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# M31-41

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Math Expert
Joined: 02 Sep 2009
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20 Jun 2015, 11:12
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Difficulty:

55% (hard)

Question Stats:

63% (01:52) correct 37% (01:58) wrong based on 41 sessions

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ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

(1) The height from point B to the hypotenuse is 120.

(2) The perimeter of the triangle is 680.

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20 Jun 2015, 11:12
Official Solution:

ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

First of all since the ration of AB to BC is $$8x:15x$$, then the hypotenuse is $$\sqrt{(8x)^2+(15x)^2}=17x$$.

(1) The height from point B to the hypotenuse is 120.

Consider the image below:

The area of a right triangle equals to $$\frac{1}{2}*leg_1*leg_2=\frac{1}{2}**AB*BC$$. But the area of a right triangle can also be found by $$\frac{1}{2}*(altitude \ from \ right \ angle)*(hypotenuse)=\frac{1}{2}*BD*AC$$.

Equate these expressions:

$$\frac{1}{2}*AB*BC=\frac{1}{2}*BD*AC$$;

$$\frac{1}{2}*8x*15x=\frac{1}{2}*120*17x$$;

$$120x=120*17$$;

$$x=17$$.

The area = $$\frac{1}{2}*AB*BC=\frac{1}{2}*8x*15x=60x=60*17$$. Sufficient.

(2) The perimeter of the triangle is 680.

The perimeter = $$8x+15x+17x=680$$. We can find x, thus we can find the area. Sufficient.

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Joined: 17 Dec 2015
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11 Nov 2016, 21:45
This question could be phrased better in prompt 1) to read the "least distance from B to the hypotenuse". Personally, I interpreted the use of "height" as a descriptor to imply the distance alongside the perimeter (or one of the sides of the triangle rather than the distance measured within the interior). Hence, I selected E for the answer since we didn't know which side it referred to (8x or 15x).
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Joined: 25 Aug 2017
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20 Oct 2017, 15:45
Given statement 1, and the explanation, is it possible to find a different area if the height from point B to the hypotenuse didn't bisect the line into 2 equal parts? I made the assumption that the height could lie on any point of the hypotenuse, not just the center. Am I completely off target here?

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20 Oct 2017, 23:09
Raffio wrote:
Given statement 1, and the explanation, is it possible to find a different area if the height from point B to the hypotenuse didn't bisect the line into 2 equal parts? I made the assumption that the height could lie on any point of the hypotenuse, not just the center. Am I completely off target here?

Hi

How did you assume that the height from the vertex is bisecting the hypotenuse into two equal parts? Thats NOT the case here.
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20 Oct 2017, 23:12
vanceag2016 wrote:
This question could be phrased better in prompt 1) to read the "least distance from B to the hypotenuse". Personally, I interpreted the use of "height" as a descriptor to imply the distance alongside the perimeter (or one of the sides of the triangle rather than the distance measured within the interior). Hence, I selected E for the answer since we didn't know which side it referred to (8x or 15x).

Hi

In a triangle, the word 'height' means perpendicular from any vertex to the opposite side. So, yes two heights here are already AB and BC. But we should know that AB is height from vertex A to opposite side BC, and BC is the height from vertex C to the opposite side AB. The third height will be from vertex B to the opposite side AC, which is what is constructed here in the diagram by Bunuel.
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Joined: 08 Jan 2018
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04 Feb 2018, 12:39
Hi,

a really stupid question, how do we get 17x out of this equation?

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04 Feb 2018, 12:55
MaxKirchmayer wrote:
Hi,

a really stupid question, how do we get 17x out of this equation?

Since the ration of AB to BC is $$8x:15x$$, then the hypotenuse is $$\sqrt{(8x)^2+(15x)^2}=17x$$.
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05 Dec 2018, 19:56
Hi Bunuel,

isnt the height from the from vertex B to hypotenuse, half of the hypotenuse?
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05 Dec 2018, 21:11
yashna36 wrote:
Hi Bunuel,

isnt the height from the from vertex B to hypotenuse, half of the hypotenuse?

Only if a right triangle is also isosceles.

Maybe you are mixing height with median. Because there is a property which says: The median on the hypotenuse of a right triangle always equals to one-half the hypotenuse.
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08 Aug 2019, 22:36
HI Bunuel,

quick question - since the sides are in ratio 8:15. couldn't the sides be 8:15:17 or 16:30:34 etc...either case , the area would be different in each scenario? I chose option B after long process. Please shed some light.
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Joined: 14 Feb 2014
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23 Nov 2019, 02:56
Hi Bunuel,

Can you please explain what happed to the X from step 2 (17x) at the end to step 3 just 17?

Quote:
Equate these expressions:

Step 1/ 12∗AB∗BC=12∗BD∗AC12∗AB∗BC=12∗BD∗AC;

Step 2/ 12∗8x∗15x=12∗120∗17x12∗8x∗15x=12∗120∗17x;

Step 3/ 120x=120∗17120x=120∗17;

Step 4/ x=17x=17.
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23 Nov 2019, 03:01
Rebaz wrote:
Hi Bunuel,

Can you please explain what happed to the X from step 2 (17x) at the end to step 3 just 17?

Quote:
Equate these expressions:

Step 1/ 12∗AB∗BC=12∗BD∗AC12∗AB∗BC=12∗BD∗AC;

Step 2/ 12∗8x∗15x=12∗120∗17x12∗8x∗15x=12∗120∗17x;

Step 3/ 120x=120∗17120x=120∗17;

Step 4/ x=17x=17.

It was reduced. We have 1/2∗8x∗15x = 1/2∗120∗17x. Now, reduce by x to get: 1/2∗8x∗15 = 1/2∗120∗17.
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Re: M31-41   [#permalink] 23 Nov 2019, 03:01
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# M31-41

Moderators: chetan2u, Bunuel