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Mathematics 15 Online
danielaaxrosas89:

Find the area of the isosceles triangle. 25m , 25m , -7m

iGreen:

Hello, do you have a picture?

Shadow:

You have three options. 1. Take a screenshot and post a link 2. Use the drawing tool and draw it out 3. Download the image/picture and post it here using the "Attach File" function

Shadow:

After we get the picture, we can help you properly

danielaaxrosas89:

iGreen:

Alright, this is the same as the other problem we did on Brainly. The only difference is that they gave you the length of the base of one of the right triangles, so it's easier.

iGreen:

So use the Pythagorean Theorem to find the height of the triangle. \(\sf a^2+b^2=c^2\) Where 'a' and 'b' are the two legs and 'c' is the hypotenuse.

iGreen:

@danielaaxrosas89 Can you plug in the values?

danielaaxrosas89:

Would it be 25^2 + 25^2=-7^2 ?

iGreen:

No, we're trying to find the other leg of the triangle. We have one leg(7m) and one hypotenuse(25m).

iGreen:

Try it again.

iGreen:

@danielaaxrosas89 Still there?

danielaaxrosas89:

Yes

iGreen:

|dw:1519440118107:dw|

iGreen:

|dw:1519440186347:dw|

danielaaxrosas89:

Would the other leg be -7 too?

iGreen:

That's not -7, it's 7...you can't have a negative measurement.

iGreen:

We don't know what the other leg is, that's what we're trying to find. We know one of them is 7m, and the hypotenuse is 25m. Plug them into the Pythagorean Theorem and solve for the missing leg.

iGreen:

Can you do that?

iGreen:

@danielaaxrosas89

danielaaxrosas89:

I keep trying but to be honest no.

iGreen:

Okay, we can put in '7' for either 'a' or 'b', it doesn't matter, but we have to put in '25' for 'c'.

danielaaxrosas89:

I did, but it didn’t match one of my 4 answers. Unless I solved it wrong...

danielaaxrosas89:

iGreen:

\(\sf a^2 + 7^2 = 25^2\) or \(\sf 7^2 + b^2 = 25^2\) Either one works.

iGreen:

Mind showing what you did?

danielaaxrosas89:

1 attachment
danielaaxrosas89:

And I keep getting the same answer, 24.

iGreen:

The one on the bottom looks right. You didn't finish it though. \(\sf 7^2 + b^2 = 25^2\) \(\sf 49 + b^2 = 625\)

iGreen:

Subtract 49 to both sides and then take the square root of both sides, what do you get?

danielaaxrosas89:

I got 24 again...

iGreen:

That's correct, we're not done though. That just gives us the height of the triangle. Now use the area formula.

iGreen:

\(\sf A=\dfrac{1}{2}bh\) We just figured out the height, what's the base?

danielaaxrosas89:

Is it 49?

iGreen:

The base? No. It's an isosceles triangle, so the distance from center to the other end must also be 7m. 7 + 7 = 14...so the base is 14. Now just plug them in and calculate.

danielaaxrosas89:

I got 84

iGreen:

I got something else, try again... \(\sf A=\dfrac{1}{2}(14)(24)\)

danielaaxrosas89:

Is it 168?

iGreen:

Yep! I got that as well.

danielaaxrosas89:

Alright, do you have time for more? If not, I appreciate your help! Now I understand these problems a bit more.

iGreen:

Yes, I do.

iGreen:

You can close this question and open a new one.

danielaaxrosas89:

Alright thanks (:

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