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30 Jun 2019, 12:36
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If parallelogram ABCD has a perimeter of 12 units and \(∠ADC = 120^{\circ}\), what is the area of the parallelogram?
(1) \(BD = 2√3\) units
(2) The height of the parallelogram is \(√3\) units
(1) \(BD = 2√3\) units
(2) The height of the parallelogram is \(√3\) units
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
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)
\(ABCD\) 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)
Archit3110
Major Poster
02 Jul 2019, 09:28
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firas92
#1
we can say that the llogram has 30:60:90 ∆ so height = √3 and base ; 3 ; perimeter given is 12 ;
area ; 1/2*3√3*2 ; 3√3
sufficient
#2
h= √3 so other side can be determined ; side 3
sufficient
IMO D
