IPMAT IndoreGeometry > Hard(33+π)k224\left(3\sqrt{3} + \pi\right) \frac{k^2}{24}(33+π)24k2(33+π)k26\left(3\sqrt{3} + \pi\right) \frac{k^2}{6}(33+π)6k2(33−π)k224\left(3\sqrt{3} - \pi\right) \frac{k^2}{24}(33−π)24k2(33+π)k26\left(3\sqrt{3} + \pi\right) \frac{k^2}{6}(33+π)6k2✅ Correct Option: 1Related questions:IPMAT Indore 2025In triangle ABC,AB=AC=x,∠ABC=θABC, AB = AC = x, ∠ABC = \thetaABC,AB=AC=x,∠ABC=θ and the circumradius is equal to yyy. Then xy\frac{x}{y}yx equalsIPMAT Indore 2022In a right-angled triangle ABC, the hypotenuse AC is of length 13 cm. A line drawn connecting the midpoints D and E of sides AB and AC is found to be 6 cm in length. The length of BC isIPMAT Indore 2022The lengths of the sides of a triangle are x,21x, 21x,21 and 404040, where xxx is the shortest side. A possible value of xxx is: