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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
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In the figure, what is the area of triangle BCD given
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10 Jan 2019, 12:59
10 Jan 2019, 12:59
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In the figure above, the area of triangle ADE is 16 and the area of triangle CDE is 12. What is the area of triangle BCD?
Answer:
Show: ::
28
Re: In the figure, what is the area of triangle BCD given
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10 Jan 2019, 19:44
10 Jan 2019, 19:44
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GreenlightTestPrep wrote:
In the figure above, the area of triangle ADE is 16 and the area of triangle CDE is 12. What is the area of triangle BCD?
Answer:
Show: ::
20
If we add ADE and CDE, we get triangle ADC, which has an area of 28, which is the sum of the areas of ADE and CDE, 16+12.
Now if we compare triangle ACD with triangle BCD, the base is CD, common to both AND the height is also the same as both triangles between two parallel lines, AB and CD.
With base and height the same, the area of the two triangles will be the same, so area of BCD is also 28.
Brent, could you please check the OA as it is given 20
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: In the figure, what is the area of triangle BCD given
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11 Jan 2019, 09:49
11 Jan 2019, 09:49
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GreenlightTestPrep wrote:
In the figure above, the area of triangle ADE is 16 and the area of triangle CDE is 12. What is the area of triangle BCD?
Answer:
Show: ::
28
∆ADC is comprised of 2 regions: ∆ADE and ∆CDE
We're told that ∆ADE and ∆CDE have areas 16 ans 12 respectively.
So, the area of ∆ADC = 16 + 12 = 28
First recognize that ∆ADC and ∆BCD both have the same base.
Also, if we let h = the height of ∆ADC, then ∆BCD also has height h
area of triangle = (base)(height)/2
So, if ∆ADC and ∆BCD have the same base and the same height, then they must have the same area.
So, if the area of ∆ADC is 28, then the area of ∆BCD must also be 28
Answer: 28
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