Vardaan Learning Institute
Mock Board Paper Solutions 2025-26
Set - 3 (Solutions)
MATHEMATICS (STANDARD)
Class - X
Time: 3 Hours
Max. Marks: 80
SECTION A (Solutions)
1.
(d) \(ab^2\)
HCF is the product of smallest powers of common prime factors. \(p=a^3b^2, q=ab^3\). Min power of a is 1, b is 2. HCF = \(ab^2\).
HCF is the product of smallest powers of common prime factors. \(p=a^3b^2, q=ab^3\). Min power of a is 1, b is 2. HCF = \(ab^2\).
2.
(d) More than 3
There are infinite polynomials with given zeroes as \(k(x^2 - (\alpha+\beta)x + \alpha\beta)\) where \(k\) is any non-zero real number.
There are infinite polynomials with given zeroes as \(k(x^2 - (\alpha+\beta)x + \alpha\beta)\) where \(k\) is any non-zero real number.
3.
(b) 2
For no solution: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\). \(\frac{1}{2} = \frac{1}{k} \Rightarrow k = 2\). Check \(\frac{1}{2} \neq \frac{-4}{-3} (4/3)\). True.
For no solution: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\). \(\frac{1}{2} = \frac{1}{k} \Rightarrow k = 2\). Check \(\frac{1}{2} \neq \frac{-4}{-3} (4/3)\). True.
4.
(c) no real roots
\(D = b^2 - 4ac = (-\sqrt{5})^2 - 4(2)(1) = 5 - 8 = -3 < 0\).
\(D = b^2 - 4ac = (-\sqrt{5})^2 - 4(2)(1) = 5 - 8 = -3 < 0\).
5.
(c) 25
\(a_{18} - a_{13} = (a+17d) - (a+12d) = 5d = 5(5) = 25\).
\(a_{18} - a_{13} = (a+17d) - (a+12d) = 5d = 5(5) = 25\).
6.
(c) 3 cm
By BPT: \(\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{3}{4.5} = \frac{2}{CE} \Rightarrow CE = \frac{2 \times 4.5}{3} = 3\).
By BPT: \(\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{3}{4.5} = \frac{2}{CE} \Rightarrow CE = \frac{2 \times 4.5}{3} = 3\).
7.
(c) (0, 0)
Midpoint \(= (\frac{2+(-2)}{2}, \frac{4+(-4)}{2}) = (0, 0)\).
Midpoint \(= (\frac{2+(-2)}{2}, \frac{4+(-4)}{2}) = (0, 0)\).
8.
(a) \(xy\)
\(\tan \theta = \frac{\sin \theta}{\cos \theta} = \sin \theta \cdot \sec \theta = x \cdot y\).
\(\tan \theta = \frac{\sin \theta}{\cos \theta} = \sin \theta \cdot \sec \theta = x \cdot y\).
9.
(b) 1
\((1+\tan^2\theta)(1-\sin^2\theta) = \sec^2\theta \cdot \cos^2\theta = (\frac{1}{\cos^2\theta})\cos^2\theta = 1\).
\((1+\tan^2\theta)(1-\sin^2\theta) = \sec^2\theta \cdot \cos^2\theta = (\frac{1}{\cos^2\theta})\cos^2\theta = 1\).
10.
(a) 7 cm
\(r = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7\).
\(r = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7\).
11.
(c) 44 cm
\(\pi r^2 = 154 \Rightarrow \frac{22}{7} r^2 = 154 \Rightarrow r^2 = 49 \Rightarrow r = 7\). Perimeter \(2\pi r = 2(22/7)(7) = 44\).
\(\pi r^2 = 154 \Rightarrow \frac{22}{7} r^2 = 154 \Rightarrow r^2 = 49 \Rightarrow r = 7\). Perimeter \(2\pi r = 2(22/7)(7) = 44\).
12.
(a) 3
Vol Sphere = Vol Cylinder Rise. Sphere \(r=9\), Cyl \(R=18\).
\(\frac{4}{3}\pi (9)^3 = \pi (18)^2 h \Rightarrow \frac{4}{3} \times 729 = 324 h \Rightarrow 972 = 324h \Rightarrow h = 3\).
Vol Sphere = Vol Cylinder Rise. Sphere \(r=9\), Cyl \(R=18\).
\(\frac{4}{3}\pi (9)^3 = \pi (18)^2 h \Rightarrow \frac{4}{3} \times 729 = 324 h \Rightarrow 972 = 324h \Rightarrow h = 3\).
13.
(a) 17
N=57. Median term \(\frac{57+1}{2} = 29\)th. Cumulative Freq: 13, 23, 38. 29th lies in 12-17. Upper limit = 17.
N=57. Median term \(\frac{57+1}{2} = 29\)th. Cumulative Freq: 13, 23, 38. 29th lies in 12-17. Upper limit = 17.
14.
(c) 3/4
Outcomes: HH, HT, TH, TT. At most one head: HT, TH, TT (3 outcomes).
Outcomes: HH, HT, TH, TT. At most one head: HT, TH, TT (3 outcomes).
15.
(b) 14
Number of bad eggs = \(0.035 \times 400 = 14\).
Number of bad eggs = \(0.035 \times 400 = 14\).
16.
(c) \(\sqrt{a^2 + b^2}\)
\(x_1 = a\cos\theta+b\sin\theta, y_1=0\). \(x_2=0, y_2=a\sin\theta-b\cos\theta\).
\(D^2 = (a\cos\theta+b\sin\theta)^2 + (a\sin\theta-b\cos\theta)^2 = a^2(\cos^2+\sin^2) + b^2(\sin^2+\cos^2) = a^2+b^2\).
\(x_1 = a\cos\theta+b\sin\theta, y_1=0\). \(x_2=0, y_2=a\sin\theta-b\cos\theta\).
\(D^2 = (a\cos\theta+b\sin\theta)^2 + (a\sin\theta-b\cos\theta)^2 = a^2(\cos^2+\sin^2) + b^2(\sin^2+\cos^2) = a^2+b^2\).
17.
(c) 4
Product = LCM \(\times\) HCF. \(x \times 18 = 36 \times 2 \Rightarrow 18x = 72 \Rightarrow x = 4\).
Product = LCM \(\times\) HCF. \(x \times 18 = 36 \times 2 \Rightarrow 18x = 72 \Rightarrow x = 4\).
18.
(b) \(150\sqrt{3}\)
\(\tan 30^\circ = \frac{150}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{x} \Rightarrow x = 150\sqrt{3}\).
\(\tan 30^\circ = \frac{150}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{x} \Rightarrow x = 150\sqrt{3}\).
19.
(c) A is true but R is false
\(\sin \theta\) increases, but \(\cos \theta\) decreases from 1 to 0 as \(\theta\) goes 0 to 90.
\(\sin \theta\) increases, but \(\cos \theta\) decreases from 1 to 0 as \(\theta\) goes 0 to 90.
20.
(b) Both A and R are true but R is not the correct
explanation of A
A is calculated using distance formula. R states section formula. Both correct, but unrelated.
A is calculated using distance formula. R states section formula. Both correct, but unrelated.
SECTION B (Solutions)
21.
Let \(7\sqrt{5}\) be rational. Then \(7\sqrt{5} = \frac{a}{b}\) where a,b are
integers, \(b \neq 0\).
\(\sqrt{5} = \frac{a}{7b}\).
Since a, b, 7 are integers, \(\frac{a}{7b}\) is rational. This implies \(\sqrt{5}\) is rational.
But this contradicts the fact that \(\sqrt{5}\) is irrational. Hence \(7\sqrt{5}\) is irrational.
\(\sqrt{5} = \frac{a}{7b}\).
Since a, b, 7 are integers, \(\frac{a}{7b}\) is rational. This implies \(\sqrt{5}\) is rational.
But this contradicts the fact that \(\sqrt{5}\) is irrational. Hence \(7\sqrt{5}\) is irrational.
22.
In \(\Delta AOB\) and \(\Delta COD\):
\(\angle OAB = \angle OCD\) (Alt. int. angles, \(AB \parallel DC\))
\(\angle OBA = \angle ODC\) (Alt. int. angles)
So \(\Delta AOB \sim \Delta COD\) (AA Similarity).
Ratios of corresponding sides are equal: \(\frac{AO}{CO} = \frac{BO}{DO} \Rightarrow \frac{AO}{BO} = \frac{CO}{DO}\).
\(\angle OAB = \angle OCD\) (Alt. int. angles, \(AB \parallel DC\))
\(\angle OBA = \angle ODC\) (Alt. int. angles)
So \(\Delta AOB \sim \Delta COD\) (AA Similarity).
Ratios of corresponding sides are equal: \(\frac{AO}{CO} = \frac{BO}{DO} \Rightarrow \frac{AO}{BO} = \frac{CO}{DO}\).
23.
\(2(\frac{2}{\sqrt{3}})^2 + x(\frac{\sqrt{3}}{2})^2 -
\frac{3}{4}(\frac{1}{\sqrt{3}})^2 = 10\).
\(2(\frac{4}{3}) + x(\frac{3}{4}) - \frac{3}{4}(\frac{1}{3}) = 10\).
\(\frac{8}{3} + \frac{3x}{4} - \frac{1}{4} = 10 \Rightarrow \frac{3x}{4} = 10 - \frac{8}{3} + \frac{1}{4}\).
\(\frac{3x}{4} = \frac{120 - 32 + 3}{12} = \frac{91}{12} \Rightarrow x = \frac{91}{9}\).
\(\cos(A-B) = \sqrt{3}/2 \Rightarrow A-B = 30^\circ\).
Adding equations: \(2A = 120^\circ \Rightarrow A = 60^\circ\).
Subtracting: \(2B = 60^\circ \Rightarrow B = 30^\circ\).
\(2(\frac{4}{3}) + x(\frac{3}{4}) - \frac{3}{4}(\frac{1}{3}) = 10\).
\(\frac{8}{3} + \frac{3x}{4} - \frac{1}{4} = 10 \Rightarrow \frac{3x}{4} = 10 - \frac{8}{3} + \frac{1}{4}\).
\(\frac{3x}{4} = \frac{120 - 32 + 3}{12} = \frac{91}{12} \Rightarrow x = \frac{91}{9}\).
OR
\(\sin(A+B) = 1 \Rightarrow A+B = 90^\circ\).\(\cos(A-B) = \sqrt{3}/2 \Rightarrow A-B = 30^\circ\).
Adding equations: \(2A = 120^\circ \Rightarrow A = 60^\circ\).
Subtracting: \(2B = 60^\circ \Rightarrow B = 30^\circ\).
24.
Radius \(r=14\). Angle swept in 15 mins \(= 360 \times \frac{15}{60} =
90^\circ\).
Area = \(\frac{90}{360} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 \text{ cm}^2\).
Area = \(\frac{90}{360} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 \text{ cm}^2\).
25.
Total outcomes = 36.
(i) Same number (doublet): (1,1), (2,2)...(6,6) = 6 outcomes. Prob = \(6/36 = 1/6\).
(ii) Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes. Prob = \(5/36\).
(i) Same number (doublet): (1,1), (2,2)...(6,6) = 6 outcomes. Prob = \(6/36 = 1/6\).
(ii) Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes. Prob = \(5/36\).
SECTION C (Solutions)
26.
\(144 = 2^4 \times 3^2\). \(180 = 2^2 \times 3^2 \times 5\). \(192 = 2^6
\times 3\).
HCF = \(2^2 \times 3^1 = 4 \times 3 = 12\).
LCM = \(2^6 \times 3^2 \times 5 = 64 \times 9 \times 5 = 2880\).
\(12=2^2 \times 3\), \(15=3 \times 5\), \(18=2 \times 3^2\).
LCM = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\) mins = 3 hours.
Next ring = 9 a.m. + 3 hours = 12:00 noon.
HCF = \(2^2 \times 3^1 = 4 \times 3 = 12\).
LCM = \(2^6 \times 3^2 \times 5 = 64 \times 9 \times 5 = 2880\).
OR
LCM of 12, 15, 18.\(12=2^2 \times 3\), \(15=3 \times 5\), \(18=2 \times 3^2\).
LCM = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\) mins = 3 hours.
Next ring = 9 a.m. + 3 hours = 12:00 noon.
27.
Sum \(\alpha + \beta = 5\). Product \(\alpha\beta = k\).
Given \(\alpha - \beta = 1\).
Solve for \(\alpha, \beta\): \((\alpha+\beta) + (\alpha-\beta) = 5+1 \Rightarrow 2\alpha = 6 \Rightarrow \alpha=3\).
\(\beta = 5-3 = 2\).
\(k = \alpha\beta = 3 \times 2 = 6\).
Given \(\alpha - \beta = 1\).
Solve for \(\alpha, \beta\): \((\alpha+\beta) + (\alpha-\beta) = 5+1 \Rightarrow 2\alpha = 6 \Rightarrow \alpha=3\).
\(\beta = 5-3 = 2\).
\(k = \alpha\beta = 3 \times 2 = 6\).
28.
Let \(\frac{1}{x+y} = u\) and \(\frac{1}{x-y} = v\).
\(10u + 2v = 4 \Rightarrow 5u + v = 2\) ... (i)
\(15u - 5v = -2\) ... (ii)
From (i) \(v = 2 - 5u\). Sub in (ii): \(15u - 5(2-5u) = -2 \Rightarrow 15u - 10 + 25u = -2\).
\(40u = 8 \Rightarrow u = 1/5\).
\(v = 2 - 5(1/5) = 1\).
\(x+y = 5\), \(x-y = 1\). Solving these: \(2x = 6 \Rightarrow x=3, y=2\).
\(10u + 2v = 4 \Rightarrow 5u + v = 2\) ... (i)
\(15u - 5v = -2\) ... (ii)
From (i) \(v = 2 - 5u\). Sub in (ii): \(15u - 5(2-5u) = -2 \Rightarrow 15u - 10 + 25u = -2\).
\(40u = 8 \Rightarrow u = 1/5\).
\(v = 2 - 5(1/5) = 1\).
\(x+y = 5\), \(x-y = 1\). Solving these: \(2x = 6 \Rightarrow x=3, y=2\).
29.
LHS = \((\sin^2 A + \text{cosec}^2 A + 2) + (\cos^2 A + \sec^2 A + 2)\).
= \((\sin^2 A + \cos^2 A) + \text{cosec}^2 A + \sec^2 A + 4\).
= \(1 + (1+\cot^2 A) + (1+\tan^2 A) + 4\).
= \(1 + 1 + 1 + 4 + \tan^2 A + \cot^2 A\).
= \(7 + \tan^2 A + \cot^2 A\) = RHS.
= \((\sin^2 A + \cos^2 A) + \text{cosec}^2 A + \sec^2 A + 4\).
= \(1 + (1+\cot^2 A) + (1+\tan^2 A) + 4\).
= \(1 + 1 + 1 + 4 + \tan^2 A + \cot^2 A\).
= \(7 + \tan^2 A + \cot^2 A\) = RHS.
30.
Tangents from external point are equal.
\(AP=AS, BP=BQ, CR=CQ, DR=DS\).
\(AB + CD = (AP+BP) + (CR+DR)\)
\(= (AS+BQ) + (CQ+DS)\)
\(= (AS+DS) + (BQ+CQ) = AD + BC\).
\(AP=AS, BP=BQ, CR=CQ, DR=DS\).
\(AB + CD = (AP+BP) + (CR+DR)\)
\(= (AS+BQ) + (CQ+DS)\)
\(= (AS+DS) + (BQ+CQ) = AD + BC\).
31.
Mean = \(\frac{\Sigma f_ix_i}{\Sigma f_i}\). Class marks: 5, 15, 25, 35,
45.
\(\Sigma f_i = 8+x+10+11+9 = 38+x\).
\(\Sigma f_ix_i = 40 + 15x + 250 + 385 + 405 = 1080 + 15x\).
\(25.2 = \frac{1080+15x}{38+x} \Rightarrow 25.2(38) + 25.2x = 1080 + 15x\).
\(957.6 + 25.2x = 1080 + 15x \Rightarrow 10.2x = 122.4 \Rightarrow x = 12\).
\(\Sigma f_i = 8+x+10+11+9 = 38+x\).
\(\Sigma f_ix_i = 40 + 15x + 250 + 385 + 405 = 1080 + 15x\).
\(25.2 = \frac{1080+15x}{38+x} \Rightarrow 25.2(38) + 25.2x = 1080 + 15x\).
\(957.6 + 25.2x = 1080 + 15x \Rightarrow 10.2x = 122.4 \Rightarrow x = 12\).
SECTION D (Solutions)
32.
Let height be \(h\). Points A, B at dist 4, 9.
Angles are \(\theta\) and \(90-\theta\).
In \(\Delta_1\): \(\tan \theta = h/4\).
In \(\Delta_2\): \(\tan(90-\theta) = h/9 \Rightarrow \cot \theta = h/9\).
Multiply: \(\tan \theta \cdot \cot \theta = (h/4)(h/9)\).
\(1 = h^2/36 \Rightarrow h^2 = 36 \Rightarrow h = 6\) m.
Angles are \(\theta\) and \(90-\theta\).
In \(\Delta_1\): \(\tan \theta = h/4\).
In \(\Delta_2\): \(\tan(90-\theta) = h/9 \Rightarrow \cot \theta = h/9\).
Multiply: \(\tan \theta \cdot \cot \theta = (h/4)(h/9)\).
\(1 = h^2/36 \Rightarrow h^2 = 36 \Rightarrow h = 6\) m.
33.
Let stream speed \(x\). Upstream \(24-x\), Downstream
\(24+x\).
\(\frac{32}{24-x} - \frac{32}{24+x} = 1\).
\(32[\frac{(24+x)-(24-x)}{576-x^2}] = 1 \Rightarrow 32(2x) = 576 - x^2\).
\(x^2 + 64x - 576 = 0\).
\((x+72)(x-8) = 0\). \(x=8\) km/h.
\(\frac{1}{x} + \frac{1}{x+3} = \frac{13}{40}\).
\(\frac{2x+3}{x^2+3x} = \frac{13}{40} \Rightarrow 80x + 120 = 13x^2 + 39x\).
\(13x^2 - 41x - 120 = 0\).
\(x = \frac{41 \pm \sqrt{1681 + 6240}}{26} = \frac{41 \pm 89}{26}\). \(x=5\) (positive).
Pipe A: 5 mins, Pipe B: 8 mins.
\(\frac{32}{24-x} - \frac{32}{24+x} = 1\).
\(32[\frac{(24+x)-(24-x)}{576-x^2}] = 1 \Rightarrow 32(2x) = 576 - x^2\).
\(x^2 + 64x - 576 = 0\).
\((x+72)(x-8) = 0\). \(x=8\) km/h.
OR
Pipe A: \(x\) min, Pipe B: \(x+3\) min.\(\frac{1}{x} + \frac{1}{x+3} = \frac{13}{40}\).
\(\frac{2x+3}{x^2+3x} = \frac{13}{40} \Rightarrow 80x + 120 = 13x^2 + 39x\).
\(13x^2 - 41x - 120 = 0\).
\(x = \frac{41 \pm \sqrt{1681 + 6240}}{26} = \frac{41 \pm 89}{26}\). \(x=5\) (positive).
Pipe A: 5 mins, Pipe B: 8 mins.
34.
Radius \(r=0.7\), Height \(h=2.4\). Slant \(l = \sqrt{0.7^2+2.4^2} =
2.5\).
TSA = CSA Cylinder + Base + CSA Cone.
\(= 2\pi rh + \pi r^2 + \pi rl = \pi r (2h + r + l)\).
\(= \frac{22}{7} \times 0.7 (4.8 + 0.7 + 2.5) = 2.2 \times 8 = 17.6 \approx 18 \text{ cm}^2\).
Inner SA = CSA Cyl + CSA Hem = \(2\pi rh + 2\pi r^2 = 2\pi r(h+r)\).
\(= 2 \times \frac{22}{7} \times 7 (6+7) = 44 \times 13 = 572 \text{ cm}^2\).
TSA = CSA Cylinder + Base + CSA Cone.
\(= 2\pi rh + \pi r^2 + \pi rl = \pi r (2h + r + l)\).
\(= \frac{22}{7} \times 0.7 (4.8 + 0.7 + 2.5) = 2.2 \times 8 = 17.6 \approx 18 \text{ cm}^2\).
OR
Hemisphere \(r=7\). Total height 13, Cyl height \(h = 13-7=6\).Inner SA = CSA Cyl + CSA Hem = \(2\pi rh + 2\pi r^2 = 2\pi r(h+r)\).
\(= 2 \times \frac{22}{7} \times 7 (6+7) = 44 \times 13 = 572 \text{ cm}^2\).
35.
\(N=40\). \(f_1 + 5 + 9 + 12 + f_2 + 3 + 2 = 40 \Rightarrow f_1 + f_2 =
9\).
Median 32.5 \(\in\) 30-40. \(l=30, h=10, f=12, cf=14+f_1\).
\(32.5 = 30 + \frac{20 - (14+f_1)}{12} \times 10\).
\(2.5 = \frac{6-f_1}{1.2} \Rightarrow 3 = 6 - f_1 \Rightarrow f_1 = 3\).
\(f_2 = 9 - 3 = 6\).
Median 32.5 \(\in\) 30-40. \(l=30, h=10, f=12, cf=14+f_1\).
\(32.5 = 30 + \frac{20 - (14+f_1)}{12} \times 10\).
\(2.5 = \frac{6-f_1}{1.2} \Rightarrow 3 = 6 - f_1 \Rightarrow f_1 = 3\).
\(f_2 = 9 - 3 = 6\).
SECTION E (Solutions)
Case Study - 1 Solution
\(a_3 = 6000 \Rightarrow a+2d=6000\). \(a_7 = 7000 \Rightarrow a+6d=7000\).
\(4d = 1000 \Rightarrow d=250\).
1. \(a = 6000 - 500 = 5500\).
2. \(a_5 = 5500 + 4(250) = 6500\).
3. \(S_7 = \frac{7}{2}(2(5500) + 6(250)) = \frac{7}{2}(11000 + 1500) = \frac{7}{2}(12500) = 43750\).
OR
\(10000 = 5500 + (n-1)250 \Rightarrow 4500 = (n-1)250 \Rightarrow n-1 = 18 \Rightarrow n=19\).
\(a_3 = 6000 \Rightarrow a+2d=6000\). \(a_7 = 7000 \Rightarrow a+6d=7000\).
\(4d = 1000 \Rightarrow d=250\).
1. \(a = 6000 - 500 = 5500\).
2. \(a_5 = 5500 + 4(250) = 6500\).
3. \(S_7 = \frac{7}{2}(2(5500) + 6(250)) = \frac{7}{2}(11000 + 1500) = \frac{7}{2}(12500) = 43750\).
OR
\(10000 = 5500 + (n-1)250 \Rightarrow 4500 = (n-1)250 \Rightarrow n-1 = 18 \Rightarrow n=19\).
Case Study - 2 Solution
1. AB = \(\sqrt{(6-3)^2 + (5-3)^2} = \sqrt{9+4} = \sqrt{13}\).
2. Midpoint \((\frac{3+6}{2}, \frac{3+5}{2}) = (4.5, 4)\).
3. \(PA^2 = PC^2 \Rightarrow (x-3)^2 + (y-3)^2 = (x-4)^2 + (y-1)^2\).
\(x^2-6x+9+y^2-6y+9 = x^2-8x+16+y^2-2y+1\).
\(-6x-6y+18 = -8x-2y+17 \Rightarrow 2x - 4y + 1 = 0\).
OR
Divides BC (6,5) and (4,1) in 2:1.
\(x = \frac{2(4)+1(6)}{3} = \frac{14}{3}\). \(y = \frac{2(1)+1(5)}{3} = \frac{7}{3}\). Point \((\frac{14}{3}, \frac{7}{3})\).
1. AB = \(\sqrt{(6-3)^2 + (5-3)^2} = \sqrt{9+4} = \sqrt{13}\).
2. Midpoint \((\frac{3+6}{2}, \frac{3+5}{2}) = (4.5, 4)\).
3. \(PA^2 = PC^2 \Rightarrow (x-3)^2 + (y-3)^2 = (x-4)^2 + (y-1)^2\).
\(x^2-6x+9+y^2-6y+9 = x^2-8x+16+y^2-2y+1\).
\(-6x-6y+18 = -8x-2y+17 \Rightarrow 2x - 4y + 1 = 0\).
OR
Divides BC (6,5) and (4,1) in 2:1.
\(x = \frac{2(4)+1(6)}{3} = \frac{14}{3}\). \(y = \frac{2(1)+1(5)}{3} = \frac{7}{3}\). Point \((\frac{14}{3}, \frac{7}{3})\).
Case Study - 3 Solution
1. Right triangle with hypotenuse as string, angle 60 with ground.
2. \(\sin 60^\circ = \frac{60}{\text{Length}} \Rightarrow \frac{\sqrt{3}}{2} = \frac{60}{L} \Rightarrow L = \frac{120}{\sqrt{3}} = 40\sqrt{3}\) m.
3. \(\sin 45^\circ = \frac{60}{L} \Rightarrow \frac{1}{\sqrt{2}} = \frac{60}{L} \Rightarrow L = 60\sqrt{2}\) m.
OR
Length \(L = 40\sqrt{3}\). New angle 30.
\(\sin 30^\circ = \frac{h}{40\sqrt{3}} \Rightarrow \frac{1}{2} = \frac{h}{40\sqrt{3}} \Rightarrow h = 20\sqrt{3}\) m.
1. Right triangle with hypotenuse as string, angle 60 with ground.
2. \(\sin 60^\circ = \frac{60}{\text{Length}} \Rightarrow \frac{\sqrt{3}}{2} = \frac{60}{L} \Rightarrow L = \frac{120}{\sqrt{3}} = 40\sqrt{3}\) m.
3. \(\sin 45^\circ = \frac{60}{L} \Rightarrow \frac{1}{\sqrt{2}} = \frac{60}{L} \Rightarrow L = 60\sqrt{2}\) m.
OR
Length \(L = 40\sqrt{3}\). New angle 30.
\(\sin 30^\circ = \frac{h}{40\sqrt{3}} \Rightarrow \frac{1}{2} = \frac{h}{40\sqrt{3}} \Rightarrow h = 20\sqrt{3}\) m.