Start learning →

Maharashtra SSC (Class 10) Mensuration: Surface Areas & Volumes — practice questions — page 2

Explore 50 free Mensuration: Surface Areas & Volumes questions in Maharashtra SSC (Class 10) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

Practise Mensuration: Surface Areas & Volumes with Quill →

Mensuration: Surface Areas & Volumes questions and exercises with answers

Showing questions 31–50 of 50.

If $z = \cos\theta + i\sin\theta$, find an expression for $z^n + \dfrac{1}{z^n}$.Read the answer and worked solution →Find the eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.Read the answer and worked solution →The foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ lie at:Read the answer and worked solution →An ellipse has foci at $(\pm 4, 0)$ and passes through the point $(5, 0)$. Find its standard equation.Read the answer and worked solution →Find the length of the latus rectum of the parabola $y^2 = 8x$.Read the answer and worked solution →Find the equation of the parabola with vertex at the origin and focus at $(0, -3)$.Read the answer and worked solution →A parabola has its vertex at the origin, its axis along the positive $x$-axis, and passes through the point $(4, 6)$. Find its equation.Read the answer and worked solution →For the parabola $y^2 = 16x$, find the equation of the directrix.Read the answer and worked solution →The standard equation of a hyperbola with centre at the origin and foci on the $x$-axis is:Read the answer and worked solution →For any hyperbola, the eccentricity $e$ satisfies:Read the answer and worked solution →What is the eccentricity of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$?Read the answer and worked solution →A hyperbola has foci at $(\pm 5, 0)$ and the length of the transverse axis is $6$. Find its equation.Read the answer and worked solution →A fair coin is tossed twice. How many outcomes are there in the sample space?Read the answer and worked solution →Area of a circle of radius r is:Read the answer and worked solution →Area of a sector subtending angle θ (deg) at the centre of a circle of radius r is:Read the answer and worked solution →Volume of a cone with radius r and height h is:Read the answer and worked solution →Volume of a sphere of radius r equals:Read the answer and worked solution →Match each solid to its total surface-area formula.Read the answer and worked solution →CLASS10 · Mathematics · Areas Related to CirclesRead the answer and worked solution →CLASS10 · Mathematics · Surface Area & VolumeRead the answer and worked solution →

Worked practice questions

If $z = \cos\theta + i\sin\theta$, find an expression for $z^n + \dfrac{1}{z^n}$.

  1. $\sin 2n\theta$
  2. $\cos n\theta$
  3. $2\cos n\theta$
  4. $2i\sin n\theta$
Show answer and solution

Correct answer: C. $2\cos n\theta$

**Apply De Moivre.** $z^n = \cos n\theta + i\sin n\theta$. **Compute the reciprocal.** Since $|z| = 1$, $\dfrac{1}{z^n} = \overline{z^n} = \cos n\theta - i\sin n\theta$ (conjugate of a unit-modulus complex number is its inverse). **Sum.** $z^n + \dfrac{1}{z^n} = 2\cos n\theta$ — the imaginary parts cancel. **Symmetric companion.** Subtracting instead gives $z^n - \dfrac{1}{z^n} = 2i\sin n\theta$ (option B) — useful for deriving multiple-angle formulae, but it's not what the question asks for.

Open this question →

Find the eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.

  1. $\dfrac{3}{4}$
  2. $\dfrac{3}{5}$
  3. $\dfrac{4}{5}$
  4. $\dfrac{5}{4}$
Show answer and solution

Correct answer: B. $\dfrac{3}{5}$

Identify: $a^2 = 25$ (the larger denominator), $b^2 = 16$. So $a = 5, b = 4$. $c^2 = a^2 - b^2 = 25 - 16 = 9 \Rightarrow c = 3$. Eccentricity $e = \dfrac{c}{a} = \dfrac{3}{5}$. For any ellipse $0 < e < 1$ (a circle is the $e = 0$ limit). Option D ($5/4 > 1$) describes a *hyperbola*, not an ellipse.

Open this question →

The foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ lie at:

  1. $(\pm 4, 0)$
  2. $(\pm 3, 0)$
  3. $(0, \pm 3)$
  4. $(\pm 5, 0)$
Show answer and solution

Correct answer: B. $(\pm 3, 0)$

For a horizontal ellipse the foci sit at $(\pm c, 0)$ on the major axis, where $c^2 = a^2 - b^2$. $c^2 = 25 - 16 = 9 \Rightarrow c = 3$. So the foci are at $(\pm 3, 0)$. Option D mistakenly picks the *vertices* option A switches $a$ and $b$.

Open this question →

How to practise

  1. Choose a Mensuration: Surface Areas & Volumes question and try to work out your answer.
  2. Open the question to check the correct answer and read its worked solution.
  3. Continue in the app for bite-sized lessons, explanations and review with Quill. You can start without creating an account.

Put it into practice

Continue with Mensuration: Surface Areas & Volumes practice for Maharashtra SSC (Class 10). Read explanations as you go. No signup form needed to start.

Practise Mensuration: Surface Areas & Volumes →

More Mathematics topics

Real Numbers and PolynomialsStatistics and ProbabilityArithmetic ProgressionQuadratic EquationsIntroduction to TrigonometryAlgebra: Simultaneous Linear Equations & QuadraticsAlgebra: Equations, Inequations & MatricesTriangles, Circles and Geometry