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4.-
In a 45°-45°-90° triangle, the length of the legs are "x" and the length of the hypotenuse is x
Then, in this problem, the length of the leg is 7 then the length of the hypotenuse will be 7
5.- In a 30°-60°-90° triangle, the length of the hypotenuse is 2x, the length of the longer leg is x and the length of the shorter leg is x.
In this problem we have the length of the hypotenuse = 20 and find x.
2x = 20
x = 20/2
x = 10
longer leg = x
6.- nonzero
7.- hypotenuse = 20 cm
Find x
2x = 20
x = 20/2
x = 10
Shoter leg = x = 10 cm
Let's carry this math sentence over to its natural, "shapey" element. We're going to look at each term not as an ordinary number, but as <em>the area of some shape</em>.
x² (read as "x <em>squared"</em>) can be seen as the area of a square with side lengths of x. 2x can similarly be seen as the area of a <em>rectangle </em>with a length of x and a width of 2. (Picture 1)
What's our question actually asking, though? Something about <em>perfect squares</em>. More specifically, we're looking for something to add on that'll <em>make this thing a perfect square</em>. We're trying to find a missing piece we can slot in to make a square, in other words. Problem is, our shapes don't look much like a square if we put them together right now. We need to do a little cutting and gluing first.
First, we're gonna cut the 2x rectangle lengthwise, getting two rectangles with an area of x, a length of 1, and a width of x. Next, we're going to attach them to the x² square, creating this shape that looks, strangely, like a square with a little bit missing from it (picture 2). What we're trying to do is <em>complete this square, </em>to find the area of that little missing chunk.
As it turns out, we have all the information we need for this. Notice that, using the lengths of the x rectangles, we can find that the square's dimensions are 1 x 1, which means that its area is 1 x 1 = 1.
If we tack this new area on to our original expression, we've "completed the square!" We now have a perfect square with side lengths of (x + 1) and an area of (x + 1)² (picture 3).
So, our final expression is x² + 2x + 1, and the missing constant - the area of the "missing square" we had to find to complete our larger one - is 1.
