CBSE Board Exam 2026 | Q.P. Code 30/8/1 (Set-1) | 4 Marks (Case Study)
Ramesh was fond of spinning/revolving a ball tied with a rope of length $2\text{ m}$. While playing, the ball slipped from the rope and hit his brother Suresh lightly. Suresh threw the ball back to Ramesh that made an angle of $60^{\circ}$ with AX along the path XYZ which is a tangent to the imaginary circle as shown in fig. 1. i.e. $\angle AXZ=60^{\circ}$.
(i) If the length of rope used by Ramesh is $2\text{ m}$, what is the diameter of the circle?
(ii) Find the length XY shown in figure.
(iii) Ramesh made the same circle on the ground using the rope he had and placed 4 sticks around it forming a circumscribing quadrilateral ABCD with $AB = 3\text{ m}$, $AD = BC = 5\text{ m}$. Find length of CD. Justify your answer.
(i) If the length of rope used by Ramesh is $2\text{ m}$, what is the diameter of the circle?
(ii) Find the length XY shown in figure.
(iii) Ramesh made the same circle on the ground using the rope he had and placed 4 sticks around it forming a circumscribing quadrilateral ABCD with $AB = 3\text{ m}$, $AD = BC = 5\text{ m}$. Find length of CD. Justify your answer.
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1
Tangent-Radius Theorem: The tangent line at any point of a circle is perpendicular to the radius. This makes $\Delta OYX$ a perfectly right-angled triangle ($\angle OYX = 90^{\circ}$).
2
Trigonometric Ratios: In right $\Delta OYX$, the tangent ratio relates the opposite side (Radius OY) to the adjacent side (Tangent length XY) via the given angle ($\angle OXY = 60^{\circ}$).
3
Circumscribed Quadrilateral Property: When a quadrilateral circumscribes a circle, the sum of the lengths of its opposite sides is always equal ($AB + CD = AD + BC$).
1
Part (i): The rope length represents the radius $r=2$. Diameter is simply $2 \times r$.
2
Part (ii): Look at $\Delta OYX$. It's a right triangle where OY is the perpendicular and XY is the base. You know $OY=2$ and angle at X is $60^{\circ}$. Apply $\tan(60^{\circ}) = \frac{\text{Perpendicular}}{\text{Base}}$.
3
Substitute $\tan(60^{\circ}) = \sqrt{3}$ and solve for XY.
4
Part (iii): Use the property $AB + CD = AD + BC$. Plug in the given values: $3 + CD = 5 + 5$. Solve for CD!
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Wrong Trig Ratio: A common error in Part (ii) is using $\sin(60^{\circ})$ or $\cos(60^{\circ})$ instead of $\tan(60^{\circ})$. Remember, you have the Perpendicular (Radius OY) and you need the Base (Tangent XY). The Hypotenuse (Line OX) is irrelevant here, making $\tan$ the only correct choice.
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The "Top+Bottom = Left+Right" Hack: For Part (iii), any shape wrapping a circle acts like a balanced scale. The sum of the vertical sides equals the sum of the horizontal sides.
Left + Right = $5 + 5 = 10$.
Top + Bottom must also be 10.
Since Bottom (AB) is 3, Top (CD) MUST be $10 - 3 = 7\text{ m}$. Done mentally in 2 seconds!
Left + Right = $5 + 5 = 10$.
Top + Bottom must also be 10.
Since Bottom (AB) is 3, Top (CD) MUST be $10 - 3 = 7\text{ m}$. Done mentally in 2 seconds!
Topper's Answer Sheet
Part (i): Diameter of the circle
The length of the rope acts as the radius of the circle.
$\therefore$ Radius $r = 2\text{ m}$.
Diameter $d = 2 \times r = 2 \times 2$
Diameter = $4\text{ m}$ [1 Mark]
Part (ii): Length of XY Path XYZ is tangent to the circle at point Y. Radius OY meets the tangent at Y. $\therefore \angle OYX = 90^{\circ}$ (Tangent $\perp$ radius)
In right-angled $\Delta OYX$, we are given $\angle OXY = 60^{\circ}$ and $OY = 2\text{ m}$. $\tan(\angle OXY) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{OY}{XY}$ $\tan(60^{\circ}) = \frac{2}{XY}$ $\sqrt{3} = \frac{2}{XY}$
Rearranging for XY: $XY = \frac{2}{\sqrt{3}}$ Rationalizing the denominator: $XY = \frac{2\sqrt{3}}{3}\text{ m}$ [1 Mark]
Part (iii): Length of CD ABCD is a quadrilateral circumscribing the circle. Justification: We know that for a circumscribed quadrilateral, the sum of opposite sides is equal. This happens because tangents drawn from external points to the circle are equal.
$\therefore AB + CD = AD + BC$ [1 Mark]
Substituting the given values ($AB=3$, $AD=5$, $BC=5$): $3 + CD = 5 + 5$ $3 + CD = 10$ $CD = 10 - 3$
$CD = 7\text{ m}$ [1 Mark]
Part (ii): Length of XY Path XYZ is tangent to the circle at point Y. Radius OY meets the tangent at Y. $\therefore \angle OYX = 90^{\circ}$ (Tangent $\perp$ radius)
In right-angled $\Delta OYX$, we are given $\angle OXY = 60^{\circ}$ and $OY = 2\text{ m}$. $\tan(\angle OXY) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{OY}{XY}$ $\tan(60^{\circ}) = \frac{2}{XY}$ $\sqrt{3} = \frac{2}{XY}$
Rearranging for XY: $XY = \frac{2}{\sqrt{3}}$ Rationalizing the denominator: $XY = \frac{2\sqrt{3}}{3}\text{ m}$ [1 Mark]
Part (iii): Length of CD ABCD is a quadrilateral circumscribing the circle. Justification: We know that for a circumscribed quadrilateral, the sum of opposite sides is equal. This happens because tangents drawn from external points to the circle are equal.
$\therefore AB + CD = AD + BC$ [1 Mark]
Substituting the given values ($AB=3$, $AD=5$, $BC=5$): $3 + CD = 5 + 5$ $3 + CD = 10$ $CD = 10 - 3$
$CD = 7\text{ m}$ [1 Mark]
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