What is the length of a diagonal of a square with sides 16 ft long? Round to the nearest tenth.
&
The length of the hypotenuse of an usosceles right traingle is 30 meters. Find the area of the triangle. Round to the nearest tenth, if necessary.
Any help?
The diagonal forms two right triangles with sides of 16 ft. Call those sides A and B and the diagonal (the hypotenuse) is C.
Then A^2 + B^2 = C^2.
For the second one, do you mean isosceles? I don't know what a usosceles right triangle is. Perhaps you can tell me.
I'm sorry, I meant Isosceles.
I done a^2+b^2=c^2 and I got 512. What do I do after that?
a^2 + b^2 = c^2
16^2 + 16^2 = c^2
256 + 256 = 512 = c^2
So extract the sqrt of both sides.
Then c = sqrt 512? Right?
I'll let you extract the sqrt.
I will look at the othrer question in a separate post.
a^2 + b^2 = c^2
But we know it is an isosceles triangle; therefore, a = b.
a^2 + a^2 = c^2
2a^2 = c^2 = 30^2
Solve for a.
Post your work if you get stuck. That will make it easier to see where you go wrong but I think you can take it from here.
i got 900.
So I'm assuming that's my answer.
900 is not right. Show me how you got it.
a^2 + a^2 = c^2
which would be 30^2 + 30^2 = 900^2.
Where am I messing up.
a^2 + a^2 = c^2
which would be 30^2 + 30^2 = 900^2.
You didn't finish. You stopped part way through.
a^2 + a^2 = c^2
2a^2 = 30^2
2a^2 = 900
a^2 = 900/2
a^2 = 450
I'll let you finish (of course 450 is not the answer.)
Check your work by substituting what you find for a into
a^2 + b^2 = and see if it is 900.
i really need help n geometry. the thing with it is i can't understand the measuring the shapes.
The diagonal derives from D^2 = 16^2 + 16^ or D = sqrt(16^2 + 15^2.
The sides of the isosceles right triangle with hypotenuse 30m derves from the same expressio. Being isosceles, the equal acute corner angles are 45º.
Since the triangle is isosceles, eaxh side is 30(sin45º)= 21.213...
The area is then A = (21.213)^2/2.
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