CBSE Class 10 Maths Most Repeated Questions with Solutions

Assume Rational Form [½ Mark]

Let us assume to the contrary that \sqrt{5} is rational. Then there exist co-prime positive integers a and b ( b \neq 0 ) such that \sqrt{5} = \frac{a}{b} , where \gcd(a, b) = 1 .

Square Both Sides & Divisibility [1 Mark]

Squaring both sides: 5 = \frac{a^2}{b^2} \implies a^2 = 5b^2 . Since 5 divides a^2 , by Fundamental Theorem of Arithmetic, 5 must also divide a (since 5 is prime).

Substitute and Deduce Contradiction [1 Mark]

Let a = 5c for some integer c . Substituting into a^2 = 5b^2 : (5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2 . This means 5 divides b^2 , which implies 5 also divides b .

Conclude Contradiction [½ Mark]

Thus 5 is a common factor of both a and b . But this contradicts our initial assumption that a and b are co-prime (having no common factor other than 1). Therefore, our assumption is false, and \sqrt{5} is irrational.

Statement & Construction [1 Mark]

**Statement:** If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. **Construction:** In \triangle ABC , line DE \parallel BC . Join BE and CD . Draw DM \perp AC and EN \perp AB .

Area Ratios on Sides AB and AC [1½ Marks]

\text{ar}(\triangle ADE) = \frac{1}{2} \times AD \times EN and \text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN . Dividing gives: \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{AD}{DB} ... (1) Similarly, with base AE and altitude DM : \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} ... (2)

Triangles on Same Base Between Parallels [1 Mark]

Notice that \triangle BDE and \triangle CDE lie on the same base DE and between the same parallel lines DE and BC . Therefore: \text{ar}(\triangle BDE) = \text{ar}(\triangle CDE) ... (3) From (1), (2), and (3), the LHS are equal \implies \frac{AD}{DB} = \frac{AE}{EC} . Hence proved.

Numerical Application [1½ Marks]

Given \frac{AD}{DB} = \frac{3}{5} . By BPT, \frac{AE}{EC} = \frac{3}{5} . Let AE = x\text{ cm} , so EC = AC - AE = 4.8 - x . \frac{x}{4.8 - x} = \frac{3}{5} \implies 5x = 3(4.8 - x) \implies 5x = 14.4 - 3x \implies 8x = 14.4 \implies x = \frac{14.4}{8} = 1.8\text{ cm} .

Theorem Proof: Congruence of Tangent Triangles [2 Marks]

Let P be an external point to a circle with center O . Tangents from P touch circle at A and B . Join OA, OB, OP . In \triangle OAP and \triangle OBP : - OA = OB (radii of the same circle) - OP = OP (common hypotenuse) - \angle OAP = \angle OBP = 90^\circ (tangent is perpendicular to radius at point of contact) By RHS Congruence Criterion, \triangle OAP \cong \triangle OBP \implies PA = PB (CPCT). Hence proved.

Identify Tangent Pairs on Quadrilateral [1½ Marks]

Let the circumscribed circle touch sides AB, BC, CD, DA at points P, Q, R, S respectively. Since lengths of tangents from an external point are equal: AP = AS ... (1) BP = BQ ... (2) CR = CQ ... (3) DR = DS ... (4)

Sum Equations and Group Opposite Sides [1½ Marks]

Adding equations (1), (2), (3), and (4): (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) Notice AP + BP = AB , CR + DR = CD , AS + DS = AD , and BQ + CQ = BC . Therefore: AB + CD = AD + BC . Hence proved.

Expand Binomial Squares [1 Mark]

Expanding LHS using (a+b)^2 = a^2 + b^2 + 2ab : LHS = (\sin^2\theta + \csc^2\theta + 2\sin\theta\csc\theta) + (\cos^2\theta + \sec^2\theta + 2\cos\theta\sec\theta)

Apply Reciprocal and Pythagorean Identities [1 Mark]

Since \sin\theta\csc\theta = 1 and \cos\theta\sec\theta = 1 : LHS = (\sin^2\theta + \cos^2\theta) + \csc^2\theta + \sec^2\theta + 2(1) + 2(1) Using \sin^2\theta + \cos^2\theta = 1 : LHS = 1 + 2 + 2 + \csc^2\theta + \sec^2\theta = 5 + \csc^2\theta + \sec^2\theta

Convert to Tangent and Cotangent [1 Mark]

Substitute \csc^2\theta = 1 + \cot^2\theta and \sec^2\theta = 1 + \tan^2\theta : LHS = 5 + (1 + \cot^2\theta) + (1 + \tan^2\theta) = 7 + \tan^2\theta + \cot^2\theta = RHS . Hence proved.

Define Variables & Relative Speeds [1 Mark]

Let the speed of the stream be x\text{ km/h} . Speed of the boat in still water = 18\text{ km/h} . - Speed of the boat upstream = (18 - x)\text{ km/h} - Speed of the boat downstream = (18 + x)\text{ km/h}

Formulate Time Difference Equation [1½ Marks]

Distance = 24\text{ km} . - Time taken upstream: T_1 = \frac{24}{18 - x}\text{ hours} - Time taken downstream: T_2 = \frac{24}{18 + x}\text{ hours} According to the question, T_1 - T_2 = 1 : \frac{24}{18 - x} - \frac{24}{18 + x} = 1

Simplify into Quadratic Form [1½ Marks]

24 \left[\frac{(18 + x) - (18 - x)}{(18 - x)(18 + x)}\right] = 1 \implies 24 \left(\frac{2x}{324 - x^2}\right) = 1 48x = 324 - x^2 \implies x^2 + 48x - 324 = 0

Solve for x & Reject Extraneous Root [1 Mark]

Factoring: x^2 + 54x - 6x - 324 = 0 \implies x(x + 54) - 6(x + 54) = 0 (x - 6)(x + 54) = 0 \implies x = 6 or x = -54 . Since speed cannot be negative, x \neq -54 . Therefore, the speed of the stream is 6\text{ km/h} .