MGMAT Geometry - p. 119, Question #6

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

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