ABCD is a trapezium in which AB is parallel to DC, AD is perpendicular to AB, and $AB = 3DC$. If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm, then the area, in sq. cm, of the trapezium is
Show Hint
The distance between two parallel tangents equals twice the radius.
In tangential quadrilaterals (those with an incircle), using coordinates with the incenter at convenient positions (like \((r, r)\)) can simplify distance calculations.
- \(54\)
- \(30\sqrt{3}\)
- \(48\)
- \(36\sqrt{2}\)
The Correct Option is C
Approach Solution - 1
Let \(ABCD\) be a trapezium such that \(AB \parallel DC\) and \(AD \perp AB\). Let \[ AB = a,\quad DC = c,\quad \text{with } a = 3c, \] and the radius of the incircle be \(r = 3\) cm. Step 1: Use tangency to the parallel sides. Since the incircle touches both parallel sides \(AB\) and \(DC\), the distance between them equals the diameter of the circle. Hence, \[ AD = 2r = 2 \times 3 = 6 \text{ cm}. \] Step 2: Choose a coordinate system. Place the trapezium on the coordinate plane as follows: \[ A(0,0), \quad B(a,0), \quad D(0,6), \quad C(c,6), \] with \(a = 3c\). The circle is tangent to \(AB\), \(DC\), and \(AD\), so its center lies 3 units above \(AB\), 3 units below \(DC\), and 3 units to the right of \(AD\). Therefore, the center of the incircle is \[ O(3,3). \] Step 3: Use the tangency condition with side \(BC\). The line \(BC\) passes through \(B(a,0)\) and \(C(c,6)\). Its slope is \[ \frac{6-0}{c-a} = \frac{6}{c-a}, \] so its equation is \[ y = \frac{6}{c-a}(x-a). \] Rewriting in standard form, \[ 6x - (c-a)y - 6a = 0. \] The perpendicular distance from the center \(O(3,3)\) to line \(BC\) must be equal to the radius \(3\): \[ \frac{|6\cdot 3 - (c-a)\cdot 3 - 6a|}{\sqrt{6^2 + (c-a)^2}} = 3. \] Simplifying the numerator, \[ 18 - 3(c-a) - 6a = 18 - 3c - 3a = 3(6 - c - a). \] Hence, \[ \frac{|6 - c - a|}{\sqrt{36 + (c-a)^2}} = 1. \] Step 4: Substitute the relation \(a = 3c\). Using \(a = 3c\), \[ a + c = 4c, \quad c - a = -2c. \] Thus, \[ \frac{|6 - 4c|}{\sqrt{36 + 4c^2}} = 1. \] Squaring both sides, \[ (6 - 4c)^2 = 36 + 4c^2. \] Expanding and simplifying, \[ 36 - 48c + 16c^2 = 36 + 4c^2, \] \[ 12c^2 - 48c = 0, \] \[ 12c(c - 4) = 0. \] Since \(c > 0\), we get \(c = 4\), and hence \[ a = 3c = 12. \] Step 5: Find the area of the trapezium. The area is given by \[ \text{Area} = \frac{1}{2}(AB + DC)\times \text{height} = \frac{1}{2}(a + c)\times AD = \frac{1}{2}(12 + 4)\times 6 = 48. \] Therefore, the area of the trapezium is \(48\ \text{cm}^2\).
Approach Solution -2
Let the trapezium \(ABCD\) have \(AB \parallel DC\), \(AD \perp AB\), and let \[ AB = a,\quad DC = c,\quad \text{with } a = 3c. \] The radius of the incircle is \(r = 3\) cm.
Step 1: Use the fact that the circle touches both parallel sides. Since the circle is tangent to both \(AB\) and \(DC\), the perpendicular distance between these two parallel lines is equal to the diameter of the circle: \[ \text{height} = AD = 2r = 2 \times 3 = 6 \text{ cm}. \]
Step 2: Set up a coordinate system. Place the trapezium as follows: \[ A = (0,0), \quad B = (a,0), \quad D = (0,6), \quad C = (c,6), \] with \(a = 3c\). The circle touches:
\(AB\) at distance 3 \(\Rightarrow\) center is 3 units above it,
\(DC\) at distance 3 \(\Rightarrow\) center is 3 units below it,
\(AD\) at distance 3 \(\Rightarrow\) center is 3 units to the right of it.
Hence, the center of the circle is \[ O = (3,3). \]
Step 3: Use the distance from the center to side \(BC\). Line \(BC\) passes through \(B(a,0)\) and \(C(c,6)\). The equation of line \(BC\) is: \[ \text{slope} = \frac{6 - 0}{c - a} = \frac{6}{c - a}, \] so \[ y = \frac{6}{c - a}(x - a). \] Rewriting in standard form: \[ (c - a)y = 6x - 6a \quad \Rightarrow \quad 6x - (c - a)y - 6a = 0. \] The perpendicular distance from \(O(3,3)\) to line \(BC\) must be equal to the radius \(3\): \[ \frac{\left|6\cdot 3 - (c - a)\cdot 3 - 6a\right|}{\sqrt{6^2 + (-(c - a))^2}} = 3. \] Simplify the numerator: \[ 6\cdot 3 - 3(c - a) - 6a = 18 - 3c + 3a - 6a = 18 - 3c - 3a = 3(6 - c - a). \] So, \[ \frac{3|6 - c - a|}{\sqrt{36 + (c - a)^2}} = 3 \quad \Rightarrow \quad \frac{|6 - c - a|}{\sqrt{36 + (c - a)^2}} = 1. \]
Step 4: Use the relation \(a = 3c\). Since \(a = 3c\), \[ a + c = 3c + c = 4c, \quad c - a = c - 3c = -2c. \] Then \[ |6 - (a + c)| = |6 - 4c|, \quad \sqrt{36 + (c - a)^2} = \sqrt{36 + (-2c)^2} = \sqrt{36 + 4c^2}. \] From \[ \frac{|6 - 4c|}{\sqrt{36 + 4c^2}} = 1 \quad \Rightarrow \quad (6 - 4c)^2 = 36 + 4c^2. \] Expanding: \[ 36 - 48c + 16c^2 = 36 + 4c^2 \Rightarrow 16c^2 - 48c = 4c^2 \Rightarrow 12c^2 - 48c = 0 \Rightarrow 12c(c - 4) = 0. \] So \(c = 4\) (since \(c>0\)), and hence \[ a = 3c = 12. \]
Step 5: Find the area of the trapezium. The area of a trapezium is \[ \text{Area} = \frac{1}{2}(AB + DC)\cdot \text{height} = \frac{1}{2}(a + c)\cdot AD = \frac{1}{2}(12 + 4)\cdot 6 = \frac{1}{2} \times 16 \times 6 = 48. \] So, the area of the trapezium is \(48\ \text{sq. cm}\).
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