📌 SECTION B
VSA · 5 Questions · 2 marks each
Very Short Answer type – internal choice in Q21 & Q24
Q21.(a) 📐 Trigonometric Equation
If \(x \cos 60^{\circ}+y \cos 0^{\circ}+\sin 30^{\circ}\) \(-\cot 45^{\circ}=5\), then find the value of \(x+2y\).
🔍 VIEW ANSWER
✅ ANSWER: \(x+2y = 11\)
📘 STEP-BY-STEP: \(\cos 60^\circ = \frac{1}{2},\; \cos 0^\circ = 1,\); \(\sin 30^\circ = \frac{1}{2},\; \cot 45^\circ = 1\) Substitute: \(x\cdot\frac{1}{2} + y\cdot1 + \frac{1}{2} – 1 = 5\) \(\Rightarrow \frac{x}{2} + y – \frac{1}{2} = 5\) \(\Rightarrow \frac{x}{2} + y = \frac{11}{2}\) Multiply by 2: \(x + 2y = 11\).
OR 21.(b) ⚡ Evaluate
Evaluate : \(\displaystyle \frac{\tan ^{2} 60^{\circ}}{\sin ^{2} 60^{\circ}+\cos ^{2} 30^{\circ}}\)
🔍 VIEW ANSWER
✅ ANSWER: 2
\(\tan 60^\circ = \sqrt{3},\; \sin 60^\circ = \frac{\sqrt{3}}{2},\) ; \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) \(\frac{(\sqrt{3})^2}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\) = \(\frac{3}{\frac{3}{4}+\frac{3}{4}} = \frac{3}{\frac{6}{4}} = 3 \times \frac{4}{6} = 2\).
Q22. 🧮 Zeroes of polynomial
Find the zeroes of the polynomial \(p(x)=x^{2}+\frac{4}{3}x-\frac{4}{3}\).
🔍 VIEW ANSWER
✅ ZEROES: \(x = \frac{2}{3},\; x = -2\)
\(a=1, b=\frac{4}{3}, c=-\frac{4}{3}\). Discriminant: \(\left(\frac{4}{3}\right)^2 – 4\cdot1\cdot\left(-\frac{4}{3}\right)\) \(= \frac{16}{9}+\frac{48}{9}=\frac{64}{9}\), \(\sqrt{D}=\frac{8}{3}\). \(x = \frac{-\frac{4}{3} \pm \frac{8}{3}}{2} = \frac{4/3}{2} = \frac{2}{3}\) and \(\frac{-12/3}{2} = -2\).
Q23. ⚫ Circle geometry
The coordinates of the centre of a circle are \((2a, a-7)\). Find value(s) of ‘\(a\)’ if the circle passes through \((11,-9)\) and has diameter \(10\sqrt{2}\) units.
🔍 VIEW ANSWER
✅ VALUES: \(a = 3\) or \(a = 5\)
Radius = \(\frac{10\sqrt{2}}{2}=5\sqrt{2}\). Distance from centre \((2a, a-7)\) to \((11,-9)\) equals radius: \(\sqrt{(11-2a)^2 + (-9-(a-7))^2} = 5\sqrt{2}\) → \((11-2a)^2 + (-2-a)^2 = 50\). \(121+4a^2-44a + 4 + a^2+4a = 50\) → \(5a^2 -40a +75 = 0\) → \(a^2 -8a+15=0\) → \((a-3)(a-5)=0\) → \(a=3,5\).
Q24.(a) 🔺 Similar Triangles
If \(\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}\) in which \(\mathrm{AB}=6\ \mathrm{cm}, \mathrm{BC}=4\ \mathrm{cm}\), \(\mathrm{AC}=8\ \mathrm{cm}\) and \(\mathrm{PR}=6\ \mathrm{cm}\), then find \((\mathrm{PQ}+\mathrm{QR})\).
🔍 VIEW ANSWER
✅ PQ + QR = 7.5 cm
\(\triangle ABC \sim \triangle PQR\) → \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}\). Given \(PR = 6\) (corresponds to AC=8). \(\frac{6}{PQ} = \frac{8}{6} \Rightarrow PQ = \frac{36}{8}=4.5\). \(\frac{4}{QR} = \frac{8}{6} \Rightarrow QR = \frac{24}{8}=3\). Sum = \(4.5+3=7.5\) cm.
OR Q24.(b) 🔷 Triangle Similarity
In the given figure, \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and \(\angle 1=\angle 2\), show that \(\Delta \mathrm{PQS} \sim \Delta \mathrm{TQR}\).
🔍 VIEW PROOF
✅ SIMILARITY: SAS criterion
\(\angle 1 = \angle 2\) ⇒ \(PR = PQ\) (sides opposite equal angles in \(\triangle PQR\)). Given \(\frac{QR}{QS} = \frac{QT}{PR}\) ⇒ \(\frac{QR}{QS} = \frac{QT}{PQ}\). Also \(\angle PQR = \angle TQR\) (common angle). ∴ \(\triangle PQS \sim \triangle TQR\) by SAS (sides proportional and included angle equal).
Q25. ⚪ Circle – Tangent
A person is standing at P outside a circular ground at a distance of 26 m from the centre. His distances from points A and B on the ground are 10 m (PA and PB are tangents). Find the radius of the circular ground.
🔍 VIEW ANSWER
✅ RADIUS = 24 m
O = centre, PA = tangent, OA ⟂ PA. In right \(\triangle OAP\): OP = 26 m, PA = 10 m. \(OA^2 = OP^2 – PA^2\) \(= 26^2 – 10^2\) \(= 676 – 100\) \(= 576\) → \(OA = 24\) m (radius).
⏹️ END OF SECTION B · VSA (2 marks each) ⏹️ ✅ 5 Questions with internal choices – fully solved, hidden answers
