Proving that EFGH is a square: points E,F,G,H
separate the sides of the square ABCD into two lines. if AE=BF=CG=DH then
AH=DG=EB=FC and it shows that if we connect the points E,F,G,H to each other with line we will have a square
HAE~EBF~FCG~GDH by rule SAS
<h3>
Answer: 10</h3>
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Explanation:
The two smaller triangles are proportional, which lets us set up this equation
5/n = n/15
Cross multiplying leads to
5*15 = n*n
n^2 = 75
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Apply the pythagorean theorem on the smaller triangle on top, or on the right.
a^2+b^2 = c^2
5^2+n^2 = m^2
25+75 = m^2
100 = m^2
m^2 = 100
m = sqrt(100)
m = 10
That's a quadratic, a nice parabola in vertex form.
The parabola has a positive x^2 term, so it's a CUP, concave up positive. It will have a minimum at the vertex, which is (2,5). Plot that point.
Now we need a couple of guide points to draw the usual parabola going up from both sides of its vertex. We try x=0 giving (0,9) and see that x=4 also gives 9, (4,9). Plot the parabola through those two points and the vertex and you're done.
Answer:
-4 i think
Step-by-step explanation:-4 i think