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3 years ago

In the right triangle shown,

Mathematics

1 answer:

Bad White [126]3 years ago

4 0

Answer:

The length of AB = $x=6\sqrt{3}$

Step-by-step explanation:

Given

θ = m∠B = 60⁰

BC = $3\sqrt{3}$

So we have to determine the length of AB which is Hypotenuse.

i.e. AB = Hypotenuse = x ?

Using the formula

Cos θ = Adjacent / Hypotenuse

Here:

Adjacent = BC = $3\sqrt{3}$

Cos θ = Adjacent / Hypotenuse

$\cos \left(60^{\circ \:}\right)=\frac{3\sqrt{3}}{x}$ ∵ AB = Hypotenuse = x

$\frac{1}{2}x=3\sqrt{3}$ ∵ $\cos \left(60^{\circ }\right)=\frac{1}{2}$

$x=6\sqrt{3}$

Therefore, the length of AB = $x=6\sqrt{3}$

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A square park measures 170 feet along each side. Two paved paths run from each corner to the opposite corner and extend 3 feet i

Cerrena [4.2K]

Answer:

The total area, in square feet, taken by the paths is 2,004

Step-by-step explanation:

see the attached figure with lines to better understand the problem

I can divide the figure into four right triangles, one small square and four rectangles

step 1

Find the area of the right triangle of each corner of the path

The area of the triangle is

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

substitute the given values

$A=(1/2)(3)(3)=4.5\ ft^2$

step 2

Find the hypotenuse of the right triangle

Applying Pythagoras Theorem

Let

d -----> hypotenuse of the right triangle

$d^{2}=3^{2}+3^{2}$

$d^{2}=18$

$d=\sqrt{18}\ ft$

simplify

$d=3\sqrt{2}\ ft$

The hypotenuse of the right triangle is equal to the width of the path

step 2

Find the area of the small square of the path

The area is

$A=b^{2}$

we have

$b=3\sqrt{2}\ ft$ ----> the width of the path

substitute

$A=(3\sqrt{2})^{2}$

$A=18\ ft^2$

step 3

Find the length of the diagonal of the square park

Applying Pythagoras Theorem

Let

D -----> diagonal of the square park

$D^{2}=170^{2}+170^{2}$

$D^{2}=57,800$

$D=\sqrt{57,800}\ ft$

simplify

$D=170\sqrt{2}\ ft$

step 4

Find the height of each right triangle on each corner

The height will be equal to the width of the path divided by two, because is a 45-90-45 right triangle

$h=1.5\sqrt{2}\ ft$

step 5

Find the area of each rectangle of the path

The area of rectangle is $A=LW$

we have

$W=3\sqrt{2}\ ft$ ----> width of the path

Find the length of each rectangle of the path

$L=(D-2h-d)/2$

where

D is the diagonal of the park

h is the height of the right triangle in the corner

d is the width of the path (length side of the small square of the path)

substitute the values

$L=(170\sqrt{2}-2(1.5\sqrt{2})-3\sqrt{2})/2$

$L=(170\sqrt{2}-3\sqrt{2}-3\sqrt{2})/2$

$L=(164\sqrt{2})/2$

$L=82\sqrt{2}\ ft$

Find the area of each rectangle of the path

$A=LW$

we have

$W=3\sqrt{2}\ ft$

$L=82\sqrt{2}\ ft$

substitute

$A=(82\sqrt{2})(3\sqrt{2})$

$A=492\ ft^2$

step 6

Find the area of the paths

Remember

The total area of the paths is equal to the area of four right triangles, one small square and four rectangles

substitute

$A=4(4.5)+18+4(492)=2,004\ ft^2$

therefore

The total area, in square feet, taken by the paths is 2,004

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3 years ago

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Elden [556K]

Answer:

$x = 8$

Step-by-step explanation:

Set up the proportion

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Step-by-step explanation:

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Now, using product law of exponent we get

$n^{6+1+1+1}\\\\=n^{9}$

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