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3 years ago

The shorter leg of a 30°- 60°- 90° right triangle is 12.5 inches. How long are the longer leg and the hypotenuse?

Mathematics

2 answers:

Elodia [21]3 years ago

8 0

A right angle with angles 30° - 60° - 90° is cut from an equilateral triangle, for example as shown below in triangle ABC

The shorter leg would be the side BD (or CD) since D is the midpoint of side BC

We are to find the length of side AD (which is the height of the triangle ABC) and the hypotenuse AB (or AC ⇒ depends on which right angle triangle that you'd want to work with)

Since we know the three angles in the right angle triangle ABD, we can use the trigonometry ratio

Side AD is the opposite of the angle 60° and we know the length of side that is adjacent to 60° so we'll use the tangent ratio

tan (x°) = opposite ÷ adjacent
tan (60°) = AD ÷ 12.5
√3 = AD ÷ 12.5
AD = 12.5 × √3
AD = 21.65

We can then use the Pythagoras theorem to work out AB

AB² = AD² + BD²
AB² = 12.5² + 21.65²
AB² = 624.9725
AB = √624.9725
AB = 25 (rounded to 2 significant figures)

dezoksy [38]3 years ago

4 0

If u know the short leg...which we do, it is 12.5......to find the hypotenuse, double it......so the hypotenuse is : 12.5 * 2 = 25 inches.....and to find the longer leg, multiply shorter leg by square root of 3.....12.5 * sqrt 3 = 21.65 (thats rounded)

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A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

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faltersainse [42]

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makkiz [27]

9514 1404 393

Answer:

420

Step-by-step explanation:

The prime factors of each denominator are ...

12 = 2²×3

14 = 2×7

5 = 5

10 = 2×5

The unique factors are 2²×3×5×7 = 420.

The least common denominator of these fractions is 420.

_____

Additional comment

5/12 = 175/420
9/14 = 270/420
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2 years ago

5x +7 y=-17 4x – 3y = -5 Substitute.

Bond [772]

Answer:

(x,y) = (-2, -1)

Step-by-step explanation:

We start off by substituting the value of "x".

5x+7y=-17

x=- $\frac{5}{4} + \frac{3}{4}y$

We then solve 5 $5(-\frac{5}{4}+\frac{3}{4}y ) +7y= -17$

y=-1 *Substitute the value of "y"

x= - $-\frac{5}{4}+\frac{3}{4}x(-1)$

x=-2

$\left \{ {{-17=-17} \atop {-5=-5}} \right.$ This works.

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