Savita Achary − Profile

b) ΔPCA Reasoning: 1. Given: PQ || BC, AB || QR, and PC intersects QR at O. 2. Since PQ || BC, ΔPQC is similar to ΔABC (by the angle-angle similarity criterion). 3. AB || QR implies that ∠BAC = ∠QCA (alternate angles). 4. In ΔPBC and ΔPCA: - ∠BPC is common - ∠PBC = ∠PCA (since PQ || BC, these are corresponding angles) - ∠PCB = ∠BAC = ∠QCA (from steps 2 and 3) 5. By the angle-angle similarity criterion, ΔPBC is similar to ΔPCA. Therefore, the correct answer is ΔPCA.

d) (II) and (III) Reasoning: In a rhombus: 1. The diagonals bisect each other at right angles. 2. The diagonals bisect the angles of the rhombus. 3. The diagonals are the perpendicular bisectors of each other. Given these properties: I. The bisector of ∠A and ∠C is not necessarily equidistant from A and C. It's only true for the diagonal that bisects these angles. II. The perpendicular bisector of AC is equidistant from A and C by definition. III. The diagonal BD is perpendicular to and bisects AC, so it is equidistant from A and C. Therefore, statements II and III are correct.

b) (p+q)/2 Reasoning: 1. In an arithmetic progression, the middle term between any two terms is their average. 2. Here, we have: (x+y)th term = p (x-y)th term = q We need to find the xth term. 3. The xth term is exactly in the middle of the (x+y)th and (x-y)th terms. 4. Therefore, the xth term will be the average of p and q: xth term = (p + q) / 2 Thus, the correct answer is (p+q)/2.

a) y = 2 Reasoning: 1. When a point is reflected over a horizontal line y = k, its x-coordinate remains the same, and its y-coordinate changes to 2k - y. 2. Here, the line of reflection is y = 3, so k = 3. 3. The original point P is (3, 4). 4. After reflection: x-coordinate of P' = 3 (unchanged) y-coordinate of P' = 2(3) - 4 = 2 5. So, P' is (3, 2). 6. A line parallel to the x-axis passing through P' would have the equation y = 2. Therefore, the equation of the line passing through P' parallel to the x-axis is y = 2.

a) A is true, R is false. Reasoning: Assertion (A) is true: In ΔPQA, sin A is indeed greatest for this triangle arrangement. This is because sin A is opposite/hypotenuse, and PQ (opposite to angle A) is the longest side in this right-angled triangle. Reason (R) is false: The value of sin A does depend on the side lengths. It is specifically the ratio of the opposite side to the hypotenuse. While it's true that sin A is a ratio, this ratio is directly related to the lengths of the sides. Therefore, A is true, but R is false.

c) 3cm Reasoning: For a right-angled triangle that circumscribes a circle: 1. Let r be the radius of the inscribed circle. 2. In a right-angled triangle, if a and b are the lengths of the two shorter sides, and c is the hypotenuse, then: r = (a + b - c) / 2 3. We know AB = 15cm, AC = 17cm. Let's find BC using the Pythagorean theorem: BC² = AC² - AB² = 17² - 15² = 289 - 225 = 64 BC = 8cm 4. Now we can apply the formula: r = (15 + 8 - 17) / 2 = 6 / 2 = 3cm Therefore, the radius of the inscribed circle is 3cm.

b) 115° Reasoning: In the diagram: 1. APB and AQB are angles in a semicircle, so they are right angles (90°). 2. ΔAPB is isosceles (AP = PB as they are radii), so ∠PAB = ∠PBA. 3. Given that ∠APB = 130°, we can find ∠PAB: ∠PAB + ∠APB + ∠PBA = 180° (sum of angles in a triangle) ∠PAB + 130° + ∠PAB = 180° 2∠PAB = 50° ∠PAB = 25° 4. In ΔACD: ∠CAD + x + 25° = 180° (sum of angles in a triangle) x = 180° - 25° - ∠CAD x = 155° - ∠CAD 5. ∠CAD = 40° (inscribed angle is half of central angle, so 80°/2 = 40°) 6. Therefore: x = 155° - 40° = 115° The correct answer is 115°.

c) y - √3x = 4 - 2√3 Reasoning: In a regular hexagon, adjacent sides are at 60° to each other. The line BE will make a 60° angle with the x-axis. 1. The slope of BE is tan(60°) = √3 2. The equation of a line with slope m passing through point (x1, y1) is: y - y1 = m(x - x1) 3. Using point B(2, 4): y - 4 = √3(x - 2) y = √3x - 2√3 + 4 y - √3x = 4 - 2√3 Therefore, the equation of BE is y - √3x = 4 - 2√3.

b) 8% Reasoning: Let's calculate the rate of interest using the given information. Principal (P) = 2500 × 24 = 60,000 (total deposit over 2 years) Interest (I) = 6250 Time (T) = 2 years Using the simple interest formula: I = (P × R × T) / 100 6250 = (60000 × R × 2) / 100 6250 = 1200R R = 6250 / 1200 = 5.208% The closest option to this calculated rate is 8%.

b) (2, -1) Reasoning: The centroid of a triangle divides each median in the ratio 2:1 (closer to the vertex). Coordinates of centroid = (x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3 Where (x1, y1) = (3, -5), (x2, y2) = (-7, 4), (x3, y3) = (10, -2) x-coordinate = (3 + (-7) + 10) / 3 = 6 / 3 = 2 y-coordinate = (-5 + 4 + (-2)) / 3 = -3 / 3 = -1 Therefore, the centroid is (2, -1).