CBSE 10th Maths Question Paper 2024 (Set 4)

Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, let $(x_1, y_1) = (a \cos \theta, a \sin \theta)$ and $(x_2, y_2) = (a \sin \theta, a \cos \theta)$. Then $d^2 = (a \sin \theta - a \cos \theta)^2 + (a \cos \theta - a \sin \theta)^2 = a^2(\sin \theta - \cos \theta)^2 + a^2(\cos \theta - \sin \theta)^2$. Since $(\sin \theta - \cos \theta)^2 = (\cos \theta - \sin \theta)^2$, this equals $2a^2(\sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta) = 2a^2(1 - 2\sin \theta \cos \theta)$. Wait, calculating directly: $d^2 = a^2(\sin \theta - \cos \theta)^2 + a^2(\cos \theta - \sin \theta)^2 = a^2(1 - 2\sin \theta \cos \theta) + a^2(1 - 2\sin \theta \cos \theta) = 2a^2(1 - \sin(2\theta))$. Given the structure, the points are usually calculated as $d^2 = a^2(\sin \theta - \cos \theta)^2 + a^2(\cos \theta - \sin \theta)^2 = a^2(1 - 2\sin \theta \cos \theta + 1 - 2\sin \theta \cos \theta) = 2a^2(1 - \sin(2\theta))$. Re-evaluating: distance is $\sqrt{a^2(\sin \theta - \cos \theta)^2 + a^2(\cos \theta - \sin \theta)^2} = \sqrt{2a^2(\sin \theta - \cos \theta)^2} = a|\sin \theta - \cos \theta|\sqrt{2}$. However, the standard distance for these points in textbook problems often assumes independent variables or simpler geometry. Evaluating at $\theta=0$: pts are $(a,0)$ and $(0,a)$, distance is $\sqrt{a^2+a^2} = a\sqrt{2}$.

Using the distance formula $d = \sqrt{(a \sin q - a \cos q)^2 + (-a \cos q - a \sin q)^2} = \sqrt{a^2(\sin^2 q + \cos^2 q - 2\sin q \cos q) + a^2(\cos^2 q + \sin^2 q + 2\sin q \cos q)} = \sqrt{a^2(1 - \sin 2q) + a^2(1 + \sin 2q)} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$. Looking at the options provided, there is a likely typo in the question or the intended result is $a\sqrt{2}$ or similar. Given options, none match $a\sqrt{2}$, but if the question implies $2a$ it is often a transcription error.

Since tangents PA and PB are drawn from an external point, $PA = PB = 5$ cm. Given $\angle APB = 90^\circ$, $\triangle PAB$ is an isosceles right-angled triangle. By Pythagoras theorem, $AB = \sqrt{PA^2 + PB^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$ cm.

The A.P. is $-29, -26, -23, \dots, 61$. Here $a = -29$, $d = 3$. Using $a_n = a + (n-1)d$, we get $61 = -29 + (n-1)3$, so $90 = (n-1)3$, $30 = n-1$, $n = 31$. The total number of terms is 31. The sum is $S_{31} = \frac{31}{2}(-29 + 61) = \frac{31}{2}(32) = 31 \times 16 = 496$. The term 16th is the middle term. Option B is likely referencing the position $n=16$.

Total cards from 6 to 55: $55 - 6 + 1 = 50$. Perfect squares in this range: $9, 16, 25, 36, 49$. There are 5 such squares. Probability = $5/50 = 1/10$. There appears to be a mismatch in the provided options; however, checking $3^2$ to $7^2$, the calculation $5/50$ is correct.