Mathematics Questions and Answers

\(\frac{1}{p^2}\) + \(\frac{1}{q^2}\) = \(\frac{q^2 + p^2}{p^2 + q^2}\)

\(\frac{\frac{2}{x}}{\frac{p^2 + q^2}{p^2 q^2}}\)

m = \(\frac{2p^2q^2}{x(p^2 + q^2)}\)

= m2p2q2 = m x (p2 + q2)

x = \(\frac{2p^2q^2}{m(q^2 + p^2)}\)

4 bags of rice - M 56 each

3 tins of milk - M 4 each

\(M 56 \times 4 = M 323\)

\(M 4 \times 3 = M 15\)

\(M (323 + 15) = M 341\)

\((m + n)^{2} = (m - n)^{2} + x^{2}\)

\(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\)

\(x^{2} = 4mn\)

\(x = \sqrt{4mn} = 2\sqrt{mn}\)

1 + tan2\(\theta\) = sec2\(\theta\)

= \(\frac{1}{cos^2\theta}\)

\(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\)

\(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\)

\(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\)

= \(\frac{(m^2 + n^2 + 2mn)}{4mn}\)

\(p : q = \frac{1}{3} : \frac{1}{2}\)

\(\frac{p}{q} = \frac{2}{3}\)

\(2q = 3p ... (1)\)

\(q : r = \frac{2}{5} : \frac{4}{7}\)

\(\frac{q}{r} = \frac{2}{5} \times \frac{7}{4} = \frac{7}{10}\)

\(10q = 7r ... (2)\)

Eliminating q, we have

(1) : \(2q = 3p \)

\(10q = 15p\)

\(\implies 15p = 7r\)

\(\therefore \frac{p}{r} = \frac{7}{15}\)

\(p : r = 7 : 15\)

Length of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r = 6

\(\theta\) x 2\(\pi\)r = 360 x 6

\(\theta\) = \(\frac{360 \times 6}{2\pi r}\)

Area of the sector = \(\frac{\theta}{360}\) x \(\pi\)r²

\(\frac{360 \times 6}{2\pi r}\) x \(\frac{1}{360}\) x \(\pi\)r² = r

= 3 x 7

= 21cm²