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vlada-n [284]
3 years ago
11

Help, please ? Thank you so much in advance and I'd appreciate it!

Mathematics
2 answers:
Darina [25.2K]3 years ago
7 0
First you calculate the area of the big rectangle (ignoring the small one the bottom):

A1 = 10 x 6 = 60.

next you calculate the small one:

A2 = 2 x 2 = 4

Now you subtract the small one from the big one and you get:

Area of the figure = A1 - A2 = 60 - 4 = 56
hoa [83]3 years ago
3 0
To start off, pretend it's a quadrilateral. It makes everything easier and uses all the numbers.

The area of a quadrilateral is base times height. 
10 = base 
6 = height 

6 * 10 = 60 

That's the value of the original, larger, quadrilateral without the mini quadrilateral that we have to subtract. 

The dimensions of the height and base of the mini quadrilateral are 2 by 2. 
2 * 2 = 4 

Subtract to get your answer: 
60 - 4 = 56


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levacccp [35]
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3 years ago
If a triangle is 12 inches high and has a 20 inch long hypotenuse then how wide is it?
Ber [7]

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The triangle should be 16 inches wide.

Step-by-step explanation:

\text {Use the Pythagorean Theorem to solve for the leg.}\\a^2 + b^2 =c^2

What we know:

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  • The hypotenuse is 20 inches.

We need to solve for b. Solving for a while b is 12 would also be acceptable.

12^2 + b^2 = 20^2  \text {  OR  }   a^2 +12^2=20^2

We will use the first equation.

12^2 + b^2 = 20^2\\12 * 12 = 144\\20 * 20 = 400\\\\\boxed {144 + b^2 = 400}

\text {Subtraction property of equality: }\\144-144+ b^2 = 400-144\\\boxed {b^2 = 256}\\\\\text {Take the square root of 256: }\\\sqrt{256}=\sqrt{b^2} \boxed {16 = b}

The triangle should be 16 inches wide.

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3 years ago
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