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3 years ago

A trapezoid has an area of 30m^2. if all of its side lengths are tripled, what will its new area be

Mathematics

1 answer:

Norma-Jean [14]3 years ago

5 0

Answer:

270

Step-by-step explanation:

Because we do not know what the side lengths are, as long as they multiply to 30m^2, it's fine

For this question, let's just say the base is 5, and the height is 6. If we triple 5, we get 15, and if we triple 6, we get 18. 15*18=270

Now what if, the sides are not 5 and 6. Will the area still be the same? Let's find out.

10*3=30 so we can say for this answer, the base is 3, and the height is 10.

10 tripled is 30, and 3 tripled is 9. 30*9=270

So as we look at these 2 answers. we can conclude that the new area, no matter the side lengths, will be 270

Hope this helpes!

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PLEASE HELP ASAP ILL GIVE BRAINLIEST!!!!!<br><br>find measure of arc MK

Gwar [14]

Answer:

Arc length MK = 15.45 units (nearest hundredth)

Arc measure = 58.24°

Step-by-step explanation:

Calculate the measure of the angle KLN (as this equals m∠KLM which is the measure of arc MK)

ΔKNL is a right triangle, so we can use the cos trig ratio to find ∠KLM:

$\sf \cos(\theta)=\dfrac{A}{H}$

where:

$\theta$ is the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Given:

$\theta$ = ∠KLM
A = LN = 8
H = KL = 15.2

$\implies \sf \cos(KLM)=\dfrac{8}{15.2}$

$\implies \sf \angle KLM=\cos^{-1}\left(\dfrac{8}{15.2}\right)$

$\implies \sf \angle KLM=58.24313614^{\circ}$

Therefore, the measure of arc MK = 58.24° (nearest hundredth)

$\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right) \quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle)}$

Given:

r = 15.2
∠KLM = 58.24313614°

$\implies \textsf{Arc length MK}=2 \pi (15.2)\left(\dfrac{\sf \angle KLM}{360^{\circ}}\right)$

$\implies \textsf{Arc length MK}=\sf 15.45132428\:units$

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2 years ago

In the figure below, L N equals 4 x minus 7 and M O equals 2 x plus 13. For which value of x is the figure a rectangle?

Georgia [21]

For this case we have the following data:

$LN = 4x-7$

$MO = 2x + 13$

So that the figure can be a rectangle, its diagonals must be equal, that is, $LN = MO$

In this way we have:

$4x-7 = 2x + 13$

Clearing x we have:

$4x-2x = 13 + 7 \\2x = 20$

$x=\frac{20}{2} \\x = 10$

Thus, x must be equal to 10 so that the figure is a rectangle.

Answer:

$x = 10$

Option A

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3 years ago

A rectangular storage container with a lid is to have a volume of 2 m3. The length of its base is twice the width. Material for

Scilla [17]

Answer:

Dimensions are 2 m by 1 meter by 1 meter,

Minimum cost is $ 18.

Step-by-step explanation:

Let w be the width ( in meters ) of the container,

Since, the length is twice of the width,

So, length of the container = 2w,

Now, if h be the height of the container,

Volume = length × width × height

2 = 2w × w × h

1 = w² × h

$\implies h=\frac{1}{w^2}$

Since, the area of the base = l × w = 2w × w = 2w²,

Area of the lid = l × w = 2w²,

While the area of the sides = 2hw + 2hl

= 2h( w + l)

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$=\frac{6w}{w^2}$

$=\frac{6}{w}$

Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,

So, the total cost,

$C(w) = 1\times 2w^2+2\times 2w^2 + 2\times \frac{6}{w}$

$=2w^2+4w^2+\frac{12}{w}$

$=6w^2+\frac{12}{w}$

Differentiating with respect to w,

$C'(w) = 12w -\frac{12}{w^2}$

Again differentiating with respect to w,

$C''(w) = 12 + \frac{24}{w^3}$

For maxima or minima,

C'(w) = 0

$\implies 12w -\frac{12}{w^2}=0$

$\implies 12w^3 - 12=0$

$w^3-1=0\implies w = 1$

For w = 1, C''(w) = positive,

Hence, for width 1 m the cost is minimum,

Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,

And, the dimension for which the cost is minimum is,

2 m by 1 meter by 1 meter.

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Answer:

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Step-by-step explanation:

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A trapezoid has an area of 30m^2. if all of its side lengths are tripled, what will its new area be​

A trapezoid has an area of 30m^2. if all of its side lengths are tripled, what will its new area be