IPMAT Indore 2024Geometry > Hardk2sinθ2+ksinθc2+k2sin2θ\dfrac{k^2 \sin \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}2k2sinθ+ksinθc2+k2sin2θk2sin2θ2+ksinθc2−k2sin2θ\dfrac{k^2 \sin 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}2k2sin2θ+ksinθc2−k2sin2θk2cos2θ2+ksinθc2−k2sin2θ\dfrac{k^2 \cos 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}2k2cos2θ+ksinθc2−k2sin2θk2cosθ2+ksinθc2+k2sin2θ\dfrac{k^2 \cos \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}2k2cosθ+ksinθc2+k2sin2θ✅ Correct Option: 2Related questions:The number of triangles with integer sides and with perimeter 15 is:Let △ABC\triangle ABC△ABC be a triangle right-angled at BBB with AB=BC=18AB = BC = 18AB=BC=18. The area of the largest rectangle that can be inscribed in this triangle and has BBB as one of the vertices is:In a triangle ABC, let D be the midpoint of BC, and AM be the altitude on BC. If the lengths of AB, BC and CA are in the ratio of 2:4:3, then the ratio of the lengths of BM and AD would be