# Oswal Practice Papers CBSE Class 10 Mathematics Solutions (Practice Paper - 7)

## Section-A

1. (a) (4, 4)

Explanation :

Given : A(2, 3), B(4, 7) and C(6, 2).

We know that,

$$\text{Centroid (G) =}\\\bigg(\frac{x_{1} + x_{2} +x_{3}}{3},\frac{y_{1} + y_{2} + y_{3}}{3}\bigg)\\=\bigg(\frac{2+4+6}{3},\frac{3+7+2}{3}\bigg)$$

= (4, 4)

2. (b) 2 : 3

Explanation :

Let P be the point that divides the line segment joining (3, – 1) and (8, 9) in ratio k : 1.

Then, coordinates of point P is given by using internal section formula,

$$\text{P(x,y)} =\bigg[\frac{8k+3}{k+1},\frac{9k-1}{k+1}\bigg]$$

Since, point P lies on line x – y – 2 = 0, so P must satisfy the above equation.

$$\text{Putting}\space x =\frac{8k+3}{k+1}, y =\frac{9k-1}{k+1}$$

in equation x – y – 2 = 0, we get

$$\frac{8k+3}{k+1}-\bigg(\frac{9k-1}{k+1}\bigg)-2 = 0\\\Rarr\space\frac{8k+3-9k+1-2k-2}{k+1} = 0\\\Rarr\space\frac{-3k+2}{k+1}= 0\\\Rarr\space -3k + 2 = 0\\\Rarr\space k =\frac{2}{3}\\\Rarr\space k:1=\frac{2}{3}:1\\\Rarr\space 2 : 3$$

Hence, the required ratio is 2 : 3.

3. (d) 70°

Explanation :

Let each base angle of isosceles triangle be x.

∴Vertical angle of an isosceles triangle = x + 15°

We know that,

∠A + ∠B + ∠C = 180°

⇒ x + 15° + x + x = 180°

⇒ 3x = 165°

⇒ x = 55°

∴ x + 15 = 55 + 15

= 70°.

4. (d) 6 cm

Explanation :

In ΔABC, if AE is the angle bisector of ∠A, then according to the angle bisector theorem,

$$\frac{\text{AB}}{\text{AC}} = \frac{\text{BE}}{\text{EC}}\\\frac{9}{\text{AC}}=\frac{3.6}{2.4}\\\text{AC} =\frac{9×2.4}{3.6}= 6\space\text{cm.}$$

5. (b) 45°

Explanation :

∠AOD = 135°

∠BOC = 180° – 135° = 45°

Since, angle subtended at the centre by a pair of opposite sides are supplementary.

$$\textbf{6.\space}(b)\space\frac{2\pi \text{R}\theta}{360\degree}$$

Explanation :

Length of arc of a sector =

$$\frac{\theta}{360\degree}×2\pi\text{R}=\frac{2\pi \text{R}\theta}{360\degree}.$$

7. (c) 9

Explanation :

f(x) = x2 – x – (2k + 2)

Substituting x for – 4, we have f(– 4) = (– 4)2 – (– 4) – (2k + 2)

= 16 + 4 – 2k – 2

= 18 – 2k = 0

Thus, 18 – 2k = 0

or 2k = 18 or k = 9.

8. (c) cos 90°

Explanation :

Given, ∆ABC is right angled at C.

i.e., ∠C = 90°

Then by triangle angle sum property

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 180° – 90°

⇒ ∠A + ∠B = 90°

Taking cos on both the sides, we get

cos (A + B) = cos 90°

cos (A + B) = 0

[ cos 90° = 0]

$$\textbf{9.\space}(b)\space\frac{2}{3}$$

Explanation :

$$\text{Given :\space}\text{tan A}=\frac{1}{\sqrt{5}}\\\Rarr\space\text{tan}^{2}\text{A} =\frac{1}{5}\\\text{...(i)}\\\Rarr\space\frac{1}{\text{cot}^{2}\text{A}}=\frac{1}{5}\\\bigg[\because\space\text{tan A}=\frac{1}{\text{cot A}}\bigg]\\\Rarr\space \text{cot}^{2}\text{A} = 5\space\text{...(ii)}$$

$$\text{Now,}\space\frac{\text{cosec}^{2}A -\text{sec}^{2}A}{\text{cosec}^{2}A + \text{sec}^{2}A}\\=\frac{1 + \text{cot}^{2}A-(1 +\text{tan}^{2}A)}{1 +\text{cot}^{2}\text{A} + 1 + \text{tan}^{2}\text{A}}\\\begin{bmatrix}\because\space\text{cosec}^{2}\text{A} = 1 + \text{cot}^{2}A\\\text{and sec}^{2}\text{A} = 1 + \text{tan}^{2}A\end{bmatrix}\\=\frac{\text{cot}^{2}\text{A} -\text{tan}^{2}\text{A}}{2 +\text{cot}^{2}\text{A} +\text{tan}^{2}\text{A}}\\=\frac{5 -\frac{1}{5}}{2 + 5 +\frac{1}{5}}$$

[From (i) and (ii)]

$$=\frac{\frac{25-1}{5}}{\frac{35+1}{5}}=\frac{24}{36}=\frac{2}{3}$$

$$\therefore\space\frac{\text{cosec}^{2}\text{A} - \text{sec}^{2}\text{A}}{\text{cosec}^{2}\text{A} + \text{sec}^{2}\text{A}}=\frac{2}{3}$$

10. (c) 9

Explanation :

We have

9 sec2 A – 9 tan2 A = 9(sec2 A – tan2 A)

= 9(1 + tan2 A – tan2 A)

= 9 × 1   [ sec2 θ = 1 + tan2 θ]

= 9

11. (a) 6

Explanation :

kx(x – 2) + 6 = 0

⇒ kx2 – 2kx + 6 = 0

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

a = k, b = – 2k, c = 6

For equal roots,

D = b2 – 4ac = 0

⇒ (– 2k)2 – 4 × k × 6 = 0

⇒ 4k2 – 24k = 0

⇒ 4k (k – 6) = 0

⇒ k = 0 (not possible) or k = 6

∴ k = 6

12. (b) 2 cm

Explanation :

An incircle is drawn with centre O which touches the sides of the triangle ABC at P, Q and R. OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle.

OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA.

OPBQ is a square. ( ∠B = 90°)

Let r be the radius of the circle

∴ PB = BQ = r

AR = AP = 8 – r

CQ = CR = 6 – r

AC = AR + CR

⇒ 10 = 8 – r + 6 – r

⇒ 10 = 14 – 2r

⇒ 2r = 14 – 10 = 4

⇒ r = 2

∴ Radius of the incircle = 2 cm.

13. (a) 5, 1

Explanation :

It has infinite many solution

$$\text{So,\space}\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\\\Rarr\space\frac{2}{\text{m+n}} =\frac{3}{\text{2m-n}}=\frac{11}{33}\\\Rarr\space\frac{2}{m+n} =\frac{3}{2m-n} =\frac{1}{3}\\\Rarr\space\frac{2}{m+n} = \frac{1}{3}\\\Rarr\space m + n=6\\\text{...(i)}\\\text{and}\space\text{2m-n} =9\\\text{...(ii)}$$

⇒ m = 5

Putting m in equation (i), 5 + n = 6

⇒ n = 1.

14. (c) – 7, 10

Explanation :

Given quadratic polynomial is x2 + 7x + 10

Here a = 1, b = 7 and c = 10

We know,

$$\text{sum of roots}\space(\alpha + \Beta)=\frac{-b}{a}\\\alpha +\beta =\frac{\normalsize-7}{1} =-7\\\text{Product of roots (ab) =}\frac{c}{a}\\\alpha\beta=\frac{10}{1}= 10$$

15. (d) – 1

Explanation :

Given, f(y) = 2y2 + 7y + 5

$$\text{Here}\space\alpha +\beta =-\frac{7}{2}\\\bigg(\because\space\alpha +\beta=\frac{-b}{a}\bigg)\\\alpha\beta =\frac{5}{2}\\\bigg(\because\space\alpha + \beta=\frac{c}{a}\bigg)\\\therefore\space(\alpha+ \beta) +(\alpha\beta)=-\frac{7}{2}+\frac{5}{2}=-1.$$

16. (d) 4

Explanation :

Given, the distance of the ladder AC from the base of the wall AB is 2 m.

$$\text{So,\space cos 60\degree}=\frac{\text{BC}}{\text{AC}}=\frac{2}{x}\\\Rarr\space\frac{1}{2} = \frac{2}{x}\\\Rarr\space\text{x = 2×2}$$

The height of the ladder = 4 m.

$$\textbf{17.}\space\text{(C)}\frac{4}{45}$$

Explanation :

Total number of possible outcomes = 90

Thus, the favourable outcomes are 2, 3, 5, 7, 11, 13, 17, 19.

Hence, the total number of favourable outcomes = 8

$$\therefore\space\text{P(E)} =\frac{8}{90}=\frac{4}{45}$$

$$\textbf{18.\space}\text{(c)}\space\frac{364}{365}$$

Explanation :

If

Total number of possible outcomes = 365 × 365

Total number of favourable outcomes = 365 × (365 – 1)

= 365 × 364

$$\therefore\space\text{P(E)} =\frac{365×364}{365×365} =\frac{364}{365}$$

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

Explanation :

Maximum frequency = 23

Hence, modal class is 12 – 15

Now,

$$\text{Mode} =l +\frac{(f_{k}-f_{k-1})}{2f_{k}-f_{k-1}-f_{k+1}}×h\\= 12 +\bigg(\frac{23-21}{46 - 21 -10}\bigg)×3\\= 12 +\frac{6}{46-31}\\= 12 +\frac{6}{15}= 12.4$$

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

Explanation :

For reason, as per the empirical formula

The sum of all the probabilities of all possible outcomes of experiment is 1.

P(Event) + P (not an Event) = 1

So, reason is true.

For assertion,

As per the description

$$\text{P(boundary) or P(E) =}\\\frac{9}{45} =\frac{1}{5}\\\text{So P(no boundary) or P(not E)}\\= 1 -\frac{1}{5} =\frac{4}{5}$$

So, assertion is also true and Reason is the correct explanation of the assertion.

## Section-B

21. On comparing the given equations with standard form of pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get

$$\text{(a)}\space\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\\\text{as}\frac{3}{1}=\frac{3}{1}=\frac{\normalsize-9}{\normalsize-3},\space\text{consistent}\\\text{(b)}\space\frac{a_{1}}{a_{2}}\neq\frac{b_{1}}{b_{2}}\space\\\text{as}\space\frac{4}{\normalsize-2}\neq =-1,\text{consistent}\\\text{(c)}\space\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\text{as}\\\frac{5}{10}=\frac{1}{2}=\frac{10}{20},\space\text{consistent}\\\text{(d)\space}\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}\text{as}\\\frac{\normalsize-2}{\normalsize-4}=\frac{1}{2}\neq\frac{3}{10},\text{inconsistent}$$

(i) So, three pair of linear equations are consistent.

(ii) Fourth pair of linear equations is inconsistent.

22. Let P be the required point on Y-axis.

∴ P = (0, y)

Now, PA = PB  [Given]

$$\therefore\space\sqrt{(0+4)^{2} + (y-3)^{2}}\\=\sqrt{(0-6)^{2} + (y-5)^{2}}\\\text{[Using distance formula]}$$

On squaring both sides,

(4)2 + (y – 3)2 = (6)2 + (y – 5)2

⇒ 16 + y2 – 6y + 9 = 36 + y2 – 10y + 25

⇒ – 6y + 25 = 61 – 10y

⇒ – 6y + 10y = 61 – 25

⇒ 4y = 36

⇒ y = 9

∴ Coordinates of the point P = (0, 9) Ans.

OR

Given : A = (3, – 4), B = (1, 2),

$$\text{P}=(p,-2)\text{and Q}=\bigg(\frac{5}{3},q\bigg)$$

Now, AP = PQ = QB

Thus AP : PB = 1 : 2 and AQ : QB = 2 : 1

$$\therefore\space p =\frac{1 + 3(2)}{1 + 2} =\frac{7}{3}\\\text{and}\space q =\frac{2(2) + 1(\normalsize-4)}{1 + 2} = 0\\\therefore\space p=\frac{7}{3}\space\text{and q = 0}.$$

Ans.

23. Total number of cards = 52

∴ n(S) = 52

(i) Number of black king or red queen= 2 + 2 = 4

∴ n(A) = 4

$$\text{P(A)} =\frac{\text{n(A)}}{\text{n(S)}}\\=\frac{4}{52}=\frac{1}{13}.$$

Ans.

(ii) Number of non-face cards = Total cards – Number of face cards

= 52 – 12 = 40

∴ n(B) = 40

$$\text{P(B)}=\frac{\text{n(B)}}{\text{n(S)}}\\=\frac{40}{52}=\frac{10}{13}.\\\textbf{Ans.}$$

24. Given sequence is 114, 109, 104, ........

Here, first term (a) = 114

Common difference (d) = 109 – 114 = – 5

Let nth term be the first negative term of the given A.P.

∴ an < 0

⇒ a + (n – 1)d < 0

⇒ 114 + (n – 1) (– 5) < 0

⇒ 114 – 5n + 5 < 0

⇒ 119 – 5n < 0

⇒ 119 < 5n

$$\Rarr\space n\gt\frac{119}{5}\\\Rarr\space n\gt 23.8$$

∴ 24th term is the first negative term of the given A.P. Ans.

OR

Given : Sn = n (4n + 1)

So, Sn – 1 = (n – 1){4(n – 1) + 1}

= (n – 1) (4n – 4 + 1)

= (n – 1) (4n – 3)

Now, nth term of A.P. is given as,

Tn = Sn – Sn – 1

= n(4n + 1) – (n – 1) (4n – 3)

= 4n2 + n – (4n2 – 3n – 4n + 3)

= 4n2 + n – 4n2 + 7n – 3

= 8n – 3

∴ T6 = 8 (6) – 3

= 48 – 3 = 45 Ans.

25. Let the coordinates of point P be (2y, y). Since, P is equidistant from Q and R

∴ PQ = PR

$$\Rarr\space\sqrt{(2y-2)^{2} + (y + 5)^{2}}\\=\sqrt{(2y+3)^{2} + (y-6)^{2}}\\\Rarr\space (2y-2)^{2} +(y +5)^{2}\\=(2y+3)^{2} +(y-6)^{2}\\\Rarr\space 4y^{2} +4 =8y +y^{2} +25+10 y$$

4y2 + 9 + 12y + y2 + 36 – 12y

⇒ 2y + 29 = 45

⇒ 2y = 45 – 29

$$\Rarr\space y =\frac{16}{2} = 8$$

Hence, the coordinates of point P are (16, 8). Ans.

## Section-C

26. Let us suppose that

$$3 -\sqrt{5}\space\text{is a rational number.}\\\text{Then, it can be represented in}\\\space\frac{p}{q}\text{form}\space i.e.,\\ 3-\sqrt{5}=\frac{p}{q}$$

[where p and q are real numbers and q ≠ 0]

$$\Rarr\space 3 -\frac{p}{q} =\sqrt{5}\\\Rarr\space \frac{3q-p}{q} =\sqrt{5}$$

p and q are real numbers.

$$\text{Therefore,}\space\frac{3q-p}{q}$$

should also be a real number.

$$\sqrt{5}\space\text{is an irrational number.}$$

So, our assumption is wrong.

$$\text{Hence, 3}-\sqrt{5}\space\text{is an irrational number.}\\\textbf{Hence Proved.}$$

27. Let, HCF be ‘H’

then LCM = 14 H

Sum of LCM and HCF is 750.

∴ 14 H + H = 750

⇒ 15 H = 750

$$\Rarr\space \text{H} =\frac{750}{15}$$

⇒ H = 50

∴ LCM = 14 H

= 14 × 50

= 700

We know,

Product of LCM and HCF= Product of two numbers

Let other number be y. 700 × 50 = 250 × y

$$y =\frac{700×50}{250}$$

y = 140

Hence, the other number is 140. Ans.

28. Given : The quadratic polynomial is f(x) = x2 – p(x + 1) – c = x2 – px – p – c And α and β are the zeros of the given polynomial.

We know that,

Sum of the zeros =

$$-\frac{\text{Coefficient of x}}{\text{Coefficient of x}^{2}}\\\text{Product of the zeros =}\\\frac{\text{Constant term}}{\text{Coefficient of x}^{2}}\\\therefore\space\alpha +\beta =-\frac{(-p)}{1}\\\alpha + \beta = p\\\text{and}\space \alpha\beta=\frac{(-p-c)}{1}\\\alpha\beta =-(p+c)\\\text{Now\space L.H.S} =(\beta + 1)(\alpha + 1)\\=\alpha\beta + (\alpha +\beta) +1$$

= – (p + c) + p + 1

= – p – c + p + 1

= 1 – c

= R.H.S. Hence Proved.

29. Let, a be the first term and d be the common difference of the given A.P. Then,

$$a_{m}=\frac{1}{n}\\a +(m-1)d=\frac{1}{n}\\\text{...(i)}\\\text{and}\space a_{n}=\frac{1}{m}\\ a + (n-1)d=\frac{1}{m}\\\text{...(ii)}$$

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

$$(m-n)d=\frac{1}{n}-\frac{1}{m}\\\text{(m-n)d}=\frac{m-n}{mn}\\ d=\frac{1}{mn}\\\text{Putting\space d}=\frac{1}{mn}\space\\\text{in equation (i), we get}\\\text{a + (m - 1)}\frac{1}{mn}=\frac{1}{n}\\a + \frac{1}{n}-\frac{1}{mn}=\frac{1}{n}\\a =\frac{1}{mn}$$

$$\text{S}_{mn}=\frac{mn}{2}\bigg[2a +(mn-1)d\bigg]\\\text{S}_{mn}=\frac{mn}{2}\begin{Bmatrix}\frac{2}{mn}+(mn-1)\frac{1}{mn}\end{Bmatrix}\\\text{S}_{mn}=\frac{1}{2}(mn+1)\\\textbf{Hence Proved.}$$

30. In ∆ABC, we have

DE || BC

$$\text{So,}\space\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\\\text{[Thales’ theorem]}\\\Rarr\space\frac{4x-3}{3x-1} = \frac{8x-7}{5x-3}$$

⇒ (4x – 3)(5x – 3) = (8x – 7)(3x – 1)

⇒ 20x2 – 15x – 12x + 9 = 24x2 – 21x – 8x + 7

⇒ 4x2 – 2x – 2 = 0

⇒ 2x2 – x – 1 = 0

⇒ 2x2 – 2x + x – 1 = 0

⇒ 2x(x – 1) + 1(x – 1) = 0

⇒ (x – 1)(2x + 1) = 0

⇒ x = 1 [as length cannot be negative]

So, the value of x = 1. Ans.

OR

Let EO || DC meet AD at E.

So, in ∆ADC, EO || DC

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

By (i) and (ii),

$$\text{So,\space}\frac{\text{BO}}{\text{OD}} = \frac{\text{AE}}{\text{ED}}\\\text{Hence},\space \frac{\text{BO}}{\text{OD}}=\frac{\text{AE}}{\text{ED}}\\\text{and BO, OD, AE and ED}$$

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.

31. Let the line QR intersects the bigger circle at S.

Join PS.

O is the mid-point of PQ.

[PQ is a diameter of the bigger circle]

QR is a tangent to the smaller circle and OR is a radius through the point of  contact R.

∴ OR ⊥ QR ⇒ OR ⊥ QS

Since, OR is perpendicular to a chord QS of the bigger circle.

∴ QR = RS

[Perpendicular from the centre to a chord bisects the chord]

⇒ R is the mid-point of QS.

∴ In ΔQSP, O is the mid-point of PQ and R is the mid-point of QS.

$$\therefore\space \text{OR} =\frac{1}{2}\text{PS}$$

[segment joining the mid-points of any two sides of a triangle is half of the third side]

⇒ PS = 2OR = 2 × 8 cm = 16 cm

In right ΔOQR, OR2 + QR2= OQ2

⇒ 82 + QR2 = 132

⇒ 64 + QR2 = 169

⇒ QR2 = 169 – 64 = 105

$$\Rarr\space \text{QR} =\sqrt{105}\\\therefore\space\text{RS = QR=}\sqrt{105}\\\text{In}\space \Delta\text{PRS},\\\space \text{PR}^{2} = \text{RS}^{2} +\text{PS}^{2}\\=(\sqrt{105})^{2} + 16^{2}$$

⇒ PR = 19 cm. Ans.

Let P be the external point from which tangents PA and PB are drawn on the circle with centre O.

Thus, OB is perpendicular to PB and OA is perpendicular to PA.

Hence, ∠OBP = ∠OAP = 90°

In ΔPOB and ΔPOA, OA = OB

OP = OP [Common side]

∠OBP = ∠OAP [Each 90°]

∴ ΔPOB @ ΔPOA [S.A.S.]

So, PA = PB [By C.P.C.T.]

Hence Proved.

## Section-D

32. Total outcomes = {1, 2, 3, 4, 5, 6, 7, 8}

⇒ n (S) = 8

(i) Let A be the event that arrow will point at 8.

A = {8}

⇒ n (A) = 1

$$\because\space\text{P(A)}=\frac{1}{8}\space\textbf{Ans.}$$

(ii) Let B be the event that arrow will point at an odd number.

∴ B = {1, 3, 5, 7}

⇒ n(B) = 4

$$\therefore\space\text{P(B)}=\frac{4}{8}=\frac{1}{2}\space\textbf{Ans.}$$

(iii) Let C be the event that arrow will point at a prime number.

C = {2, 3, 5, 7}

⇒ n (C) = 4

$$\because\space\text{P(C)} =\frac{4}{8}=\frac{1}{2}\\\textbf{Ans.}$$

OR

Total number of coins = (100 + 50 + 20 + 10) = 180

(i) Possible number of outcomes of a 50 p coin falling out = 100

$$\text{Hence,\space}\text{P(E)}=\frac{100}{180}=\frac{5}{9}$$

Ans.

(ii) Possible number of outcomes of a coin falling out having value more than ₹1 = (20 + 10) = 30

$$\text{Hence,\space P(E)}=\frac{30}{180}=\frac{1}{6}\\\textbf{Ans.}$$

(iii) Possible number of outcomes of a coin falling out having value less than ₹5 = (100 + 50 + 20) = 170

$$\text{Hence,}\space\text{P(E)}=\frac{170}{180}=\frac{17}{18}\\\textbf{Ans.}$$

(iv) Possible number of outcomes of a coin falling out having value either ₹1 or ₹2 = (50 + 20) = 70

$$\text{Hence,}\space \text{P(E)}=\frac{\text{70}}{\text{180}}=\frac{7}{18}\\\textbf{Ans.}$$

33. Given, ΔABC ~ ΔPQR

$$\Rarr\space\frac{\text{AB}}{\text{PQ}} =\frac{\text{BC}}{\text{QR}}=\frac{\text{AC}}{\text{PR}}$$

(From the side-ratio property of similar triangles)

⇒ ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ...(i)

BC = 2BD and QR = 2QM

(P and M are the mid-points of BC and QR)

$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{2\text{BD}}{2\text{QM}}=\frac{\text{AC}}{\text{PR}}\\\text{...(ii)}$$

Now in ΔABD and ΔPQM

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}\\\text{...[\text{from} (ii)]}\\\angle B =\angle Q\\\text{...[from (i)]}\\\Rarr\space\Delta\text{ABD}∼\Delta\text{PQM}$$

(By SAS similarity criterion)

$$\therefore\space\frac{\text{AB}}{\text{PQ}} = \frac{\text{AD}}{\text{PM}}\\\textbf{Hence proved.}$$

34. Let ACB be the given arc subtending an angle of 60° at the centre.

Here, r = 14 cm and θ = 60°.

Area of the minor segment ACBA = (Area of the sector OACBO) – (Area of ∆OAB)

$$=\frac{\pi r^{2}\theta}{360\degree} -\frac{1}{2}r^{2}\text{sin}\theta\\\=\frac{22}{7}×14×14×\\\frac{60\degree}{360\degree}-\frac{1}{2}×14×14×\text{sin}\space 60\degree\\=\frac{308}{3}-7×14×\frac{\sqrt{3}}{2}\\=\frac{308}{3}-49\sqrt{3} = 17.89\text{cm}^{2}$$

Area of the major segment BDAB = Area of circle – Area of minor segment ACBA

= πr2 – 17.89

$$=\frac{22}{7}×14×14-17.89$$

= 616 – 17.89 = 598.11 ≈ 598 cm2. Ans.

35. Let XOX′ and YOY′ be the X-axis and Y-axis respectively.

Now, 3x – 2y + 2 = 0 ⇒ 3x = 2y – 2

$$\Rarr\space x =\frac{2(y-1)}{3}$$

If y = – 2, x = – 2

If y = 1, x = 0

If y = 4, x = 2

Therefore,

 x – 2 0 2 y – 2 1 4

Thus, plotting the points P(– 2, – 2), Q(0, 1) and R(2, 4) on the graph paper, we get the graph of 3x – 2y + 2 = 0, which is represented by PR.

$$\text{Now,\space}\frac{3}{2}x - y + 3 = 0\\\Rarr\space 3x-2y+6 =0\\\Rarr\space y =\frac{3(x+2)}{2}$$

If x = 0, y = 3

If x = – 2, y = 0

If x = – 4, y = – 3

Therefore,

 x 0 -2 – 4 y 3 0 – 3

Thus, plotting the points S(0, 3), T(– 2, 0) and U(– 4, – 3) on the graph paper, we get the graph of

$$\frac{3}{2}x - y + 3 = 0,$$

which is represented by SU.

Since, the lines do not intersect each other at any point, so the pair of equations is inconsistent. Also the lines representing the graphs of the equations, meet the Y-axis at points (0, 1) and (0, 3). Ans.

OR

Let XOX′ and YOY′ be the X-axis and Y-axis respectively.

Now, x – y + 1 = 0 ⇒ x = y – 1

If y = 1, x = 0

If y = 0, x = – 1

If y = 3, x = 2

Therefore,

 x 0 -1 2 y 1 0 3

Thus, plotting the points P(0, 1), Q(– 1, 0) and R(2, 3) on the graph paper, we get the graph of x – y + 1 = 0, which is represented by PR.

Now, 3x + 2y – 12 = 0 ⇒ 2y = 12 – 3x

$$\Rarr\space y =\frac{3(4-x)}{2}$$

If x = 4, y = 0

If x = 2, y = 3

If x = 0, y = 6

Therefore,

 x 4 2 0 y 0 3 6

Thus, plotting the points S(4, 0), R(2, 3) and T(0, 6) on the graph paper, we get the graph of 3x + 2y – 12 = 0, which is represented by ST.

The solution of the pair of equations is given by (2, 3).

Hence, the coordinates of the triangle formed by these two lines and the X-axis are Q(– 1, 0), R(2, 3) and S(4, 0).

$$\text{Area} =\frac{1}{2}×\text{Base × Height}$$

Now, Base = 4 – (– 1)

= 5

Height = 3

$$\text{Thus\space Area}=\frac{1}{2}×5×3$$

= 7.5 sq. units. Ans.

## Section-E

36. (i) Radius of a dome, r = 2.5 m

The dome is hemispherical in shape.

Then, Cloth material required = 2 × Surface area of hemisphere

= 2 × 2πr2

$$= 4×\frac{22}{7}×2.5×2.5$$

= 78.57 m2 Ans.

(ii) Height of each pillar, h = 7 m

Radius of base, r = 1.4 m

Lateral surface area or curved surface area of 2 pillars = 2 × 2πrh

$$= 4×\frac{22}{7}×1.4×7$$

= 123.2 m2 Ans.

OR

Volume of 2 hemispheres of radius 1 cm

$$= 2×\frac{2}{3}\pi r^{3}=\frac{4}{3}\pi(1)^{3}\\=\frac{4}{3}\pi\text{cm}^{3}$$

Volume of a sphere of radius 2 cm =

$$\frac{4}{3}\pi(2)^{3}=\frac{32}{3}\pi\space\text{cm}^{3}\\\text{Then,\space Required ratio =}\frac{\frac{4}{3}\pi}{\frac{32}{3}\pi}\\=\frac{1}{8}\space = 1:8\space\textbf{Ans.}$$

(iii) Radius of hemisphere, r = 3.5 m

Then, volume of a hemisphere,

$$\text{V}=\frac{2}{3}\pi r^{3}\\=\frac{2}{3}×\frac{22}{7}×(3.5)^{3}$$

= 89.83 m3 Ans.

37. (i) Let be distance between building and tower, CB = DE = x

AE = h

$$\text{In} \Delta \text{ADE,}\\\space\text{tan\space} 30\degree = \frac{\text{AE}}{\text{DE}}=\frac{\text{h}}{\text{x}}\\=\frac{1}{\sqrt{3}}=\frac{h}{x}\\x =\sqrt{3}h\\\text{In} ∆\text{ACB,}\\\text{tan\space} 60\degree=\frac{\text{AB}}{\text{CB}}\\=\frac{15+h}{x}\\\sqrt{3} =\frac{15 + h}{\sqrt{3}h}$$

[Using (i)]

⇒ 3h = 15 + h

⇒ 2h = 15

$$h =\frac{15}{2} = 7.5\space\text{m}$$

Height of tower = AB = BE + EA

= 15 + 7.5 = 22.5 m Ans.

OR

We have,

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

$$\text{tan}\space 30\degree =\frac{\text{AE}}{\text{DE}}=\frac{h}{x}\\\frac{1}{\sqrt{3}}=\frac{7.5}{x}$$

x = 7.5 × 1.73

= 12.975 m

Ans.

$$\text{sin}\space 30\degree =\frac{\text{AE}}{\text{AD}}\\(\because\space\text{Length of wire = AD})\\\frac{1}{2}=\frac{7.5}{\text{AD}}$$

D = 7.5 × 2 = 15 m Ans.

38. (i) Let AB denote the lamp-post and CD be the girl after walking for 4 seconds away from the lamp-post.

Now, her distance from the base of the lamp

BD = 1.2 m × 4 = 4.8 m Ans.

(ii) ∆ABE ~ ∆CDE

$$\frac{\text{BE}}{\text{DE}}=\frac{\text{AB}}{\text{CD}}\\\text{(Let DE = x)}\\\Rarr\space\frac{4.8 +x}{x}=\frac{3.6}{0.9}$$

⇒ 4.8 + x = 4x

⇒ x = 1.6 m Ans.

So, the shadow of the girl after walking for 4 seconds.

OR

$$\frac{\text{AE}}{\text{CE}}=\frac{\text{BE}}{\text{DE}}\\=\frac{4.8 +1.6}{1.6}=\frac{6.4}{1.6} = 4$$

⇒ AE = 4CE

⇒ AC + CE = 4CE

⇒ AC = 3CE

$$\Rarr\space\frac{\text{AC}}{\text{CE}}=\frac{3}{1}\space\textbf{Ans.}$$

(iii) Since ratio of the area of two similar triangles is equal to the ratio of the square of their corresponding sides, Ratio of areas of similar triangles = (9)2 : (16)2 = 81 : 256.

#### CBSE Practice Paper Mathematics Class 10

All Practice Paper for Class 10 Exam 2024