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प्रश्न
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Jagdish has a field which is in the shape of a right-angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O.
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Based on the above information, answer the following questions.
- Taking O as the origin, the coordinates of P are (–200, 0) and that of Q are (200, 0). PQRS being a square, what are the coordinates of R and S?
- What is the area of square PQRS?
- What is the length of diagonal PR in square PQRS?
- If S divides CA in the ratio k : 1, what is the value of k, where point (A) is (200, 800)?
- What are the coordinates of C ?
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उत्तर
i. Coordinates of points R and S:
Given:
Coordinates of \(P = (-200, 0)\)
Coordinates of \(Q = (200, 0)\)
Side length of square PQRS: \[\text{Side } PQ = |x_Q - x_P|\]
\[= 200 - (-200)\]
\[= 400\text{ units}\]
Since PQRS is a square with its base PQ on the x-axis:
\[QR = PS = 400\text{ units}\]
Point R lies vertically above Q(200, 0):
\[x_R = 200, \quad y_R = 0 + 400 = 400\]
\[\implies R(200, 400)\]
Point S lies vertically above \(P(-200, 0)\):
\[x_S = -200, \quad y_S = 0 + 400 = 400\]
\[\implies S(-200, 400)\]
\[R(200, 400), \quad S(-200, 400)\]
ii. Area of square PQRS:
Formula: \[\text{Area} = (\text{side})^2\]
Calculation: \[\text{Area} = (400)^2\]
\[160000\text{ sq units}\]
iii. Length of diagonal PR:
Formula: In a square of side a, diagonal \(d = a\sqrt{2}\)
Calculation: \[PR = \sqrt{(200 - (-200))^2 + (400 - 0))^2}\]
\[PR = \sqrt{(400)^2 + (400)^2}\]
\[PR = \sqrt{160000 + 160000}\]
\[= \sqrt{320000}\]
\[= 400\sqrt{2}\text{ units}\]
iv. Value of (k):
Given:
Point C lies on the x-axis, so its coordinates are \((-x, 0)\).
Point \(A = (200, 800)\).
Point \(S(-200, 400)\) divides the line segment CA in the ratio k : 1.
Section Formula for the y-coordinate: \[y = \frac{k y_2 + 1 \cdot y_1}{k + 1}\]
\[400 = \frac{k(800) + 1(0)}{k + 1}\]
\[400(k + 1) = 800k\]
\[400k + 400 = 800k\]
\[400k = 400\]
\[\implies k = 1\]
v. Coordinates of point C:
Since \(k = 1\), \(S(-200, 400)\) is the midpoint of CA.
Midpoint Formula for (x)-coordinate: \[x_S = \frac{x_C + x_A}{2}\]
\[-200 = \frac{-x + 200}{2}\]
\[-400 = -x + 200\]
\[-x = -600\]
\[\implies x = 600\]
Since point C lies on the negative x-axis, its coordinates are (\(-600, 0)\).
