Visual Lens | Coordinate Geometry | MEDIUM
The (x, y) coordinates of vertices P, Q and R of a parallelogram PQRS are (-3, -2), (1, -5) and (9, 1), respectively. If the diagonal SQ intersects the x-axis at (a, 0), then the value of a is
- 10/3
- 27/7
- 13/4
- 29/9
Answer
For a parallelogram P-Q-R-S in that order, S = P + R − Q.
P = (−3, −2), Q = (1, −5), R = (9, 1) ⇒ S = (−3+9−1, −2+1−(−5)) = (5, 4).
Line SQ passes through Q(1, −5) and S(5, 4). Slope = (4 − (−5)) / (5 − 1) = 9/4.
Equation: y + 5 = (9/4)(x − 1). Put y = 0 to find x-intercept: 5 = (9/4)(x − 1).
x − 1 = 20/9 ⇒ x = 1 + 20/9 = 29/9.
Answer: 29/9 (option 4).
Concepts of Rhombus + Algebra
If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is ____
Answer
Let the diagonals be d1 and d2.
Area of a rhombus = 1/2 × d1 × d2 = 396 → d1 × d2 = 792.
In a rhombus, diagonals bisect each other at right angles.
So by Pythagoras: (d1/2)^2 + (d2/2)^2 = 36^2.
Multiply by 4: d1^2 + d2^2 = 4 × 1296 = 5184.
Use identity: (d1 − d2)^2 = (d1^2 + d2^2) − 2(d1 × d2).
Substitute: (d1 − d2)^2 = 5184 − 2×792 = 5184 − 1584 = 3600.
So |d1 − d2| = √3600 = 60.
Final Answer: 60
Concepts of Circles Trigonometry | HARD | CAT 2025 Slot 1
In a circle with center O and radius 6√2 cm, PQ and SR are two parallel chords separated by one of the diameters. If ∠PQC = 45°, and the ratio of the perpendicular distance of PQ and SR from C is 3:2, then the area, in sq cm, of the quadrilateral PQRS is 1. 20(3 + √14) 2. 4(3√3 + 14) 3. 20(√2 + √7) 4. 4(3√2 + 7)
Answer
Radius R = 6√2. Assume ∠POQ = 90° so PQ subtends 90° at center.
For PQ: d1 = OM = R cos(45°) = 6√2 · 1/√2 = 6.
PQ = 2R sin(45°) = 2·6√2·1/√2 = 12.
Given d1:d2 = 3:2, so d2 = 6·(2/3) = 4.
For SR: SN = sqrt(R^2 − d2^2) = sqrt((6√2)^2 − 4^2) = sqrt(72 − 16) = sqrt(56) = 2√14.
Hence SR = 2·SN = 4√14.
The chords PQ and SR are parallel, so PQRS is a trapezium with parallel sides PQ and SR.
Height h = d1 + d2 = 6 + 4 = 10.
Area = 1/2 · (PQ + SR) · h = 1/2 · (12 + 4√14) · 10.
Area = 5 · (12 + 4√14) = 60 + 20√14.
Therefore area = 20(3 + √14), which is Option A.
