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1 year ago

What is the area of the trapezoid? If necessary, round your answer to the nearest tenth.

Mathematics

1 answer:

Katyanochek1 [597]1 year ago

5 0

The area of the trapezoid is calculated as 45.5 feet squared.

What does Area of Trapezoid mean?

The entire area occupied by the sides of a trapezoid is its area. It's important to notice that if we know the lengths of all the sides, we can divide the trapezoid into smaller polygons, such as triangles and rectangles, calculate their areas, and then sum them all together to determine the area of the trapezoid.

<h3>The formula for Trapezoid Area:</h3>

If the parallel sides of the trapezoid are known, along with their height, it is possible to determine their area. It is expressed as,

A= $1/2$ (a+b) h

where, A= the area of the trapezoid

h= height

'a' & 'b' = parallel side bases

<h3>Solution:</h3>

a=5.5 feet , b= 8.5 feet & h=6.5 feet

A= $1/2$ (a+b)h

∴A= $1/2$ (5.5+8.5)*6.5

⇒A= $1/2$ (14)*6.5

⇒A=7*6.5

⇒A=45.5 feet squared

Therefore, the solution is (b) 45.5 feet squared.

Learn more about Trapezoid:

brainly.com/question/12451654?referrer=searchResults

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Answer:

<u>graphically</u> :

The graph is symmetrical about the origin.

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1 year ago

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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

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Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

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