#bff

23 messages · Page 1 of 1 (latest)

rigid badger
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Ok I have to use Pythagorean

mint juniper
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Yh

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It’s diagonal so you can find the value of x as it’ll be a right angled triangle

rigid badger
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But I’m lost after it

mint juniper
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Have u got a x value?

rigid badger
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No

mint juniper
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U got working?

rigid badger
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So I just did 25 squared +x squared

mint juniper
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Ok so wait

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25 is the diagonal length so it’s the hypotenuse of a right angled triangle with a base of 2x and height of x

rigid badger
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Or u can do 2xsquared

rigid badger
mint juniper
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So a^2 + b^2 = c^2

rigid badger
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I did 2x + x

mint juniper
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(2x)^2 + (x)^2 = (25)^2

rigid badger
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Yh

mint juniper
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Then
4x^2 + x^2 = 25^2
5x^2 = 25^2
root(5x^2) = 25
x root(5) = 25
x = 25/root(5)

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For the area it’s
2x * x so just find 2x and multiply it by x

rigid badger
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5/5 x2

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250

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5/5 x2 x5/5

keen cairn
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/solved