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Cloud [144]
3 years ago
12

On a boat, a cabin's window is in the shape of an isosceles trapezoid, as shown below. What is the area of the window?

Mathematics
1 answer:
babunello [35]3 years ago
4 0

If you read the question you can find that the question is incomplete. By googling it you can find the complete question here: brainly.com/question/11917098.

The area of the window is 182 square inches.

Option: B.

<u>Step-by-step explanation:</u>

From the given data, we are given with the length of the upper side of the trapezoid, the right side's bottom excluded the length upper side's length and height of the trapezoid.

Area of an isosceles trapezoid = \frac{1}{2}( a+b) h.

where a is the length of upper side of the trapezoid, b is the length of lower side of the trapezoid and h is the height of the trapezoid.

a= 12 inches.

h= 13 inches.

We are given only with the right side's length of base excluding the length of upper side.

So we have to calculate b= left side + upper side+ right side.

                                        b= 2+12+2.

                                        b= 16 inches.

Area of an isosceles trapezoid = \frac{1}{2} (12+16)(13).

                                                   = \frac{1}{2} (28)(13).

                                                   = 14(13).

                                                   = 182 square inches.

∴ Area of the window= 182 square inches.

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2 years ago
Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible
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Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

pq =    \sqrt{( {x2 - x1)}^{2}  + ( {y2 - y1)}^{2} }

or \:  \sqrt{( {5  + 3)}^{2} + ( {2 - 2)}^{2}  }

By simplifying it we get,

8

Now,

again finding the distance between PR,

pr = \sqrt{( {x2 - x1}^{2}  + ( {y2 - y1)}^{2} }

or,

\sqrt{( {x + 3)}^{2}  + ( {y - 2)}^{2} }

By simplifying it we get,

=  \sqrt{ {x}^{2} +  {y}^{2} + 6x  -  4y + 13  }

now, finding the distance of QR,

qr =  \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2}  }

or, by simplification we get,

\sqrt{ {x}^{2} +  {y}^{2}  - 10x - 4y + 29 }

now, equating PR and QR,

\sqrt{ {x}^{2} +  {y}^{2}  + 6x  -  4y  + 13}  =  \sqrt{ {x}^{2}  +  {y}^{2} - 10x - 4y + 29 }

we cancelled the root ,

{x}^{2} +  {y}^{2} + 6x - 4y + 13 =  {x}^{2}   +  {y}^{2}   -10x - 4y + 29

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

\sqrt{ {x}^{2} +  {y}^{2}  + 6x - 4y + 13 }  =  8}

cancel the roots,

{x}^{2} +  {y}^{2}   + 6x - 4y + 13  = 8

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

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y=(8,-12)

So, i think it's (1,8) is your answer..

<em>Hope it helps</em><em>.</em><em>.</em><em>.</em>

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Hello there

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hope this helps

Best Regards Queen Z

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Leno4ka [110]

Answer:

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