Oswal Model Specimen Papers ICSE Class 10 Mathematics Solutions (Specimen Paper - 9)

Section-A

1. (i) (c) ₹ 1300

Explanation :    

Monthly deposit (P) = ₹ 2,500

Time Period (n) = 1 year = 12 months

Rate (r) = 8% p.a.

since, we know

$$\text{I} =\text{P}×\frac{n(n+1)}{2×12} ×\frac{r}{100}\\= 2500×\frac{12(13)}{2×12}×\frac{8}{100}$$

= ₹ 1,300

$$(\text{ii})\space (c)\space\frac{2}{3},4$$

Explanation :    

3x2 – 14x + 8 = 0

⇒ 3x2 – 12x – 2x + 8 = 0

3x(x – 4) – 2(x – 4) = 0

⇒ (3x – 2)(x – 4) = 0

⇒ 3x – 2 = 0, or x – 4 = 0

$$\Rarr\space x =\frac{2}{3}\space\text{or x = 4}$$

(iii) (d) 16

Explanation :    

Let f(x) = 3x3 – x2 – px – 4 ...(i)

Since, (x + 2) is a factor of f(x), f(–2) = 0

⇒ 3(–2)3 – (–2)2 – p(–2) – 4 = 0

⇒ –24 – 4 + 2p – 4 = 0

⇒ 2p = 32

⇒ p = 16

(iv) (a) 6

Explanation :    

By comparing elements of equal matrices. we get,

x = 6

(v) (c) – 77

Explanation :    

Given,

A.P. = 10, 7, 4, ....

First term, a = 10

Common difference, d = a2 – a1 = 7 – 10 = –3

As we know, for an A.P.,

an = a + (n – 1)d

Putting the values;

a30 = 10 + (30 – 1)(–3)

⇒ a30 = 10 + (29)(–3)

⇒ a30 = 10 – 87 = –77

(vi) (b) Quadratic equation

(vii) (a) ΔPQR ~ ΔCAB

Explanation :    

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

∴ ΔPQR ~ ΔCAB

[By SSS similarity axiom]

(viii) (a) 19,404 cm3

Explanation :    

$$\text{V}=\frac{1}{3}\pi r^{2}h\\-\frac{1}{3}×\frac{22}{7}×21×21×42 \\= 19,404\space\text{cm}^{3}.$$

(ix) (d) 0

Explanation :    

The possible outcomes are {(1, 1), (1, 2), (1, 3), … (6, 6)}

Total No. of possible outcomes = 36

The maximum number as sum is 6 + 6 = 12

Hence, there is no favourable outcome.

P(getting sum more than 12) =

$$\frac{0}{36} = 0.$$

(x) (a) ax2 + bx + c < 0

Explanation :    

In ax2 + bx + c < 0, the highest power of variable x is 2. So, it is quadratic inequality.

(xi) (c) 1 : 4

Explanation :    

Let the ratio be k : 1.

Hence by section formula,

$$\frac{k×4 + 1×\normalsize -1}{\text{k+1}}= 0$$

(as x-coordinate on the y-axis is zero)

i.e., 4k – 1 = 0

$$\Rarr\space k =\frac{1}{4}$$

∴ Ratio is 1 : 4.

(xii) (a) 100°

9ds_science_m1(xii)

OA = OB (Radii of the same circle)

Therefore, ΔOAB is an isosceles triangle.

Hence, ∠OBA = ∠OAB = 30°

Similarly, ΔOAC is an isosceles triangle.

Hence, ∠OCA = ∠OAC = 20°

Hence, ∠BAC = ∠BAO +∠CAO

⇒ ∠BAC = 30° + 20° = 50°

Hence, ∠BOC = 2 ∠BAC = 2 × 50° = 100°

(Angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining part of the circle.)

(xiii) (b) 7

Explanation :    

Let x1 and x2 be the lower and upper limit of the class.

Hence, x2 – x1 = 6

$$\text{and}\qquad \frac{x_{1} +x_{2}}{2}= 10$$

8ds_math_m4(1xiii)

$$\text{(xiv)\space(d)\space}\frac{1}{13}$$

Explanation :    

Total no. of possible outcomes n(S) = 52

Number of favourable outcomes n(E) = 4

∴ Probability of drawing a Jack =

$$\frac{n(E)}{n(S)}=\frac{4}{52} =\frac{1}{13}$$

(xv) (b) A is false, R is true

Explanation :    

(4a + 5b) (4c – 5d) = (4a – 5b) (4c + 5d)

$$\Rarr\space\frac{4a + 5b}{4a - 5b}=\frac{4c + 5d}{4c - 5d}$$

Applying componendo and dividendo, we get

$$\frac{8a}{10b} =\frac{8c}{10 d}\\\Rarr\space \frac{a}{b} = \frac{c}{d}$$

∴ a, b, c and d are in proportion.

2. (i) (a) GST paid by the manufacturer

= 8% of 15,000

$$=\frac{8}{100}×15,000 = ₹ 1,200$$

GST paid by the wholesaler

= 8% of 1,200

$$\frac{8}{100}×1,200 = ₹96$$

Total GST received by the state government on the sale of machine from the manufacturer and the wholesaler

= ₹ (1,200 + 96)

= ₹ 1,296 Ans.

(b) Amount paid by the consumer

= ₹ (15,000 + 1,200 + 1,800) + GST @ 8%

$$ =₹\bigg(18,000 + \frac{8}{100}×18,000\bigg)$$

= ₹(18,000 + 1,440)

= ₹ 19,440

Ans.

(ii) x, 3, 12, y are in continued proportion.

$$\therefore\space\frac{x}{3} =\frac{3}{12} =\frac{12}{y}\\\Rarr\space\frac{x}{3} =\frac{3}{12}\text{and}\frac{3}{12}= \frac{12}{y}\\\Rarr\space x=\frac{3×12}{12}\\\text{and y}=\frac{12×2}{3}\\\Rarr\space x=\frac{3}{4}\space\text{and y = 48.}\space\textbf{Ans.}$$

$$\text{(iii)\space}\frac{1}{\text{cosec A - cot A}}-\frac{1}{\text{sin A}}\\=\frac{1}{\text{sin A}}-\frac{1}{\text{cosec A + cot A}}\\\text{L.H.S} =\\\frac{1}{\text{cosec A - cot A}}-\frac{1}{\text{sin A}}\\=\frac{1}{\text{cosec A - cot A}}×\\\frac{(\text{cosec A + cot A})}{(\text{cosec A + cot A})}-\frac{1}{\text{sin A}}\\=\frac{\text{cosec A + cot A}}{\text{cosec}^{2}\text{A} -\text{cot}^{2}\text{A}}-\\\text{cosec A}$$

= cosec A + cot A – cosec A

[cosec2 A – cot2 A = 1]

= cot A.

$$\text{R.H.S.} =\\\frac{1}{\text{sin A}} - \frac{1}{\text{cosec A + cot A}}\\=\frac{1}{\text{sin A}}-\frac{1}{\text{cosec A + cot A}}×\\\frac{(\text{cosec A - cot A})}{(\text{cosec A - cot A})}\\=\text{cosec A} -\frac{\text{cosec A - cot A}}{\text{cosec}^{2}\text{A} -\text{cot}^{2}A}$$

= cosec A – (cosec A – cot A)

[ cosec2 A – cot2 A = 1]

= cosec A – cosec A + cot A

= cot A

∴ L.H.S. = R.H.S.

Hence Proved.

3. (i) Let R = External radius = 9 cm

h = Length of pipe = 14 cm

r = Internal radius

V = Volume of pipe = 1408 cm3.

Now, V = πh (R2 – r2)

$$\Rarr\space 1408 =\frac{22}{7}×14(9^{2} - r^{2})$$

⇒ 1408 = 44 (81 – r2)

⇒ 32 = 81 – r2

⇒ r2 = 81 – 32

⇒ r2 = 49

⇒ r = 7 cm

[∵ radius can't be negative]

Therefore, thickness of pipe is R – r = 9 – 7 = 2 cm. Ans.

(ii) Let the required line be AB cut off positive intercepts OA and OB on X-axis and Y-axis respectively.

OA : OB = 2 : 3, let OA = 2a and OB = 3a.

∴ A = (2a, 0) and B = (0, 3a)

Equation of a line is given as

$$y - y_{1}=\frac{y_{2}-y_{1}}{x_{2} - x_{1}}(x - x_{1})\\\Rarr\space y-0=\frac{3a -0}{0-2a}(x-2a)\\\Rarr\space y =-\frac{3a}{2a}(x- 2a)$$

⇒ 2y = – 3(x – 2a)

⇒ 2y = – 3x + 6a

⇒ 3x + 2y – 6a = 0 ...(i)

It passes through P(1, 2),

On putting, x = 1 and y = 2, we get

3 × 1 + 2 × 2 – 6a = 0

⇒ 7 – 6a = 0

⇒ 6a = 7

$$\Rarr\space a=\frac{7}{6}$$

∴ From equation (i),

$$3x + 2y - 6×\frac{7}{6} = 0$$

⇒ 3x + 2y – 7 = 0,

which is the required equation of line

$$\Rarr\space \text{Slope of AB =}\\-\frac{\text{Coefficient of x}}{\text{Coefficient of y}}=-\frac{3}{2}\space\textbf{Ans.}$$

(iii) (a) P(4, 6), Q(1, 2)

(b) Coordinates of P′ = (4, – 6)

(c) Coordinates of Q′ = (7, 2)

(d) PQP′Q′ is a kite

Scale : At X axis : 1 cm = 1 unit

At Y axis : 1 cm = 1 unit

Section-B

4. (i) Given,

M.P. = ₹18,000

Discount = 20% for shop-owner.

Discount = 10% for consumer

GST = 8%

Discount for shop-owner = 20% of 18,000

$$= \frac{20 × 18000}{100}$$

= ₹3600

Cost for shop–owner

= ₹18000 – ₹3600

= ₹14400

Cost for consumer = ₹ 18000 – 10% of ₹ 18000

$$= ₹ 18000 - ₹\frac{10×18000}{100}$$

= ₹ 18000 – ₹ 1800

= ₹ 16,200

(a) ∴ GST paid by the shopkeeper

= Tax on the value added by shopkeeper

= 8% of (16,200 – 14,400)

$$=\frac{8}{100}× 1800$$

= ₹ 144 Ans.

(b) Tax charged by the shopkeeper

$$= ₹ 16200×\frac{8}{100} = ₹1296$$

Amount paid by consumer

= ₹ (16200 + 1296)

= ₹ 17496. Ans.

(ii) Given equation is

x2 + (p – 3) x + p = 0

Here, a = 1, b = p – 3 and c = p

For real and equal roots,

⇒ D = b2 – 4ac = 0

⇒ (p – 3)2 – 4 × 1 × p = 0

⇒ p2 – 6p + 9 – 4p = 0

⇒ p2 – 10p + 9 = 0

⇒ p2 – p – 9p + 9 = 0

⇒ p (p – 1) – 9 (p – 1) = 0

⇒ (p – 1) (p – 9) = 0

⇒ p – 1 = 0 or p – 9 = 0

⇒ p = 1 or p = 9 Ans.

(iii)

Height in cm No. of Boys c.f.
135 – 140 4 4
140 – 145 8 12
145 – 150 20 32
150 – 155 14 46
155 – 160 7 53
160 – 165 6 59
165 – 170 1 60
n = 60

$$\text{(a)\space \text{Median}}=\frac{n}{2}\text{th observation}\\=\frac{60}{2}\text{th observation}$$

= 30th observation

= 150 cm (from ogive) Ans.

(b) Lower quartile =

$$\frac{n}{4}\text{th observation}\\=\frac{60}{4}\text{th observation}$$

= 15th observation

= 146 cm (from ogive) Ans.

(c) No. of boys whose height is less than 158 cm = 51. (from ogive)

∴ No. of tall boys = 60 – 51 = 9. Ans.

$$\textbf{5.\space}(i)\space\text{A}=\begin{bmatrix}1 &2\\2 &1\end{bmatrix},\text{B =}\begin{bmatrix}2 &1\\1 &2\end{bmatrix}\\\therefore\space\text{BA} =\begin{bmatrix}2 &1\\1 &2\end{bmatrix}\begin{bmatrix}1 &2\\2 &1\end{bmatrix}\\=\begin{bmatrix}2 + 2 &4+1\\1 +4 &2+2\end{bmatrix} =\begin{bmatrix}4 &5\\5 &4\end{bmatrix}\\\therefore\space\text{A(BA)}=\begin{bmatrix}1 &2\\2 &1\end{bmatrix}\begin{bmatrix}4 &5\\5 &4\end{bmatrix}\\=\begin{bmatrix}4 +10 &5 +8\\ 8+5 &10 +4\end{bmatrix}\\=\begin{bmatrix}14 &13\\13 &14\end{bmatrix}\space\textbf{Ans.}$$

(ii) (a) In ∆ PAM,

∠APM = ∠AMP … (i)

PA = AM (Given)

by alternate segment property of tangent

∠ABM = ∠AMP

∴ ∠APM = ∠ABM [from (i) and (ii)]

∴ PM = MB

i.e. ∆ PMB is an isosceles. Hence Proved.

(b) By rectangle property of tangent and chord

PM2 = PA × PB

∴ MB2 = PA × PB [ PM = MB]

Hence Proved.

(iii) Let f (x) = x3 + 5x2 – ax + 6.

When f (x) is divided by (x – 1),

Remainder = 2a

⇒ f (1) = 2a ⇒ 13 + 5.12 – a.1 + 6 = 2a

⇒ 1 + 5 – a + 6 = 2a ⇒ 3a = 12 ⇒ a = 4. Ans.

6. (i) m = tan θ = tan 45°

m = 1

(ii) Equation of line PQ

y – y1 = m (x – x1)

⇒ y – 3 = 1 (x – 5)

⇒ y – 3 = x – 5

⇒ x – y – 2 = 0

(iii) Equation of PQ is

x – y – 2 = 0

Put x = 0 (coordinates of Q)

– y – 2 = 0

⇒ y = – 2

So, co-ordinates of Q (0, – 2).

(ii) Consider,

$$\text{L.H.S} =\\\frac{\text{cos}^{2}\theta}{\text{1 - tan}\theta} + \frac{\text{sin}^{2}\theta}{\text{sin}\space\theta - \text{cos}\space\theta}\\=\frac{\text{cos}^{2}\theta}{1- \frac{\text{sin}\theta}{\text{cos}\theta}} +\frac{\text{sin}^{3}\theta}{\text{sin}\space\theta - \text{cos}\space\theta}\\=\frac{\text{cos}^{3}\theta}{\text{cos}\theta - \text{sin}\theta} + \frac{\text{sin}^{3}\theta}{\text{sin}\theta - cos\theta}\\=\frac{\text{cos}^{3}\theta}{\text{cos}\theta - \text{sin}\theta}-\frac{\text{sin}^{3}\theta}{\text{cos}\theta - \text{sin}\theta}\\=\frac{\text{cos}^{3}\theta -\text{sin}^{3}\theta}{\text{cos}\theta - \text{sin}\theta}$$

$$=\\\frac{(\text{cos}\space \theta -\text{sin}\space\theta)(\text{cos}^{2}\theta +\text{sin}\theta\text{cos}\theta +\text{cos}^{2}\theta)}{\text{cos}\space\theta - \text{sin}\space\theta}$$

[ a3 – b3 = (a – b)(a2 + b2 + ab)]

= 1 + sin θ cos θ

[∵ sin2 θ + cos2 θ = 1]

= R.H.S. Hence Proved.

$$\text{(iii) Given A.P. is 20,}19\frac{1}{3},18\frac{2}{3},.......$$

Here, a = 20 and

$$d = 19\frac{1}{3} - 20\\=\frac{58 -60}{3}=-\frac{2}{3}$$

Let the number of terms taken be n.

$$\therefore\space\text{S}_{n}=\frac{n}{2}[2a + (n-1)d]\\\Rarr\space 300 =\\\frac{n}{2}\bigg[2×20 +(n-1)×-\frac{2}{3}\bigg]\\\Rarr\space 600 =n\begin{bmatrix}40-\frac{2}{3}n + \frac{2}{3}\end{bmatrix}$$

⇒ 1800 = 120n – 2n2 + 2n

⇒ 2n2 – 122n + 1800 = 0

⇒ n2 – 61n + 900 = 0

⇒ n2 – 25n – 36n+ 900 = 0

⇒ n(n – 25) – 36 (n – 25) = 0

⇒ (n – 25) (n – 36) = 0

⇒ n = 25 or n = 36

∴ Required number of terms = 25 or 36. Ans.

7. (i) Let present age of man be x and that of his son be y.

∴ x – y = 25 ⇒ y = x – 25 ...(i)

10 years back, man’s age = x – 10

Son’s age = y – 10 = x – 25 – 10 = x – 35

[Using (i)]

According to question

(x – 10)(x – 35) = 350

⇒ x2 – 35x – 10x + 350 – 350 = 0

⇒ x2 – 45x = 0

⇒ x(x – 45) = 0

⇒ x = 0 or x – 45 = 0

⇒ x = 0 or x = 45

∴ x = 45 [x cannot be 0]

∴ Man’s age = x = 45 years

Son’s age = x – 25 = 45 – 25 = 20 years. Ans.

(ii)

x f Cumulative Frequency
1 3 3
2 5 8
3 9 17
4 15 32
5 20 52
6 16 68
7 10 78
8 2 80
n = 20

(a) Lower quartile (Q1) =

$$\text{The value of}\bigg(\frac{n}{4}\bigg)^{\text{th}}\space\text{observation}\\=\text{The value of}\bigg(\frac{80}{4}\bigg)^{\text{th}}\space\text{observation}$$

= The value of 20th observation

Q1 = 4. Ans.

(b) Upper quartile (Q3) =

$$\text{The Value of}\bigg(\frac{3n}{4}\bigg)^{\text{th}}\space\text{observation}\\=\\\text{The value of}\\\bigg(\frac{3×80}{4}\bigg)^{\text{th}}\space\text{observation}$$

∴ Q3 = 6. Ans.

(c) Inter quartile range = Q3 – Q1

= 6 – 4 = 2. Ans.

(d) Semi-quartile range =

$$\frac{Q_{3} - Q_{1}}{2}\\=\frac{2}{2}=1\space\textbf{Ans.}$$

8. (i) We have,

$$-\frac{2}{3}\lt-\frac{x}{3} +2\leq\frac{2}{3}x\epsilon\text{R}\\\Rarr\space -\frac{2}{3}×3\lt-\frac{x}{3}×3 + 2×3\leq\frac{2}{3}×3$$

$$\Rarr\space-2\lt-x +6\leq 2\\\Rarr\space -2-6\lt-x +6-6\leq 2-6\\\Rarr\space-8\lt-x\leq -4\\\Rarr\space8\gt x\geq4\\\Rarr\space 4\leq x\lt 8$$

∴ Solution set = {x : 4 ≤ x < 8, x ∈ R}.

9ds_math_m8(i)

(ii)

Class interval Frequency (fi) Mid value (xi) $$u_{i}=\frac{x_{i}- A}{h};$$h = [upper limit – lower limit] fi×ui
50 – 70 18 60 –2 –36
70 – 90 12 80 –1 –12
90 – 110 13 100 = A 0 0
110 – 130 27 120 1 27
130 – 150 8 140 2 16
150 – 170 22 160 3 66
Σfi = 100 Σ(fi × ui) = 61

Thus, A = 100, h = 20, Σfi = 100, Σ(fi × ui ) = 61

$$\text{Mean} =\\\text{A +}\begin{Bmatrix}h×\frac{\Sigma(f_{i}×u_{i})}{\Sigma f_{i}}\end{Bmatrix}\\=100 +\begin{Bmatrix}20×\frac{61}{100}\end{Bmatrix}$$

= 100 + 12.2 = 112.2 Ans.

(iii) Given, DE || BC

∴ ∠ADE = ∠ABC

and ∠AED = ∠ACB (Corresponding angles)

∴  ∆ADE ~ ∆ABC (By AA similarity)

$$\therefore\space\frac{\text{AD}}{\text{AB}}=\frac{\text{DE}}{\text{BC}} =\frac{\text{AE}}{\text{AC}}\\\text{...(i)}$$

9ds_math_m8(iii)

$$\text{(a)\space Given,\space}\frac{\text{AD}}{\text{DB}} =\frac{3}{2}$$

⇒ 2AD = 3DB

⇒ 2AD = 3(AB – AD)

⇒ 2AD = 3AB – 3AD

⇒ 5AD = 3AB

$$\therefore\space \frac{\text{AD}}{\text{AB}}=\frac{3}{5}\\\textbf{Ans.}$$

From equation (i),

$$\frac{\text{AD}}{\text{AB}} =\frac{\text{DE}}{\text{BC}}\\\Rarr\space\frac{\text{DE}}{\text{BC}}=\frac{3}{5}$$

Ans.

(b) DE || BC

∴ ∠EDF = ∠FCB

and ∠DEF = ∠FBC (Alternate angles)

∴  ∆DEF ~ ∆CBF (By AA similarity)

$$\therefore\space \frac{\text{DE}}{\text{CB}}=\frac{\text{EF}}{\text{BF}}=\frac{\text{DF}}{\text{CF}}\\\Rarr\space\frac{\text{DE}}{\text{CB}} =\frac{\text{EF}}{\text{BF}}\\\Rarr\space \frac{\text{EF}}{\text{BF}}=\frac{3}{5}\space\textbf{Ans.}$$

9. (i) Here, Sample Space,

S = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

∴ n(S) = 10

(a) Let A be the event of getting a prime number.

A = {2}

∴ n(A) = 1

$$\therefore\space\text{P(A)} =\frac{n(A)}{n(S)}=\frac{1}{10}$$

Ans.

(b) Let B be the event of getting a number divisible by 4.

∴ B = {4, 8, 12, 16, 20}

∴ n(B) = 5

$$\therefore\space\text{P(B)}=\frac{\text{n(B)}}{\text{n(S)}}\\=\frac{5}{10}=\frac{1}{2}$$

Ans.

(c) Let C be the event of getting a number which is multiple of 6.

∴ C = {6, 12, 18}

∴ n(C) = 3

$$\therefore\space\text{P(C)}=\frac{\text{n(C)}}{\text{n(S)}}=\frac{3}{10}\\\textbf{Ans.}$$

(d) Let D be the event of getting an odd number.

∴ D = { }

∴ n(D) = 0

$$\therefore\space\text{P(D)}=\frac{n(D)}{n(S)}-\frac{0}{10}=0$$

Ans.

(ii) For cylinder, r1 = 18 cm, h1 = 32 cm

$$\therefore\space \text{Volume} =\pi r^{2}_{1}\\=\frac{22}{7}×(18)^{2}×32\space\text{cm}^{3}$$

For cone, h2 = 24 cm, r2 = ?

(a)  Volume of cone = Volume of cylinder

$$\frac{1}{3}\pi r^{2}_{2}h_{2}\\=\frac{22}{7}×18×18×32\\\Rarr\space \frac{1}{3}×\frac{22}{7}×r^{2}_{2}×24\\=\frac{22}{7}×18×18×32\\\Rarr\space r^{2}_{2}=\frac{\frac{22}{7}×18 ×18×32}{\frac{1}{3}×\frac{22}{7}×24}\\= 1296\\\Rarr\space r_{2} =\sqrt{1296} = 36\space\text{cm.}\\\textbf{Ans.}$$

$$\text{(b)\space Slant height =}\sqrt{r^{2}_{2} + h^{2}_{2}}\\=\sqrt{36^{2} + 24^{2}}\\=\sqrt{1296 + 576}\\=\sqrt{1872}\\= 12\sqrt{13}\space\text{cm.}$$

Ans.

(iii) Given : ∠CAB = 40°, ∠CBD = 70°

Now, ∠CAD = ∠CBD

[Angles in same segment are equal]

= 70°

∴ ∠BAD = ∠BAC + ∠CAD

⇒ ∠BAD = 40° + 70° = 110° ...(i)

9ds_math_m9(iii)

Also, ∠BAD + ∠BCD = 180°

[Opposite angles of cyclic quadrilateral are supplementary]

⇒ 110° + ∠BCD = 180° [Using (i)]

⇒ ∠BCD = 180° – 110° = 70°. Ans.

10. (i) Given, a : b :: c : d

$$\Rarr\space\frac{a}{b} =\frac{c}{d}\\\Rarr\space\frac{2a}{3b} =\frac{2c}{3d}\\\Rarr\space\frac{2a +3b}{2a - 3b} =\frac{2c + 3d}{2c - 3d}$$

[Using componendo and dividendo]

⇒ (2a + 3b) (2c – 3d) = (2a – 3b) (2c + 3d).

Hence Proved.

(ii) Interior angle = 120°

(a) Draw a line segment AB = 4 cm.

(b) At A and B, draw angles equal to 120° each.

(c) From A and B, cut off AF and BC respectively, each equal to 4 cm.

(d) Draw 120° at F and C and cut off FE and CD respectively, each equal to 4 cm.

(e) Join E and D to complete hexagon ABCDEF.

(f) Bisect the sides AB and BC such that the two bisectors meet at O.

(g) Taking O as centre, draw a circumcircle passing through all the vertices.

(iii) (a) In right angled ΔCDE, we have

$$\frac{\text{DE}}{\text{EC}} = \text{cot x\degree}\\\Rarr\space \frac{\text{AB}}{\text{BC - AE}}=\frac{5}{2}\\\Rarr\space \frac{\text{AB}}{\text{10-2}}=\frac{5}{2}\\\Rarr\space \text{AB} =\bigg(\frac{5}{2}×8\bigg)m = 20\space m.\\\textbf{Ans.}$$

(b) When AB = 15 m, then DE = 15 m.

In right angled ΔCDE, we have

$$\text{tan}\space \angle\text{EDC}=\frac{\text{EC}}{\text{DE}}\\=\frac{8}{15} = 0.5333$$

From tables of natural tangents, we have

∠EDC = 28°2´ nearest

≈ 28° (nearest degree) Ans.

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