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?
- s≥6
- s≠6
- s≤6
- 5≤s≤7
The Correct Option is A
Solution and Explanation
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:
\[ \boxed{AD \geq 6} \]
Top Questions on Quadrilaterals
Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.
Given: PQRS is a rhombus. Diagonals PR and SQ intersect each other at point T.
To prove: PS\(^2\) + SR\(^2\) + QR\(^2\) + PQ\(^2\) = PR\(^2\) + QS\(^2\)
Activity: Diagonals of a rhombus bisect each other.
In \(\triangle\)PQS, PT is the median and in \(\triangle\)QRS, RT is the median.
\(\therefore\) by Apollonius theorem,
\[\begin{aligned} PQ^2 + PS^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(I)} \\ QR^2 + SR^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(II)} \\ \text{Adding (I) and (II),} \quad PQ^2 + PS^2 + QR^2 + SR^2 &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \\ &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \quad (\text{RT = PT}) \\ &= 4PT^2 + 4QT^2 \\ &= (\boxed{\phantom{X}})^2 + (2QT)^2 \\ \therefore \quad PQ^2 + PS^2 + QR^2 + SR^2 &= PR^2 + \boxed{\phantom{X}} \\ \end{aligned}\]
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