Question

The figure formed by joining the mid points of the adjacent sides of a rectangle is :

• A. Square
• B. Rhombus
• C. Trapezium
• D. None of these

Hint:

Key theorem: The quadrilateral formed by joining the midpoints of the sides of any quadrilateral is a parallelogram (Varignon's Theorem).
For a general quadrilateral \(\rightarrow\) Parallelogram
For a rectangle \(\rightarrow\) {Rhombus} (because diagonals of a rectangle are equal, making all sides of the midpoint figure equal).
For a rhombus \(\rightarrow\) Rectangle (because diagonals of a rhombus are perpendicular).
For a square \(\rightarrow\) Square.

Answer 1

The Correct Option is B

Concept: This question refers to a well-known geometrical theorem called Varignon's Theorem, or a specific case of it. Varignon's Theorem states that the figure formed by joining the midpoints of the sides of any quadrilateral is a parallelogram. We need to determine the specific type of parallelogram formed when the original quadrilateral is a rectangle. Step 1: Properties of a Rectangle A rectangle is a parallelogram with four right angles. Its diagonals are equal in length and bisect each other. Step 2: Consider a rectangle ABCD Let P, Q, R, and S be the midpoints of sides AB, BC, CD, and DA respectively. We need to determine the type of quadrilateral PQRS. Step 3: Using the Midpoint Theorem In \(\triangle ABD\), S is the midpoint of AD and P is the midpoint of AB. By the Midpoint Theorem, SP is parallel to DB and \(SP = \frac{1}{2}DB\). Similarly, in \(\triangle CBD\), R is the midpoint of CD and Q is the midpoint of BC. By the Midpoint Theorem, RQ is parallel to DB and \(RQ = \frac{1}{2}DB\). Thus, SP is parallel to RQ and \(SP = RQ\). Also, in \(\triangle ABC\), P is the midpoint of AB and Q is the midpoint of BC. By the Midpoint Theorem, PQ is parallel to AC and \(PQ = \frac{1}{2}AC\). Similarly, in \(\triangle ADC\), S is the midpoint of AD and R is the midpoint of CD. By the Midpoint Theorem, SR is parallel to AC and \(SR = \frac{1}{2}AC\). Thus, PQ is parallel to SR and \(PQ = SR\). Since opposite sides are equal and parallel (SP=RQ and PQ=SR), PQRS is a parallelogram. Step 4: Properties of the sides of parallelogram PQRS We have \(SP = RQ = \frac{1}{2}DB\) and \(PQ = SR = \frac{1}{2}AC\). In a rectangle, the diagonals are equal, i.e., \(AC = DB\). Therefore, \(SP = PQ = QR = RS = \frac{1}{2} \times (\text{length of diagonal of rectangle})\). Since all four sides of the parallelogram PQRS are equal, PQRS is a rhombus. Step 5: When would it be a square? A rhombus becomes a square if its angles are \(90^\circ\). The angles of the rhombus PQRS would be \(90^\circ\) if the diagonals of the original rectangle (AC and DB) are perpendicular. However, the diagonals of a general rectangle are not necessarily perpendicular (they are only perpendicular if the rectangle is a square). So, for a general rectangle, the figure formed is a rhombus. If the original rectangle is a square, then the figure formed by joining midpoints is also a square (which is a special type of rhombus). But the general case for a rectangle is a rhombus.