times 170}\phantom{\rule{0ex}{0ex}}=\sqrt{416160000}\phantom{\rule{0ex}{0ex}}=20400\mathrm{sq}\mathrm{m}$

Area of PQRS = Ar($\u2206$PSR) + Ar($\u2206$PQR)

= 12000 + 20400

= 32400 m^{2 }

1 hectare = 10,000 sq.m

$32400{\mathrm{m}}^{2}=3.24\mathrm{hec}$

#### Page No 120:

#### Question 12:

In a library, 50% of total number of books is of Marathi. The books of English are $\frac{1}{3}$rd of Marathi books. The books on mathematics are 25% of the English books. The remaining 560 books are of other subjects. What is the total number of books in the library?

#### Answer:

Let the total number of books be *x.*

50% books are in Marathi = $\frac{50}{100}x=\frac{x}{2}$

Books of English = $\frac{1}{3}$rd of Marathi books = $\frac{1}{3}\times \frac{x}{2}=\frac{x}{6}$

Books on mathematics are 25% of the English books = $\frac{25}{100}\times \frac{x}{6}=\frac{x}{24}$

Remaining books = 560

So,

$\frac{x}{2}+\frac{x}{6}+\frac{x}{24}+560=x\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{17x}{24}+560=x\phantom{\rule{0ex}{0ex}}\Rightarrow x-\frac{{\displaystyle 17x}}{{\displaystyle 24}}=560\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7x}{24}=560\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{560\times 24}{7}=1920$

Total books in the library are 1920.

#### Page No 120:

#### Question 13:

Divide the polynomial (6*x*^{3}+11*x*^{2}−10*x*−7) by the binomial (2*x*+1). Write the quotient and the remainder.

#### Answer:

Given polynomial = (6*x*^{3}+11*x*^{2}−10*x*−7)

Binomial = 2*x*+1

Thus, Quotient = $3{x}^{2}+4x-7$

Remainder = 0

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