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3 years ago

What is the exact perimeter of kite ? Show your work.

Mathematics

1 answer:

Afina-wow [57]3 years ago

7 0

Answer:

The perimeter of the kite is 29 units.

Step-by-step explanation:

1. Let's use the Pythagorean Theorem to find the perimeter of the kite:

With the information given, we have four right triangles with a new point A, intersecting UW and VX. Those triangles are:

Δ UVA

Δ VWA

Δ UXA

Δ XWA

As we can see in the figure, length of UV and VW is the same and the length of UX and WX is also the same.

2. Let's find the value of UV and VW

UV is the hypotenuse of Δ UVA and it sides are 3 and 4 units length. So, we can calculate the length of UV this way:

Length of UV ² = UA ² + VA ²

Replacing with the real values:

Length of UV ² = 3 ² + 4 ²

Length of UV ² = 9 + 16

Length of UV ² = 25

√Length of UV ² = √25

Length of UV = 5 units ⇒ Length of VW = 5 units

3. Let's find the value of UX and WX

UX is the hypotenuse of Δ UXA and it sides are 3 and 9 units length. So, we can calculate the length of UX this way:

Length of UX ² = UA ² + XA ²

Replacing with the real values:

Length of UX ² = 3 ² + 9 ²

Length of UX ² = 9 + 81

Length of UV ² = 90

√Length of UV ² = √90

Length of UV = 9.5 units (rounding to the nearest tenth) ⇒ Length of WX= 9.5 units

4. Let's calculate the perimeter of the kite:

Perimeter of the kite = UV + VW + WX + UX

Perimeter of the kite = 5 + 5 + 9.5 + 9.5

Perimeter of the kite = 29 units

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Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

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$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

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$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

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