Question

A parallelogram ABCD has area 48 sqcm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?

• A. s≥6
• B. s≠6
• C. s≤6
• D. 5≤s≤7

Answer 1

The Correct Option is A

Step 1: Area of Parallelogram and Triangle 

We are given that the area of parallelogram \( ABCD \) is 48. It is also given that:

\[ \text{Area of parallelogram } ABCD = 2 \times \text{Area of } \triangle ACD \Rightarrow 48 = 2 \times \text{Area of } \triangle ACD \]

\[ \Rightarrow \text{Area of } \triangle ACD = 24 \]

Step 2: Area of Triangle ACD Using Trigonometry

The area of triangle \( ACD \) using sine formula is:

\[ \text{Area} = \frac{1}{2} \cdot CD \cdot AD \cdot \sin(\angle ADC) \]

Given area = 24:

\[ \frac{1}{2} \cdot CD \cdot AD \cdot \sin(\angle ADC) = 24 \]

Assume \( CD = 2 \) (without loss of generality, since we are solving for minimum \( AD \)), then:

\[ CD \cdot AD \cdot \sin(\angle ADC) = 48 \Rightarrow AD \cdot \sin(\angle ADC) = 6 \tag{1} \]

Step 3: Use the Property of Sine Function

We know that \( \sin(\theta) \leq 1 \). To minimize \( AD \), we want \( \sin(\angle ADC) \) to be as large as possible. Maximum value of \( \sin(\angle ADC) = 1 \), so:

\[ AD \cdot 1 = 6 \Rightarrow AD = 6 \]

If \( \sin(\angle ADC) < 1 \), then \( AD > 6 \).\ Hence, the minimum possible value of \( AD \) is:

\[ \boxed{AD \geq 6} \]