CBSE Class 10 Maths Sample Paper 2026: Solved Section C & D (High-Weightage Questions)

While Sections A and B build your foundation, Section C (3 Marks) and Section D (5 Marks) are the true tests of a topper. These sections demand more than just the final answer—they require a clear "Given," "To Prove," "Formula Used," and a logical derivation.

Following the latest CBSE 2026 blueprint, we have curated the most expected long-answer questions that will help you secure those high-weightage marks.

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Section C: Short Answer Questions (3 Marks Each)

Q26. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 3x^2 - 4x + 1$, find a quadratic polynomial whose zeroes are $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$.

Q27. The sum of the $4^{th}$ and $8^{th}$ terms of an AP is $24$ and the sum of the $6^{th}$ and $10^{th}$ terms is $44$. Find the first three terms of the AP.

Q28. Prove the following identity:

$$\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta} = \tan \theta$$

Q29. Prove that the parallelogram circumscribing a circle is a rhombus.

Q30. Find the area of the shaded region in the figure, where a circular arc of radius $6 \text{ cm}$ has been drawn with vertex $O$ of an equilateral triangle $OAB$ of side $12 \text{ cm}$ as centre.

Q31. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears:

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5

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Section D: Long Answer Questions (5 Marks Each)

Q32. State and Prove the Basic Proportionality Theorem (Thales Theorem).

Q33. A motorboat whose speed is $18 \text{ km/h}$ in still water takes 1 hour more to go $24 \text{ km}$ upstream than to return downstream to the same spot. Find the speed of the stream.

Q34. A building is in the form of a cylinder surmounted by a hemispherical dome. The total height of the building is $19 \text{ m}$ and the diameter of the cylinder is $7 \text{ m}$. Find the total surface area and the volume of the building.

Q35. The angles of depression of the top and the bottom of an $8 \text{ m}$ tall building from the top of a multi-storeyed building are $30^\circ$ and $45^\circ$, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

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✅ Answer Key & Detailed Solutions

Section C Solutions (3 Marks)

• Q26: Sum $(\alpha + \beta) = 4/3$, Product $(\alpha\beta) = 1/3$. New sum $= \frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{28/27}{1/3} = 28/9$. New Product $= \alpha\beta = 1/3$. Polynomial: $9x^2 - 28x + 3$.

• Q27: Equations: $2a + 10d = 24$ and $2a + 14d = 44$. Solving gives $d = 5$ and $a = -13$. AP: $-13, -8, -3$.

• Q28: LHS $= \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)}$. Since $(1 - 2\sin^2 \theta) = \cos 2\theta$ and $(2\cos^2 \theta - 1) = \cos 2\theta$, they cancel out. LHS $= \tan \theta$.

• Q29: Since $ABCD$ is a parallelogram, $AB=CD$ and $AD=BC$. As it circumscribes a circle, $AB+CD = AD+BC$. $2AB = 2AD \implies AB=AD$. A parallelogram with adjacent sides equal is a rhombus.

• Q30: Area $= \text{Area of major sector} + \text{Area of equilateral triangle}$.

  Area $= \frac{300}{360} \times \pi \times 6^2 + \frac{\sqrt{3}}{4} \times 12^2 = (30\pi + 36\sqrt{3}) \text{ cm}^2$.

• Q31: (i) $81/90 = 9/10$ (ii) $9/90 = 1/10$ (iii) $18/90 = 1/5$.

Section D Solutions (5 Marks)

• Q32: (Refer to standard NCERT proof for BPT). Essential points: Area of triangles on the same base and between same parallels.

• Q33: Let speed of stream be $x$. $\frac{24}{18-x} - \frac{24}{18+x} = 1$. Solving the quadratic $x^2 + 48x - 324 = 0$ gives $x = 6 \text{ km/h}$.

• Q34: Radius $r = 3.5 \text{ m}$. Height of cylinder $H = 19 - 3.5 = 15.5 \text{ m}$.

  Volume: $\pi r^2 H + \frac{2}{3}\pi r^3 \approx 685.17 \text{ m}^3$.

  TSA: $2\pi r H + 2\pi r^2 + \pi r^2 \text{ (base)} = 341 + 77 + 38.5 = 456.5 \text{ m}^2$.

• Q35: Let height be $H$ and distance be $x$. $\tan 45^\circ = H/x \implies H = x$.

  $\tan 30^\circ = (H-8)/x \implies 1/\sqrt{3} = (H-8)/H$.

  Height $= 4(3+\sqrt{3}) \approx 18.93 \text{ m}$. Distance is the same.