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2 years ago

In ΔDEF, m∠D = 103° and m∠E = 14°. Which statement about the sides of ΔDEF must be true?

Mathematics

1 answer:

UkoKoshka [18]2 years ago

4 0

The statement that we can make that must be true about the sides of ΔDEF is: EF > DE > DF.

<h3>Triangle Sides and angles Relationship</h3>

The length of the three sides of a triangle is relative to the measure of the size of the angle opposite to each of he sides.
That is, the longest side will be directly opposite the largest angle and vice versa.

Thus, the statements about the sides of ΔDEF are missing, however, we can make the following deductions:

m∠D = 103° is opposite to the longest side EF.
m∠F = 63° (180 - 103 - 14) is opposite to the medium side DE
m∠E = 14° is opposite to the shortest side DF.

Therefore, the statement that we can make that must be true about the sides of ΔDEF is: EF > DE > DF.

Learn more about the relationship between sides and angles of a triangle on:

brainly.com/question/10695554

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Answer:

The base and height of the solid is 5.5cm

Step-by-step explanation:

Given

$Surface\ Area = 181.5cm^2$

Required

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Let the base dimension be x and the height be h

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$Volume =x^2 * h$

$Volume =x^2h$

181.5 =x^2h

The surface area (S) is calculated as this:

$S = 2(x^2 + xh + xh)$

$S = 2(x^2 + 2xh)$

$S = 2x^2 + 4xh$

Substitute 181.5 for S

$181.5 = 2x^2 + 4xh$

Make h the subject:

$4xh = 181.5 - 2x^2$

$h = \frac{181.5 - 2x^2}{4x}$

Substitute $h = \frac{181.5 - 2x^2}{4x}$ in $Volume =x^2h$

$V = x^2(\frac{181.5 - 2x^2}{4x})$

$V = x(\frac{181.5 - 2x^2}{4})$

$V = \frac{1}{4}(x)(181.5 - 2x^2)$

$V = \frac{181.5x}{4} - \frac{2x^3}{4}$

$V = \frac{181.5x}{4} - \frac{x^3}{2}$

To get the maximum, we differentiate V with respect to t and set the differentiation to 0

$dV = \frac{181.5}{4} - \frac{3x^2}{2}$

Set to 0

$0 = \frac{181.5}{4} - \frac{3x^2}{2}$

$\frac{3x^2}{2} = \frac{181.5}{4}$

Multiply through by 4

$4 * \frac{3x^2}{2} = \frac{181.5}{4} * 4$

$2*3x^2 = 181.5$

$6x^2 = 181.5$

$x^2 = \frac{181.5}{6}$

$x^2 = 30.25$

$x = \sqrt{30.25$

$x = 5.5$

Recall that:

$h = \frac{181.5 - 2x^2}{4x}$

$h = \frac{181.5 - 2 * 5.5^2}{4 * 5.5}$

$h = \frac{181.5 - 60.5}{22}$

$h = \frac{121}{22}$

$h = 5.5$

So, we have:

$h = 5.5$

$x = 5.5$

<em>Hence, the base and height of the solid is 5.5cm</em>

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Step-by-step explanation:

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