2.19 repeating as a fraction
1 answer:
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Answer:
7 1/17
Step-by-step explanation:
A figure can be helpful.
The inscribed semicircle has its center at the midpoint of th base. It is tangent to the side of the isosceles triangle, so a radius makes a 90° angle there.
The long side of the isosceles triangle can be found from the Pythagorean theorem to be ...
BC² = BD² +CD²
BC² = 8² +15² = 289
BC = √289 = 17
The radius mentioned (DE) creates right triangles that are similar to ∆BCD. In particular, we have ...
(long side)/(hypotenuse) = DE/BD = CD/BC
DE = BD·CD/BC = 8·15/17
DE = 7 1/17 ≈ 7.059
Answer:
The expressions that show the value of q are
1)
2)
3)
4)
5)
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
case A)
In the right triangle of the figure
Applying the Pythagoras Theorem
case B)
In the right triangle of the figure
solve for q
case C)
In the right triangle of the figure
solve for q
case D)
In a right triangle
if
then
therefore
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