How did we know to just multiply the 18 and the 8 by 0.5 and 6 and be done?
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Module 2 Week 3 Day 12 Your Turn Question
Uhhhhhhhh... answer was never really stated. How did we know to just multiply the 18 and the 8 by 0.5 and 6 and be done?
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@TSS-Graviser I think you are asking about the Your Turn question, am I right? (Not the mini-question at the end.) How about we go over the thing one more time, from beginning to end, with the help of some little diagrams and our Expii colors?
We are trying to find the area of this tangential quadrilateral (note that this is different from a cyclic quadrilateral, which is inside a circle). All we know about this shape is that two opposite sides have length \(8\) and \(18,\) and the radius of the circle is \(6.\)
Let's color these four triangles like so. Since the radii are altitudes of each triangle, all we have to do in order to find the area of the entire thing is to find out the sum of the bases of all the triangles.
$$ \text{ Area of triangle } = \frac{1}{2} \times \text{ base } \times \text{ height} $$
$$ \text{ Area of whole thing } = \text{ Area of } \left( \textcolor{green}{\text{green triangle}} + \textcolor{purple}{\text{ purple triangle }} + \textcolor{blue}{\text{ blue triangle }} + \textcolor{orange}{\text{ orange triangle }} \right) $$$$ \text{ Area of whole thing } = \frac{1}{2} \times 6 \times \left( \textcolor{green}{\text{green base}} + \textcolor{purple}{\text{ purple base }} + \textcolor{blue}{\text{ blue base }} + \textcolor{orange}{\text{ orange base }} \right) $$
Note that we only need to know the sum of the bases, not necessarily the length of each base!
Now, let's form kites by drawing all the altitudes to each side. We can use the fact that kites have two pairs of sides the same length to show that both of these \( \textcolor{red}{\text{ red segments}}\) have the same length (let's call it \(a.\))
Similarly, let's call the remaining segment of the left side \(b,\) and we know that there is a matching segment of length \(b\) on the other side of the bottom-left kite:
This leaves us with a segment of length \(18 -b,\) but we know that there is another matching segment of length \(18-b\) on the other side, leaving us with the last section of length \(8-a.\)
Taking a look at just these two green triangles, we notice that the sum of the green bases is equal to
$$ a + b + ( 8-a) + (18-b) = 8 + 18 $$
$$= 26$$
This is exactly the same as the sum of the bases of the two other triangles, which we'll color in purple.
In fact, if we remember that this fact is true for tangential quadrilaterals, then we can do the entire question in just one step: we know that two opposite bases sum to \(26,\) so the other two opposite bases sum to \(26,\) and the area of the entire quadrilateral is just
$$ \text{ Area of whole thing } = \frac{1}{2} \times 6 \times \left( \textcolor{green}{\text{green bases}} + \textcolor{purple}{\text{ purple bases }} \right) $$
$$ = \frac{1}{2} \times 6 \times (26 + 26) $$
$$ = 6 \times 26 $$
$$ = \boxed{156} $$
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uhhh I think it was the mini question sorry
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@TSS-Graviser Are you talking about the mini-question from Day 12, Your Turn Part 2?
Your original question was, "How did we know to just multiply the 18 and the 8 by 0.5 and 6 and be done?" Could you please explain a little more about what you are wondering about?
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Oh wait I'm really confused never mind sorry
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@TSS-Graviser
It's all good!
