MGMAT Geometry - p. 119, Question #6
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- Brent@GMATPrepNow
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I added some letters to help guide the solution.
Area of triangle = (1/2)(base)(height)
IMPORTANT CONCEPT: we can use ANY of the three sides as our base.
So, for example, if we want to find the area of triangle ABC, we can use side AB as the base, or we can use side AC as the base, or we can use side BC as the base.
If we use side AB as the base, then the base has length 12 and the height is 3
So, area of triangle ABC = (1/2)(12)(3)
If we use side AC as the base, then the base has length 7 and the height is x
So, area of triangle ABC = (1/2)(7)(x)
IMPORTANT: If we use side AB as the base, the area of the triangle will be the same as the area we get if we use side AC as the base.
So, (1/2)(12)(3) = (1/2)(7)(x) [solve for x]
Divide both sides by 1/2 to get: (12)(3) = (7)(x)
Divide both sides by 7 to get: [spoiler]36/7 = x[/spoiler]
Cheers,
Brent
Last edited by Brent@GMATPrepNow on Thu Apr 19, 2018 1:45 pm, edited 1 time in total.
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As Brent mentioned, any side of a triangle can be deemed the base, and each base has a corresponding height.
More practice with this concept:
https://www.beatthegmat.com/geometry-hel ... 17047.html
https://www.beatthegmat.com/geometry-t271788.html
https://www.beatthegmat.com/need-to-unde ... 68314.html
More practice with this concept:
https://www.beatthegmat.com/geometry-hel ... 17047.html
https://www.beatthegmat.com/geometry-t271788.html
https://www.beatthegmat.com/need-to-unde ... 68314.html
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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- bml1105
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Thanks Brent & GMATGuru!
I think I was getting mixed up because I was trying to find the area of the entire diagram (included the dotted lines) without realizing that I didn't actually need to do that to find x.
I think I was getting mixed up because I was trying to find the area of the entire diagram (included the dotted lines) without realizing that I didn't actually need to do that to find x.
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stuffs88 wrote:Hi,
Can you please explain the concept with similar triangles? Which 2 triangles are they comparing in the book?
Thank You!
∆ABE is similar to ∆ACD
We know this because ∠BAE = ∠CAD, and because ∠AEB = ∠ADC (both are 90º)
If two triangles share 2 equal angles, their 3rd angles must also be equal.
If the triangles share 3 equal angles, they are SIMILAR TRIANGLES, which means the ratios of their respective sides must be equal.
The hypotenuse of ∆ABE has length 12. The hypotenuse of ∆ACD has length 7.
In ∆ABE, the side opposite ∠A has length x. In ∆ACD, the side opposite ∠A has length 3.
So, we can write: 12/7 = x/3
Cross multiply to get: (12)(3) = 7x
Divide both sides by 7 to get: x = [spoiler]36/7[/spoiler]
Cheers,
Brent