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24 Mar 2020, 09:41
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
90% (01:48) correct
10%
(02:33)
wrong
based on 20
sessions
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Time
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Not Attempted Yet
Attachment:
7654.jpg [ 11.87 KiB | Viewed 2213 times ]
What is the area of polygon ABCDE shown above?
(A) \(4 + 2\sqrt{3}\)
(B) \(3 + 3\sqrt{2}\)
(C) \(6\sqrt{3}\)
(D) \(2 + 6\sqrt{2}\)
(E) \(8\sqrt{2}\)
25 Mar 2020, 23:50
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SajjadAhmad
Attachment:
7654.jpg
What is the area of polygon ABCDE shown above?
(A) \(4 + 2\sqrt{3}\)
(B) \(3 + 3\sqrt{2}\)
(C) \(6\sqrt{3}\)
(D) \(2 + 6\sqrt{2}\)
(E) \(8\sqrt{2}\)
Solution:
We can divide this polygon in three triangles:
- • Triangle ABE, Triangle EBD, and Triangle CBD.
- o Area of polygon = area of triangle ABE+ area of triangle CBD + area of triangle EBD.
- o \(AB = AE = 2\)
- ABE is a 45- 45-90 triangle
- o Area of triangle ABE = \(\frac{1 }{2} *AB *AE = \frac{1}{ 2}*2*2 = 2\) units.
- o \(CB = CD = 2\)
- CBD is a 45-45-90 triangle
- o Area of triangle CBD = \(\frac{1}{ 2} *CB*CD = \frac{1}{ 2}*2*2 = 2\) units
- o ED = \(2 \sqrt{2}\)
- BE = BD = \(2 \sqrt{2}\) [From (i) and (ii)]
All the sides of EBD are equal.
EBD is an equilateral triangle
• Area of polygon = \(2 + 2 + 2 \sqrt{3} = 4 + 2 \sqrt{3}\)
26 Mar 2020, 00:18
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This Pentagon may be divided into three triangles. We see right angles at A and C; we also see that AB=AE and BC = CD. This is a clear clue that we are dealing with isosceles right-angled triangles.
However, we will have to create the triangle by joining B to E and D respectively. When we do this, we see that BE and BD represent the hypotenuse of triangles ABE and BCD respectively. The figure looks like the one shown below:
26th Mar 2020 - Reply 1.jpg [ 24.51 KiB | Viewed 2041 times ]
Since the triangles are isosceles right-angled triangles, the length of the hypotenuse = √2*side. Therefore, BE = BD = 2 √2.
This means triangle BDE is an equilateral triangle.
So, the area of the pentagon = Sum of areas of triangles BDE, BCD and ABE.
Area of triangle BCD = Area of triangle ABE = ½ * 2*2 = 2 sq.units.
Area of triangle BDE = \(\frac{√3}{4}\)*(2√2)^2 = \(\frac{√3}{4}\)*8 = 2√3 sq.units.
Therefore, area of the pentagon = 2+2+2√3 = 4+2√3 sq.units.
The correct answer option is A.
In a question on finding the area of a polygon like a pentagon or a hexagon, the idea is to divide the given figure into a combination of triangles/quadrilaterals since we know the area formulas for these figures. Look out for clues that will help you do this. In this problem, the clues were the right angles and the sides being equal.
Hope that helps!
However, we will have to create the triangle by joining B to E and D respectively. When we do this, we see that BE and BD represent the hypotenuse of triangles ABE and BCD respectively. The figure looks like the one shown below:
Attachment:
26th Mar 2020 - Reply 1.jpg [ 24.51 KiB | Viewed 2041 times ]
Since the triangles are isosceles right-angled triangles, the length of the hypotenuse = √2*side. Therefore, BE = BD = 2 √2.
This means triangle BDE is an equilateral triangle.
So, the area of the pentagon = Sum of areas of triangles BDE, BCD and ABE.
Area of triangle BCD = Area of triangle ABE = ½ * 2*2 = 2 sq.units.
Area of triangle BDE = \(\frac{√3}{4}\)*(2√2)^2 = \(\frac{√3}{4}\)*8 = 2√3 sq.units.
Therefore, area of the pentagon = 2+2+2√3 = 4+2√3 sq.units.
The correct answer option is A.
In a question on finding the area of a polygon like a pentagon or a hexagon, the idea is to divide the given figure into a combination of triangles/quadrilaterals since we know the area formulas for these figures. Look out for clues that will help you do this. In this problem, the clues were the right angles and the sides being equal.
Hope that helps!
