CBSE 10th Maths Question Paper 2023 (Set 1)

For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is given by $-b/a$. In option (A), $-(-4)/2 = 4/2 = 2$. Wait, re-evaluating the provided options: Option A: $-(-4)/2 = 2$. Option B: $-4/1 = -4$. Option C: $-(-4)/2 = 2$. Option D: $-(-4)/4 = 1$. It appears there is a typo in the question or options provided, but based on the formula $-b/a = 4$, none of the standard equations match. However, if the question intended for $-b/a = 4$, then $b/a = -4$. If we divide by 2, $-b/a$ should be 4. Re-calculating: for (A) $2x^2-8x+8=0$ would be 4. Given the standard MCQ source constraints, A is often marked if $2x^2-8x+8$ was intended. Given the provided text, none yield 4.

The length of an arc is given by the formula $l = \frac{\theta}{360^\circ} \times 2\pi r$. Substituting the values: $l = \frac{90^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 2 \times 22 \times 2 = \frac{88}{4} = 22$ cm.

Since $\triangle ABC \sim \triangle PQR$, the corresponding angles are equal: $\angle A = \angle P = 32^\circ$, $\angle B = \angle Q$, and $\angle C = \angle R = 65^\circ$. In $\triangle ABC$, the sum of angles is $180^\circ$. Thus, $\angle B = 180^\circ - (32^\circ + 65^\circ) = 180^\circ - 97^\circ = 83^\circ$.

Similar to question 16677, $\angle A = 32^\circ$ and $\angle C = \angle R = 65^\circ$. The angle $\angle B = 180^\circ - (32^\circ + 65^\circ) = 83^\circ$.

A rectangle is formed by vertices. B(0,0), C(3,0) lies on the x-axis, and D(0,4) lies on the y-axis. For ABCD to be a rectangle, the vertex A must be the corner opposite to the origin, which is (3,4) because the sides must be parallel to the axes.