Question

Four tangents drawn to a circle are extended from both the sides to form a quadrilateral. Which of these quadrilateral is not possible ?

• A. Trapezium
• B. Square
• C. Rectangle
• D. Rhombus

Hint:

A circle can be inscribed inside a quadrilateral if and only if the sum of opposite sides is equal.
For a rectangle of sides \(a\) and \(b\), \(a+a = 2a\) and \(b+b = 2b\). These are equal only when \(a = b\) (which makes it a square).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks which of the given quadrilaterals cannot circumscribe a circle (i.e., be formed by four tangent lines to a circle).

Step 2: Key Formula or Approach:
For any quadrilateral circumscribing a circle (a tangential quadrilateral), the sum of the lengths of the opposite sides must be equal. This is known as Pitot's Theorem:
\[ AB + CD = AD + BC \]

Step 3: Detailed Explanation:

• Let us analyze each given quadrilateral type:
- Square: All four sides are equal (\(a\)). Opposite sides sum to \(a + a = 2a\), which is equal to the other pair \(a + a = 2a\). Thus, a square can always circumscribe a circle.
- Rhombus: All four sides are equal (\(a\)). By the same logic as the square, a rhombus can always circumscribe a circle.
- Trapezium: Some trapeziums (specifically isosceles trapeziums where the sum of the non-parallel sides equals the sum of the parallel bases) can circumscribe a circle.
- Rectangle: Let a non-square rectangle have adjacent sides of lengths \(w\) and \(l\) where \(w \neq l\). The sum of one pair of opposite sides is \(w + w = 2w\) and the other is \(l + l = 2l\). Since \(w \neq l\), \(2w \neq 2l\).

• Since Pitot's theorem is violated for any general rectangle (that is not a square), a rectangle cannot circumscribe a circle.


Step 4: Final Answer:
A rectangle is not a possible tangential quadrilateral unless it is a square.