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°
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
$$\text{and}\qquad \frac{x_{1} +x_{2}}{2}= 10$$
$$\text{(xiv)\space(d)\space}\frac{1}{13}$$
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}.
(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)
$$\therefore\space\frac{\text{AD}}{\text{AB}}=\frac{\text{DE}}{\text{BC}} =\frac{\text{AE}}{\text{AC}}\\\text{...(i)}$$
$$\text{(a)\space Given,\space}\frac{\text{AD}}{\text{DB}} =\frac{3}{2}$$
⇒ 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)
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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