Question

Consider a square $ABCD$ of side $60$ cm. It contains arcs $BD$ and $AC$ drawn with centres at $A$ and $D$ respectively. A circle is drawn such that it is tangent to side $AB$, and the arcs $BD$ and $AC$. What is the radius of the circle?

• A. 9 cm
• B. 10 cm
• C. 12 cm
• D. 15 cm
• E. None of the above

Hint:

In problems involving tangency with arcs of circles drawn from square vertices, always convert the condition into distance equations using Pythagoras. Visualize the distances as $60 \pm r$ depending on whether the circle is inside or outside relative to the arc.

Answer 1

The Correct Option is B

Step 1: Represent the square and arcs.
We are given a square $ABCD$ with side length $60$ cm. - Arc $BD$ is drawn with centre at $A$ and radius $AB=60$ cm. - Arc $AC$ is drawn with centre at $D$ and radius $DC=60$ cm. A circle of radius $r$ is drawn tangent to side $AB$, tangent to arc $BD$, and tangent to arc $AC$. Let $O$ be the centre of the required circle, and $P, Q$ be points of tangency as shown in the diagram.

Step 2: Set up relations.
From the geometry: - Distance $AO = 60 - r$ (since arc centred at $A$ has radius 60). - Distance $DO = 60 + r$ (since arc centred at $D$ has radius 60, but circle of radius $r$ is inside). - Distance $OP = r$. Also, $\triangle AOQ$ and $\triangle DOQ$ will be right-angled due to tangency.

Step 3: Apply Pythagoras in $\triangle AOQ$.
\[ (OQ)^2 = (AO)^2 - (AQ)^2 \] But $AQ = r$, $AO = 60 - r$. So, \[ (OQ)^2 = (60 - r)^2 - r^2 \] \[ (OQ)^2 = 3600 - 120r \]

Step 4: Apply Pythagoras in $\triangle DOQ$.
Here, $DO = 60 + r$, $DQ = 60 - r$. So, \[ (OD)^2 = (DQ)^2 + (OQ)^2 \] \[ (60 + r)^2 = (60 - r)^2 + (3600 - 120r) \] Expand: \[ 3600 + 120r + r^2 = 3600 - 120r + r^2 + 3600 - 120r \] \[ 3600 + 120r + r^2 = 7200 - 240r + r^2 \] Cancel $r^2$: \[ 3600 + 120r = 7200 - 240r \] \[ 360r = 3600 \] \[ r = 10 \; \text{cm} \]

Step 5: Final Answer.
Thus, the radius of the required circle is: \[ \boxed{10 \text{ cm}} \]