If parallelogram ABCD has a perimeter of 12 units and ∠ADC = 120

OFFICIAL EXPLANATION

\(ABCD\) is a parallelogram.

Let \(AB\) and \(CD\) measure \(p\) units and \(AD\) and \(BC\) measure \(q\) units and height of the parallelogram be \(h\) units

Therefore, Area of the parallelogram = \(ph\)



\(2(p + q) = 12\)

\(p + q = 6\)-------------(I)

\(∠ADC = 120^{\circ}\)

This implies \(∠BAD = ∠BCD = 60^{\circ}\)

Since ΔAED is a 30-60-90 triangle, applying the angle-side property (1:√3:2)

\(h=\frac{q√3}{2}\) and \(AE=\frac{q}{2}\)--------------(II)

So, \(Area=\frac{pq√3}{2}\)------------------(III)

Substituting the value of q from (I) in (III)

\(Area=\frac{p(6−p)√3}{2}\)

Thus, in order to find the area, we need to know the value of \(pq\) or of \(p\)

Statement (1) : \(BD = 2√3\) units



In right \(ΔBED\),

\(BE^2+h^2=(2√3)^2\)

\((p−AE)^2+(\frac{q√3}{2})^2=12\)

\((p−\frac{q}{2})^2+(\frac{q√3}{2})^2=12\)

\(p^2+\frac{q^2}{4}−pq+\frac{3q^2}{4}=12\)

\(⇒p^2+q^2−pq=12\)

⇒\((p+q)^2−3pq=12\)

⇒\(6^2−3pq=12\) (Using Equation I)

⇒\(3pq=24\)

⇒\(pq=8\)

⇒\(Area=\frac{8√3}{2}=4√3\)

Since we get a unique value of Area, Statement 1 is sufficient to answer the question.

Statement (2) : \(h= √3\) units

From Equation II, \(q = 2\)

From Equation I, \(p = 4\)

Since we now know the value of \(p\) and \(q\), we can find the value of product \(pq\), and hence of the area of the parallelogram, using Equation III

Hence, Statement 2 is sufficient to answer the question.

Therefore, EACH statement ALONE is sufficient to answer the question asked

Answer is (D)