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In the figure above, BC || AD. What is the area of quadrilat
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24 Mar 2020, 00:10
24 Mar 2020, 00:10
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In the figure above, BC || AD. What is the area of quadrilateral ABCD ?
(A) 5\(\sqrt{2}\)
(B) [9\(\sqrt{3}\)]/2
(C) [27\(\sqrt{3}\)]/4
(D) \(\frac{27}{2}\)
(E) 16
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In the figure above, BC || AD. What is the area of quadrilat
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12 Feb 2021, 06:21
12 Feb 2021, 06:21
2
Quadrilateral ABCD is a trapezium, where AD is parallel to BC, BC = 5 and AD = 4
Drop a perpendicular from D to line segment BC (Let us name this point as E)
Now, in triangle DEC;
Angle DAC = 90
Angle EDC = 30 .......... (120 - 90)
Angle ECD = 60
Also, triangle DEC is a 30-60-90 triangle. Ratio of sides in 30-60-90 triangle is \(x : \sqrt{3}x : 2x\)
i.e.
opposite to 30 = \(x\)
opposite to 60 = \(\sqrt{3}x\)
opposite to 90 = \(2x\)
Opposite to 90 is side CD = 3
So, \(2x = 3\)
\(x = \frac{3}{2}\)
Therefore, CE = \(\frac{3}{2}\) and DE = \(\frac{3\sqrt{3}}{2}\)
Area of Trapezium = \(\frac{1}{2}\) x (sum of the parallel sides) x Height
= \(\frac{1}{2}\) x (5 + 4) x \(\frac{3\sqrt{3}}{2}\)
= \(\frac{27\sqrt{3}}{4}\)
Hence, option C
Drop a perpendicular from D to line segment BC (Let us name this point as E)
Now, in triangle DEC;
Angle DAC = 90
Angle EDC = 30 .......... (120 - 90)
Angle ECD = 60
Also, triangle DEC is a 30-60-90 triangle. Ratio of sides in 30-60-90 triangle is \(x : \sqrt{3}x : 2x\)
i.e.
opposite to 30 = \(x\)
opposite to 60 = \(\sqrt{3}x\)
opposite to 90 = \(2x\)
Opposite to 90 is side CD = 3
So, \(2x = 3\)
\(x = \frac{3}{2}\)
Therefore, CE = \(\frac{3}{2}\) and DE = \(\frac{3\sqrt{3}}{2}\)
Area of Trapezium = \(\frac{1}{2}\) x (sum of the parallel sides) x Height
= \(\frac{1}{2}\) x (5 + 4) x \(\frac{3\sqrt{3}}{2}\)
= \(\frac{27\sqrt{3}}{4}\)
Hence, option C
Attachments
Quadrilateral ABCD is a trapezium, where AD is parallel to BC, BC = 5 and AD = 4.png [ 19.19 KiB | Viewed 2021 times ]
General Discussion
Senior Manager
Joined: 23 Jan 2021
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Concentration: , International Business
Re: In the figure above, BC || AD. What is the area of quadrilat
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05 Jun 2021, 18:16
05 Jun 2021, 18:16
KarunMendiratta wrote:
Quadrilateral ABCD is a trapezium, where AD is parallel to BC, BC = 5 and AD = 4
Drop a perpendicular from D to line segment BC (Let us name this point as E)
Now, in triangle DEC;
Angle DAC = 90
Angle EDC = 30 .......... (120 - 90)
Angle ECD = 60
Also, triangle DEC is a 30-60-90 triangle. Ratio of sides in 30-60-90 triangle is \(x : \sqrt{3}x : 2x\)
i.e.
opposite to 30 = \(x\)
opposite to 60 = \(\sqrt{3}x\)
opposite to 90 = \(2x\)
Opposite to 90 is side CD = 3
So, \(2x = 3\)
\(x = \frac{3}{2}\)
Therefore, CE = \(\frac{3}{2}\) and DE = \(\frac{3\sqrt{3}}{2}\)
Area of Trapezium = \(\frac{1}{2}\) x (sum of the parallel sides) x Height
= \(\frac{1}{2}\) x (5 + 4) x \(\frac{3\sqrt{3}}{2}\)
= \(\frac{27\sqrt{3}}{4}\)
Hence, option C
Drop a perpendicular from D to line segment BC (Let us name this point as E)
Now, in triangle DEC;
Angle DAC = 90
Angle EDC = 30 .......... (120 - 90)
Angle ECD = 60
Also, triangle DEC is a 30-60-90 triangle. Ratio of sides in 30-60-90 triangle is \(x : \sqrt{3}x : 2x\)
i.e.
opposite to 30 = \(x\)
opposite to 60 = \(\sqrt{3}x\)
opposite to 90 = \(2x\)
Opposite to 90 is side CD = 3
So, \(2x = 3\)
\(x = \frac{3}{2}\)
Therefore, CE = \(\frac{3}{2}\) and DE = \(\frac{3\sqrt{3}}{2}\)
Area of Trapezium = \(\frac{1}{2}\) x (sum of the parallel sides) x Height
= \(\frac{1}{2}\) x (5 + 4) x \(\frac{3\sqrt{3}}{2}\)
= \(\frac{27\sqrt{3}}{4}\)
Hence, option C
Sir, I do understand in calculating height of the trapezoid. Would you mind to add figure , Please. TIA
