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September 18th, 2015, 09:35 AM   #1
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Mum Needing Help - Find the length of the perpendicular height

Good evening all

My daughter is only 11 and wants my help with homework but I am afraid I'm struggling. I have tried to google but I still do not understand. Rather than getting her to speak to her teacher I want to try and learn myself so I can help her.

This is an example of what she needs to find out. I am not using the figures in her homework so she can work it out herself.

Untitled.jpg

So we have one big triangle with a triangle inside the triangle like in this picture.

The left side of the large triange is 15cm
The base is 8 cm
The right side of the triangle is 17cm
The black line inside the triangle is y (isn't very clear but it goes from the bottom right side of big triangle and across to the other side

The question is find the length of the perpendicular height U in this triangle, write your answer to one decimal place

I don't know if it matters but the top of the mini triangle is a right angle

I hope you don't mind me posting to this forum. I use Mr Excel forum all the time so I am hoping this is the same type of thing

Thank you
From struggling mother!
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Last edited by alethea2000; September 18th, 2015 at 09:45 AM. Reason: to make it clearer
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September 18th, 2015, 09:38 AM   #2
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I'm not sure what you mean by "perpendicular height U". The diagram is not labelled.
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September 18th, 2015, 09:39 AM   #3
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let me see if I can get a clearer picture, sorry
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September 18th, 2015, 09:47 AM   #4
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is the picture above in the post any better?

that is the question, find the length of the perpendicular height y in this triangle. I don't know if it means find the length of y or find out how tall the smaller triangle is
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September 18th, 2015, 09:58 AM   #5
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You may set up two equations:

$\displaystyle a+b=17$ and $\displaystyle 225-a^2=64-b^2$, where $\displaystyle a$ and $\displaystyle b$ are the segments of the hypotenuse of the large triangle, as divided by y.
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September 18th, 2015, 10:01 AM   #6
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I need to find a forum where I can understand the answers
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September 18th, 2015, 10:11 AM   #7
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Do you know the Pythagorean theorem?
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September 18th, 2015, 12:00 PM   #8
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So we have one triangle ABC,

$\displaystyle m(\hat{A})=90^0, AB=8cm, AC=15cm, BC=17cm
\\\;\\
If \ AD\perp BC,\ D\in BC\ and\ AD=y,$

we must find y.

We use formula:

$\displaystyle AD=\dfrac{AB\cdot AC}{BC}=\dfrac{8\cdot15}{17}=\dfrac{120}{17} \approx 7.058 \approx 7.1cm
\\\;\\
So,\ \ y \approx 7.1 cm$
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September 18th, 2015, 02:42 PM   #9
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The formula aurel5 gave may be derived using the formula for the area of a triangle: the area of a triangle is one half of the base times the height - for the given triangle, the formula for area may be written in two ways. Setting them equal gives

$\displaystyle \dfrac{17}{2}y=\dfrac{15\cdot8}{2}$

Solving for $\displaystyle y$: $\displaystyle y=\dfrac{120}{17}\approx7.1$

My apologies for making this problem more complicated that it needed to be.

Last edited by greg1313; September 18th, 2015 at 02:45 PM.
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September 19th, 2015, 09:40 AM   #10
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The crucial point is that, since the non-right angles in a right triangle are complementary, the "large" triangle and two interior triangles are similar triangles- they have the same angles and the various sides are in the same proportion.

In particular, the ratio of shorter leg to longer is 8/15. The hypotenuse is divided by that perpendicular into two parts- call the longer "y" so the shorter is 17- y.

Equating the ratios of the smaller leg to longer in all three triangles, (17- y)x= x/y= 8/15.

From x/y= 8/15, 15x= 8y. From (17- y)/x= 8/15, 15(17- y)= 8x.

From 15x= 8y, y= (15/8 )x. Putting that into the second equation 15(17- (5/8 )x)= 255- (75/8 )x= 8x. Solve that equation for x.
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