Oswal Practice Papers CBSE Class 10 Mathematics Solutions (Practice Paper - 8)

Section-A

1. (b) 150 cm²

Explanation :

$$\text{Diagonal of cube =}\space5\sqrt{3}\\\therefore\space a\sqrt{3} = 5\sqrt{3}$$

Where a is the side of cube.

⇒ a = 5 cm

Total surface area of cube = 6a²

= 6 × (5)²

= 6 × 25

= 150 cm²

2. (b) 25 m

Explanation :

Here, width of the river = DC

In ∆ ABC,

$$\text{tan}\space 30\degree =\frac{\text{AB}}{\text{BC}}\\\Rarr\space\text{BC} =\frac{9}{\text{tan} 30\degree}=9\sqrt{3}\space \text{m}$$

Now, in ∆ ABD

$$\text{tan 45\degree}=\frac{\text{AB}}{\text{BD}}\\\Rarr\space 1 =\frac{\text{AB}}{\text{AD}}$$

⇒ AB = BD

⇒ BD = 9 m

Hence, DC = DB + BC

$$= 9 + 9\sqrt{3}\\=9(\sqrt{3} + 1)\\\lbrack\sqrt{3} = 1.732\rbrack$$

= 9 (1.732 + 1)

= 9 × 2.732

= 24.588 m

≃ 25 m.

$$\textbf{3.\space}(a)\space 8\sqrt{3}\space \text{m}$$

Explanation :

Let AB and DC be two poles and AD be the wire

In ∆ ADE,

$$\text{sin}\space 60\degree=\frac{\text{DE}}{\text{AD}}\\\lbrack\text{AD = Length of wire}\rbrack\\\Rarr\space \text{AD}=\frac{\text{DE}}{\text{sin}\space 60\degree}=\frac{12}{\frac{\sqrt{3}}{2}}\\=\frac{12×2}{\sqrt{3}}=\frac{24×\sqrt{3}}{\sqrt{3}×\sqrt{3}}$$

$$\lbrack\text{On multiplying and}\\\text{dividing by}\space\sqrt{3}\rbrack\\=\frac{24\sqrt{3}}{3}=8\sqrt{3}\space m$$

4. (d) 100°

Explanation :

In ΔAPB and ΔCPD,

∠APB = ∠CPD = 50°

[vertically opposite angles]

$$\frac{\text{AP}}{\text{PD}}=\frac{6}{5}\\\text{and}\space\frac{\text{BP}}{\text{CP}}=\frac{3}{2.5}=\frac{6}{5}\\\text{...(ii)}$$

From equations (i) and (ii),

$$\frac{\text{AP}}{\text{PD}} = \frac{\text{BP}}{\text{CP}}$$

∴ ΔAPB ~ ΔCPD

[by SAS similarity criterion]

∴ ∠A = ∠D = 30°

[corresponding angles of similar triangles]

In ΔAPB, ∠A + ∠B + ∠APB = 180°

[sum of angles of a triangle = 180°]

⇒ 30° + ∠B + 50° = 180°

∴ ∠B = 180° – (50° + 30°)

= 100°

i.e., ∠PBA = 100°.

5. (d) – 1

Explanation :

By section formula,

$$x =\frac{mx_{2} + mx_{1}}{m + n},\\ y =\frac{my_{2} + ny_{1}}{m + n}$$

Here x = k, m = 1, n = 2, x₁ = 2, x₂ = – 7

∴ We get

$$k =\frac{(1)(\normalsize-7) + (2) (2)}{1 + 2}\\\Rarr\space k=\frac{-7 + 4}{3}\\\Rarr\space k=\frac{\normalsize-3}{3}\\\Rarr\space k = -1$$

6. (d) 1 : 8 cm

Explanation :

Let the height of the bigger cone be 2x cm.

Thus, height of the smaller cone = x cm

Also let the radius of the larger cone be 2r cm.

Thus, radius of the smaller cone = r cm (by mid-point theorem)

Now, volume of the large cone

$$\text{V} =\frac{1}{3}×\pi(2r)^{2}(2x)\\=\frac{8\pi}{3}r^{2}x$$

and volume of the smaller cone =

$$\frac{1}{3}\pi(r)^{2}(x)\\=\frac{\pi}{3}r^{2}x\\\text{Thus,}\\\frac{\text{Volume of the smaller cone}}{\text{Volume of the larger cone}}\\=\frac{\frac{\pi}{3}r^{2}x}{\frac{8\pi}{3}r^{2}x}=\frac{1}{8}$$

7. (a) 16.39 min

Explanation :

Let h be the height of vertical tower. Time taken by car to reach point A to B is 12 min.

Let v be the velocity of car.

AB = distance travelled by car in 12 min.

So, AB = 12v

Suppose, the car takes t minutes to reach from B to C

Then BC = vt

In ΔBCD, we have

$$\text{tan}\space45\degree =\frac{\text{DC}}{\text{BC}}\\\Rarr\space\ 1 =\frac{h}{vt}$$

h = vt

In ΔACD, we have

$$\text{tan\space} 30\degree =\frac{\text{DC}}{\text{AC}}\\\frac{1}{\sqrt{3}}=\frac{h}{12 v + vt}\\\text{...(ii)}$$

From (i) and (ii), we get

$$\sqrt{3}\space\text{vt} = 12 v + vt\\\sqrt{3t} = t + 12\\t(\sqrt{3}-1) = 12\\t =\frac{12}{\sqrt{3}-1}\\=\frac{12(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\=\frac{12(\sqrt{3} + 1)}{2}\\t = 6(\sqrt{3} +1)$$

= 6 × 2.732 = 16.39 min

8. (a) 12.6 cm

Explanation :

Let r be the length of pendulum length of arc = 6.6 cm

∠AOB = 30°

$$\text{Length of an arc =}\\\frac{\theta}{360\degree}×2\pi r\\6.6=\frac{30\degree}{360\degree}×2×\frac{22}{7}×r\\\Rarr\space r =\frac{6.6×360×7}{2×30×22}\\=\frac{6.6×12×7}{44}\\=\frac{6×12×7}{40}$$

r = 12.6 cm.

9. (b) 12 and 14

Explanation :

Let x and (x + 2) be two consecutive even integers.

x² + (x + 2)² = 340

⇒ x² + x² + 4x + 4 = 340

⇒ 2x² + 4x + 4 = 340

⇒ x² + 2x + 2 = 170

⇒ x² + 2x – 168 = 0

⇒ x² + 14x – 12x – 168 = 0

⇒ x(x + 14) – 12 (x + 14) = 0

⇒ (x – 12) (x + 14) = 0

⇒ x = 12 or x = – 14

So, given number can be 12, 14 or – 12, – 14.

$$\textbf{10.\space}(a)\space -1\text{and}\frac{1}{2}$$

Explanation :

2x² + x – 1 = 0

⇒ 2x² + 2x – x – 1 = 0

⇒ 2x (x + 1) – 1 (x + 1) = 0

⇒ (x + 1) (2x – 1) = 0

$$\Rarr\space \text{x = -1 or x}=\frac{1}{2}$$

Hence roots of equation are

$$– 1 \text{and}\space\frac{1}{2}.$$

11. (d) 5 units

Explanation :

$$\text{Given, tan}\space\theta =\frac{3}{4}$$

Length of greatest side = Hypotenuse of triangle

$$\text{As},\space\text{tan}\space\theta =\frac{\text{Perpendicular}}{\text{Base}}\\=\frac{3}{4}$$

Let, P = 3 and B = 4.

By Pythagoras theorem, H² = P² + B²

⇒ H² = (3)² + (4)²

⇒ H² = 9 + 16

⇒ H² = 25

⇒ H = 5 units

Hence length of greatest side is 5 units.

12. (a) 78

Explanation :

Class Interval	f	x	f × x
50 – 60	8	55	440
60 – 70	6	65	390
70 – 80	12	75	900
80 – 90	11	85	935
90 – 100	13	95	1235
	Σf = 50		Σf_ix_i = 3900

$$\text{Mean} =\bar{x} =\frac{3900}{50}\\=78$$

13. (b) 60

Explanation :

Between 10 and 250 multiples of 4 are
12, 16, 20,..................., 248

The given numbers are in A.P.

So, a = 12, d = 4

a_n = 248

[∵ a_n = a + (n – 1)d]

Let the number of terms = n

∴ 248 = 12 + (n – 1)4

⇒ 236 = 4n – 4

⇒ 240 = 4n

$$n =\frac{240}{n} = 60$$

Hence, there are 60 multiples of 4 between 10 and 250.

14. (a) 3

Explanation :

$$\text{Given,}\\\space\text{cosec}\space\theta -\text{cot}\space\theta=\frac{1}{3}$$

We know that

cosec θ + cot θ =

$$\frac{(\text{cosec}\space\theta +\text{cot}\space\theta)×(\text{cosec}\space\theta +\text{cot}\space\theta)}{(\text{cosec}\space\theta -\text{cot}\space\theta)}$$

$$=\frac{\text{cosec}^{2}\theta + \text{cot}^{2}\theta}{\text{cosec}\space\theta - \text{cot}\space\theta}\\=\frac{1}{\text{cosec}\space\theta -\text{cot}\theta}\\\Rarr\space\text{cosec}\space\theta + \text{cot}\space\theta =\\\frac{1}{\frac{1}{3}} = 3$$

15. (b) irrational number.

Explanation :

The product of a non-zero rational number and an irrational number is always irrational number.

16. (c) intersecting or coincident

Explanation :

If a pair of linear equations in two variables is consistent, then the lines represented by two equations are intersecting or coincident, i.e., either it has a unique solution or infinitely many solutions.

$$\textbf{17.\space}\pm3\sqrt{3}$$

Explanation :

x² – 27 = 0

x² = 27

$$x =\sqrt{27}\\x =\pm 3\sqrt{3}$$

18. (d) infinite solution

Explanation :

Given equation of system

3x+ y = 4

and 6x + 2y = 8

Here,

$$\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} =\frac{1}{2}$$

Therefore, the system of equation has infinite solutions because it is a coincident line.

19. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation :

As per the reason statement,

Sum of n terms of A.P. is

$$\text{S}_{n}=\frac{n}{2}[2A + (n-1)D]$$

Where A = first term, D = common difference. Hence, the sum of n terms of an A.P. is always in the form of quadratic expression. Hence, it is proved that reason is true.

As per the assertion given, S_n = 2n² + 3n + 1

∴ t_n = S_n – S_n–1

= [2n² + 3n + 1] – [2(n – 1)² + 3(n – 1) + 1]

= 2[n² – (n – 1)²] + 3[n – n + 1]

= 2[2n – 1] + 3

= 4n + 1

∴ D = t_n+1 – t_n = 4(n + 1) + 1 – 4n – 1 = 4

So, both Assertion and Reason are true and Reason is correct explanation of Assertion.

20. (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).

Explanation :

Given, f(1) + f (2) = 0

For f (x) = ax² + bx + c, we have

a + b + c + 4a + 2b + c = 0

⇒ 5a + 3b + 2c = 0 ...(1)

Since – 1 is a root of the equation

f (– 1) = a – b + c = 0 ...(2)

By eliminating c and b in turn by solving equations (1) and (2), we have

$$\frac{-b}{a} =\frac{3}{5},\frac{c}{a}=\frac{-8}{5}$$

Since – 1 is one root, the other root is

$$\frac{8}{5}.$$

Section-B

21. Let P(x, y) be the required point which lie on the line segment AB.

$$x =\frac{m_{2}x_{1} + m_{1}x_{2}}{m_{1} + m_{2}}\\=\frac{2×7 +1×3}{1 + 2}\\=\frac{14 + 3}{3} =\frac{17}{3}\\\text{and}\\\space y =\frac{m_{2}y_{1} +m_{1}y_{2}}{m_{1} + m_{2}}\\=\frac{2×(-6)+ 1×4}{1 + 2}\\=\frac{-12 +4}{3} =\frac{-8}{3}\\\text{Thus, point of P}\bigg(\frac{17}{3},\frac{\normalsize-8}{3}\bigg)$$

lies in the 4th quadrant. Ans

22. It is given that on dividing 70 by the required number, there is a remainder 5. This means that 70 – 5 = 65 is exactly divisible by the required number. Similarly, 125 – 8 = 117 is also exactly divisible by required number.

Now, required number = H.C.F. (65, 117)

∴ 65 = 5 × 13

and 117 = 3² × 13

∴ H.C.F. (65, 117) = 13

∴ Required number = 13 Ans.

Prime factors of 23150 = 2 × 5 × 5 × 463.

As per the fundamental theorem of Arithmetic, every number has a unique factorisation. Ans.

23. Number ‘x’ is selected from {1, 2, 3}.

Number ‘y’ is selected from {1, 4, 9}.

Therefore, two numbers can be selected in 9 ways. {(1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (2, 9), (3, 1), (3, 4), (3, 9)}

So, Total number of outcomes = 9

Product of xy less than ‘9’ = {(1, 1), (1, 4), (2, 1), (2, 4), (3, 1)}

$$\therefore\space\text{P(Product xy less than 9)}\\=\frac{5}{9}\space\textbf{Ans.}$$

24. Given quadratic equation is

kx² + 11x – 4 = 0

On comparing with ax² + bx + c = 0, we get

a = k, b = 11, c = – 4

It has real roots.

∴ D ≥ 0

⇒ b² – 4ac ≥ 0

⇒ (11)² – 4 × k × (– 4) ≥ 0

⇒ 121 + 16k ≥ 0

$$\Rarr\space k\geq\frac{-121}{16}\\\textbf{Ans.}$$

25. We have,

S_n = 3n² + 5n

⇒ S_{n – 1} = 3(n – 1)² + 5(n – 1)

= 3(n² + 1 – 2n) + 5n – 5

= 3n² + 3 – 6n + 5n – 5

= 3n² – n – 2

General term of A.P. is given as, T_n = S_n – S_{n – 1}

= 3n² + 5n – 3n² + n + 2

= 6n + 2

Putting n = 1, 2, 3, we get T₁ = 6 × 1 + 2 = 8

T₂ = 6 × 2 + 2 = 14

T₃ = 6 × 3 + 2 = 20

Now, T₂ – T₁ = 14 – 8 = 6

and T₃ – T₂ = 20 – 14 = 6

∴ Common difference = 6 Ans.

Let S_m and S_n be the sum of the first m and n terms of the A.P., respectively.

$$\text{S}_{m}=\frac{m}{2}[2a +(m-1)d]\\\text{S}_{n}=\frac{n}{2}[2a +(n-1)d]$$

Here, a is the first term and d is the common difference.

$$\text{Given,\space}\frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}\\\therefore\space\frac{\frac{m}{2}[2a +(m-1)d]}{\frac{n}{2}[2a + (n-1)d]}=\frac{m^{2}}{n^{2}}\\\Rarr\space\frac{m[2a +(m-1)d]}{n[2a +(n-1)d]}=\frac{m^{2}}{n^{2}}\\\Rarr\space\frac{2a + (m-1)d}{2a + (n-1)d}=\frac{m}{n}$$

⇒ [2a + (m – 1)d]n = [2a + (n – 1)d]m

⇒ 2an + (m – 1)dn = 2am + (n – 1)dm

⇒ d [(m – 1)n – (n – 1)m] = 2am – 2an

⇒ d[mn – n – mn + m] = 2a (m – n)

⇒ d(m – n) = 2a (m – n)

⇒ d = 2a

Now, T_m = a + (m – 1)d

= a + (m – 1)2a

= a [1 + 2(m – 1)]

= a [2m – 1]

T_n = a + (n – 1)d

= a + (n – 1)2a

= a [1 + (n – 1)2]

= a (2n – 1)

$$\text{Thus,\space}\\\frac{\text{T}_{m}}{\text{T}_{n}}=\frac{a(2m-1)}{a(2n-1)}\\\Rarr\space\frac{\text{T}_{m}}{\text{T}_{n}} =\frac{2m-1}{2n-1}\\\textbf{Hence Proved.}$$

Section-C

26. Given, A = (– 1, 0) and B = (5, 0).

Let the point on the X-axis be P(x, 0).

It is given that AP = PB

$$\Rarr\space\sqrt{(x+1)^{2} +(0-0)^{2}}\\=\sqrt{(5-x)^{2} + (0-0)^{2}}\\\Rarr\space (x +1)^{2} =(5-x)^{2}$$

⇒ x + 1 = 5 – x

⇒ 2x = 4

⇒ x = 2

Thus, the point is (2, 0). Ans.

27. The minimum distance that each should walk is the LCM of 54, 60 and 48.

Now, 54 = 2 × 33

60 = 22 × 3 × 5

48 = 24 × 3

∴ LCM (54, 60, 48) = 24 × 33 × 5

= 2160

∴ Required distance = 2160 cm

= 21 m 60 cm Ans.

28. Given, Width of cloth = 5 m

Base diameter of tent = 14 m

$$\therefore\space\text{Radius}=\frac{14}{2}= 7\space\text{cm}\\\text{Height = 24 m}\\\text{Slant height of cone,}\\\space l =\sqrt{h^{2} + r^{2}}\\=\sqrt{(24)^{2} + (7)^{2}}\\=\sqrt{576 + 49}\\=\sqrt{625}$$

= 25 cm

Let the length of the cloth be l’ m.

∴ Curved surface area of cone = Area of cloth

$$\Rarr\space\pi rl = l'×b\\\Rarr\space\frac{22}{7}×7×25 = l'×5\\\Rarr\space 22×25 =l'×5\\\Rarr\space l'=\frac{22×25}{5}\text{m}$$

⇒ l’ = 110 m

Thus, total cost of the cloth = ₹(25 × 110) = ₹2750. Ans.

Let the radius of base and height of a solid cylinder be r and h respectively.

Now, we have

r + h = 37 cm ...(i)

and T.S.A. of solid cylinder = 2πr (r + h)

= 1628 cm²

⇒ 2πr (37) = 1628 [Using (i)]

$$\Rarr\space r =\frac{1628}{37×2×\frac{22}{7}}$$

⇒ r = 7 cm

∴ Volume of the cylinder = πr²h

$$=\frac{22}{7}×7×7×30$$

[Using eq. (i), h = 30]

= 4620 cm³ Ans.

29. Given,

$$\text{S}_{n} =\frac{3n^{2} + 13n}{2}$$

We know that, a_n = S_n-S_n-1

or a₂₅ = S₂₅ – S₂₄

$$=\frac{3(25)^{2} + 13(25)}{2}-\\\frac{3(24)^{2} + 13(24)}{2}\\=\frac{1}{2}\lbrace3(25^{2} - 24^{2}) + 13(25-24)\rbrace\\=\frac{1}{2}(3×49 + 13)\\=\frac{1}{2}×160 =80$$

Ans.

Let number of students living in slums area be x

$$\therefore\space\text{Share per student = }₹\frac{18000}{x}$$

When number of students = x + 20

$$\text{Then, share per student =}\\₹\frac{18000}{x + 20}$$

According to question,

$$\frac{18000}{x}-\frac{18000}{x +20} = 240\\\Rarr\space\frac{18000x + 360000 - 18000x}{(x + 20)(x)}\\ = 240$$

⇒ 360000 = 240(x² + 20x)

⇒ 1500 = x² + 20x

⇒ x² + 20x – 1500 = 0

⇒ x² + 50x – 30x – 1500 = 0

⇒ x(x + 50) – 30(x + 50) = 0

⇒ (x + 50) (x – 30) = 0

⇒ x = 30 or x = – 50 (Rejected)

⇒ x = 30

∴ Number of students living in the slum area = 30. Ans.

30. Given class interval is of inclusive form.

We have to convert it to exclusive form by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class interval.

Class Interval	Corrected class interval	Frequency (f_i)	Mid-value (x_i)	u_i = (x_i – A)	f_iu_i
210 – 229	209.5 – 229.5	8	219.5	–40	–320
230 – 249	229.5 – 249.5	9	239.5	–20	–180
250 – 269	249.5 – 269.5	10	259.5 = A	0	0
270 – 289	269.5 – 289.5	7	279.5	20	140
290 – 309	289.5 – 309.5	2	299.5	40	80
		∑f_i = 36			∑f_iu_i = –280

$$\text{Mean} = A + \frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}}\\=259.5 +\frac{-280}{36}$$

= 259.5 – 7.778

= 251.722 Ans.

31. We have,

$$\text{tan}\theta +\frac{1}{\text{tan}\theta} = 2\\\Rarr\space\bigg(\text{tan}\space\theta + \frac{1}{\text{tan}\space\theta}\bigg)^{2} = 2^{2}\\\Rarr\space\text{tan}^{2}\theta +\frac{1}{\text{tan}^{2}\theta} +\\2×\text{tan}\space\theta×\frac{1}{\text{tan}\theta} = 4\\\Rarr\space \text{tan}^{2}\theta +\frac{1}{\text{tan}^{2}\theta} + 2 = 4\\\Rarr\space\text{tan}^{2}\theta +\frac{1}{\text{tan}^{2}\theta} = 2$$

Ans.

Section-D

32. (a) Let a and d be the first term and common difference of given A.P. respectively.

Given : 3rd term of A.P. = 7

⇒ T₃ = 7

⇒ a + 2d = 7 ...(i)

and T₇ – 3T₃ = 2

⇒ T₇ – 3 × 7 = 2

⇒ T₇ = 2 + 21

⇒ T₇ = 23

⇒ a + 6d = 23 ...(ii)

Subtracting equation (ii) from (i), we get

$$\text{a + 2d = 7}\\\text{a + 6d = 23}\\--\space\space\space -\\\begin{matrix}\hline-4d =-16\end{matrix}$$

⇒ d = 4

Putting the value of d in equation (i), we get

a + 2 × 4 = 7

⇒ a = 7 – 8 = – 1

Now, Sum of first 20 terms =

$$\frac{20}{2}[2a + (20-1)d]$$

= 10 [2 × (– 1) + 19 × 4]

= 10 [– 2 + 76]

= 10 × 74 = 740 Ans.

(b) Numbers which are multiples of 2 and 9 are
18, 36, 54, 72, 90, 108, 126, ...

Since, 36 – 18 = 54 – 36 = 18

So, above series is an A.P. with a = 18, d = 18

Sum of first seven terms of the A.P. =

$$\frac{7}{2}[2a + (7-1)d]\\=\frac{7}{2}[2 × 18 + 6×18]\\=\frac{7}{2}[36 + 108]\\=\frac{7}{2}×144$$

= 7 × 72 = 504 Ans.

33. Since the arrow can stop in any one of the six sectors. So, a and b both can assume values from 1 to 6.

Thus, the ordered pair (a, b) can assume the following values :

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Clearly, there are total 36 elementary events, out of which we have 6 elementary events (a, b) in the diagonal for which a = b and 15 elementary events (a, b) below the diagonal for which a > b.

∴ Favourable number of events = 15

Hence, Required probability =

$$\frac{15}{36}=\frac{5}{12}\space\textbf{Ans.}$$

34. Since, E divides BC in the ratio 2 : 1

$$\therefore\space\frac{\text{BE}}{\text{EC}}=\frac{2}{1}\\\text{...(i)}$$

∠BED = ∠BDE [Given]

⇒ BD = BE [Sides opposite to equal angles] ...(ii)

In ∆CBG,

DE || CG [By construction]

$$\Rarr\space\frac{\text{BD}}{\text{DG}}=\frac{\text{BE}}{\text{EC}}\\\text{[Thale’s theorem]}\\\Rarr\space\frac{\text{BD}}{\text{DG}} =\frac{\text{BE}}{\text{EC}}=\frac{2}{1}\\\text{[From eq. (i)]}\\\Rarr\space\frac{\text{BD}}{\text{DG}}=\frac{2}{1}$$

⇒ 2DG = BD

⇒ 2DG = BE [From eq. (ii)]

$$\Rarr\space\text{DG}=\frac{1}{2}\text{BE}\\\text{...(iii)}$$

In ∆ADF,

CG || DF [By construction]

$$\Rarr\space\frac{\text{AG}}{\text{GD}}=\frac{\text{AC}}{\text{CF}}\\\text{[Thales’ theorem]}\\\Rarr\space\frac{\text{AG}}{\text{GD}} + 1 =\frac{\text{AC}}{\text{Cf}}+1\\\text{[Adding 1 on both sides]}\\\Rarr\space\frac{\text{AG + GD}}{\text{GD}} =\frac{\text{AC + CF}}{ \text{CF}}\\\Rarr\space\frac{\text{AD}}{\text{GD}} =\frac{\text{AF}}{\text{CF}}$$

⇒ AF × GD = AD × CF

$$\Rarr\space \text{AF}×\frac{\text{BE}}{2} =\\\text{AD × CF}\\\text{[From (iii)]}$$

⇒ AF × BE = 2AD × CF. Hence Proved.

Let EO || DC meet AD at E.

So, in ∆ADC, EO || DC

$$\text{So,}\space\frac{\text{AO}}{\text{OC}} =\frac{\text{AE}}{\text{ED}}\\\text{[Thales’ theorem] …(i)}\\\text{But}\space\frac{\text{AO}}{\text{OC}} =\frac{\text{BO}}{\text{OD}}\\\text{[Given] …(ii)}\\\text{By (i) and (ii),}\\\text{So,}\space\frac{\text{BO}}{\text{OD}}=\frac{\text{AE}}{\text{ED}}$$

$$\text{Hence},\space\frac{\text{BO}}{\text{OD}}=\frac{\text{AE}}{\text{ED}}\space\\\text{and BO, OD, AE}\\\text{and ED are segments of ∆DAB.}$$

So, EO || AB [By converse of Thales’ theorem]

But EO || DC

Thus, AB || DC

As there are only two parallel sides in this quadrilateral, it is a trapezium.

Hence Proved.

35. Given, r = 10 cm and length of chord PQ = 16 cm.

As TO bisects PQ at R, so PR = RQ = 8 cm

Let TP = x cm and TR = y cm

Now OR² = PO² – PR²

[By Pythagoras theorem as ∠PRO = 90°]

⇒ OR² = (10)² – (8)²

⇒ OR² = 100 – 64

⇒ OR² = 36

⇒ OR = 6 cm

Thus, TO = TR + RO = ( y + 6) cm

Also TP² = TR² + PR²

⇒ x² = y² + 8²

⇒ x² = y² + 64 …(i)

Again TO² = OP² + TP²

⇒ (y + 6)² = 10² + x²

⇒ (y + 6)² = 10² + y² + 64 [From (i)]

⇒ y² + 36 + 12y = y² + 164

⇒ 36 + 12y = 164

⇒ 12y = 128

⇒ 3y = 32

$$\Rarr\space y =\frac{32}{3}\text{cm}$$

Hence TP² = x² = y² + 64

$$\Rarr\space\text{TP}^{2} = x^{2}=\bigg(\frac{32}{3}\bigg)^{2} + 64\\\Rarr\space x^{2}=\frac{1024}{9} + 64\\\Rarr\space x^{2} =\frac{1024 + 576}{9}\\=\frac{1600}{9}\\\Rarr\space x =\frac{40}{3}$$

Hence TP = 13.33 cm. Ans.

Let O be the centre of the circle. Join OC.

Since angle between the radius and the tangent is 90°.

∴ ∠OCT = 90°

⇒ ∠OCA = ∠OCT – ∠ACT

= 90° – 48° = 42°

and OA = OC = OB [Radii of the same circle]

⇒ ∠OAC = ∠OCA = 42° [Angles opposite to equal sides are equal]

In ΔACT, exterior ∠CAB = ∠ACT + ∠CTA = 48° + 36° = 84°

∴ ∠OAB = ∠CAB – ∠OAC

⇒ ∠OAB = 84° – 42° = 42°

⇒ ∠OBA = ∠OAB = 42° [OA = OB = radii]

In isosceles ΔAOB,

∠OBA + ∠OAB + ∠AOB = 180°

⇒ 42° + 42° + ∠AOB = 180°

⇒ ∠AOB = 180° – 84° = 96°

Hence, the angle subtended by AB at centre is 96°. Ans.

Section-E

36. (i) Distance covered = Speed × Time

= 20 × 10 × 60 m

= 12,000 m

= 12 km Ans.

(ii) Let the height of the aeroplane from the ground be ‘h’ km

Then,

$$\text{tan 60\degree}=\frac{h}{\text{PQ}}\\\Rarr\space\text{PQ} =\frac{h}{\sqrt{3}}\\\text{and}\space\text{tan}\space 30\degree=\frac{h}{\text{PR}}\\\Rarr\text{PR}=\sqrt{3}h$$

But PR – PQ = QR = AB

$$\Rarr\space\sqrt{3}h -\frac{h}{\sqrt{3}} = 12\space\text{km}\\\Rarr\space\text{3h-h = 12}\sqrt{3}\\\Rarr\space 2h = 12\sqrt{3}\\\Rarr\space h =6\sqrt{3}$$

= 6 × 1.73 = 10.38 km Ans.

$$\text{In\space}\Delta\text{PQB,}\space\text{sin}\space 60\degree=\\\frac{\text{BQ}}{\text{PB}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{h}{\text{PB}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{10.38}{\text{PB}}$$

⇒ PB = 12 km Ans.

(iii) Similarly, in ΔAPR

$$\text{sin 30\degree}=\frac{\text{AR}}{\text{AP}}\\\frac{1}{2} =\frac{h}{\text{AP}}$$

⇒ AP = 2 × 10.38

= 20.76 km Ans.

37. (i) For cylinder A,

h = 25 cm

and circumference of base, 2πr = 14 cm

$$\Rarr\space\frac{2×22}{7} r = 14\\\Rarr\space r =\frac{7×14}{44}=\frac{49}{22}$$

Then V_A = πr²h

$$=\frac{22}{7}×\frac{49}{22}×\frac{49}{22}×25$$

= 389.8 cm³ Ans.

For cylinder B, h’ = 14 cm

Circumference of base, 2πr’ = 25

$$r' =\frac{25}{2\pi}$$

V_B = πr’²h’

$$=\frac{\pi×25}{2\pi}×\frac{25}{2\pi}×14\\=\frac{625×7×7}{2×22}$$

= 696.02 cm³ Ans.

(ii) Radius of cylinder A, 2πr = 14

$$r =\frac{14}{2\pi}\space\text{...(i)}$$

Radius of cylinder B, 2pr’ = 25

$$r' =\frac{25}{2\pi}\space\text{...(ii)}$$

From (i) and (ii), we get

$$\frac{r}{r'}=\frac{\frac{14}{2\pi}}{\frac{25}{2\pi}}=\frac{14}{25}$$

Ans.

(iii) 2πrh Ans.

38. (i) Let ΔABC is the triangle formed by both hotels and mountain top. ΔCDE is the triangle formed by both huts and mountain top.

Clearly DE ‖AB and so ΔABC ~ ΔDEC

[By AA-similarity criterion]

Now, required ratio = Ratio of their corresponding sides =

$$\frac{\text{BC}}{\text{EC}}=\frac{10}{7}\space\text{i.e. 10 : 7.}$$

Ans.

(ii) Since, DE‖AB, therefore

$$\frac{\text{CD}}{\text{AD}} =\frac{\text{CE}}{\text{EB}}\\\Rarr\space \frac{10}{\text{AD}} =\frac{7}{3}\\\Rarr\space\text{AD} =\frac{10×3}{7}\\= 4.29\space\text{miles.}$$

(iii) Since, ΔABC ~ ΔDEC

$$\therefore\space\frac{\text{BC}}{\text{EC}}=\frac{\text{AB}}{\text{DE}}$$

[∵ Corresponding sides of similar triangles are proportional]

$$\Rarr\space\frac{10}{7}=\frac{\text{AB}}{8}\\\Rarr\space \text{AB} =\frac{80}{7}= 11.43\space\text{miles.}$$

Ans.

Given, DC = 5 + BC

Clearly, BC = 10 – 5 = 5 miles

$$\text{Now,\space}\text{CE}=\frac{7}{10}×\text{BC}\\=\frac{7}{10} × 5$$

= 3.5 miles Ans.