A semicircle is inscribed in an isosceles triangle with base 16 and height 15 so that the
1 answer:
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
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Take the Laplace transform of both sides:
L[y'' - 4y' + 8y] = L[δ(t - 1)]
I'll denote the Laplace transform of y = y(t) by Y = Y(s). Solve for Y :
(s²Y - s y(0) - y'(0)) - 4 (sY - y(0)) + 8Y = exp(-s) L[δ(t)]
s²Y - 4sY + 8Y = exp(-s)
(s² - 4s + 8) Y = exp(-s)
Y = exp(-s) / (s² - 4s + 8)
and complete the square in the denominator,
Y = exp(-s) / ((s - 2)^2 + 4)
Recall that
L⁻¹[F(s - c)] = exp(ct) f(t)
In order to apply this property, we multiply Y by exp(2)/exp(2), so that
Y = exp(-2) • exp(-s) exp(2) / ((s - 2)² + 4)
Y = exp(-2) • exp(-s + 2) / ((s - 2)² + 4)
Y = exp(-2) • exp(-(s - 2)) / ((s - 2)² + 4)
Then taking the inverse transform, we have
L⁻¹[Y] = exp(-2) L⁻¹[exp(-(s - 2)) / ((s - 2)² + 4)]
L⁻¹[Y] = exp(-2) exp(2t) L⁻¹[exp(-s) / (s² + 4)]
L⁻¹[Y] = exp(2t - 2) L⁻¹[exp(-s) / (s² + 4)]
Next, we recall another property,
L⁻¹[exp(-cs) F(s)] = u(t - c) f(t - c)
where F is the Laplace transform of f, and u(t) is the unit step function
To apply this property, we first identify c = 1 and F(s) = 1/(s² + 4), whose inverse transform is
L⁻¹[F(s)] = 1/2 L⁻¹[2/(s² + 2²)] = 1/2 sin(2t)
Then we find
L⁻¹[Y] = exp(2t - 2) u(t - 1) • 1/2 sin(2 (t - 1))
and so we end up with
y = 1/2 exp(2t - 2) u(t - 1) sin(2t - 2)
Answer:
The center of the circle is .
Step-by-step explanation:
The center of the circle is the midpoint of the segment between the endpoints. We can determine the location of the center by this vectorial expression:
(1)
Where:
- Center.
,
- Location of the endpoints.
If we know that and
, then the location of the center of the circle is:
The center of the circle is .