My Math Forum Mum Needing Help - Find the length of the perpendicular height

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September 18th, 2015, 08:35 AM   #1
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Joined: Sep 2015
From: Dorset

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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!
Attached Images
 triangle.jpg (4.2 KB, 7 views)

Last edited by alethea2000; September 18th, 2015 at 08:45 AM. Reason: to make it clearer

 September 18th, 2015, 08:38 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond I'm not sure what you mean by "perpendicular height U". The diagram is not labelled.
 September 18th, 2015, 08:39 AM #3 Newbie   Joined: Sep 2015 From: Dorset Posts: 4 Thanks: 0 let me see if I can get a clearer picture, sorry
 September 18th, 2015, 08:47 AM #4 Newbie   Joined: Sep 2015 From: Dorset Posts: 4 Thanks: 0 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
 September 18th, 2015, 08:58 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 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.
 September 18th, 2015, 09:01 AM #6 Newbie   Joined: Sep 2015 From: Dorset Posts: 4 Thanks: 0 I need to find a forum where I can understand the answers
 September 18th, 2015, 09:11 AM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Do you know the Pythagorean theorem?
 September 18th, 2015, 11:00 AM #8 Senior Member     Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 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: $\displaystyle AD=\dfrac{AB\cdot AC}{BC}=\dfrac{8\cdot15}{17}=\dfrac{120}{17} \approx 7.058 \approx 7.1cm \\\;\\ So,\ \ y \approx 7.1 cm$ Thanks from greg1313
 September 18th, 2015, 01:42 PM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 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 01:45 PM.
 September 19th, 2015, 08:40 AM #10 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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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