What is the length of a diagonal of a square with sides 16 ft long? Round to the nearest tenth.
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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.
To find the length of a diagonal of a square with sides 16 ft long, you can use the Pythagorean theorem. The Pythagorean theorem states that for any right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.
In this case, the two sides of the right triangle formed by the diagonal are both 16 ft long. So, we can call each side "A" and "B", and the diagonal (the hypotenuse) as "C".
Using the Pythagorean theorem, we have A^2 + B^2 = C^2. Substituting the values, we get 16^2 + 16^2 = C^2.
Simplifying, we have 256 + 256 = C^2. Adding them together, we get 512 = C^2.
To find C, we need to take the square root of both sides of the equation. So, C = √512.
Using a calculator, you can find that √512 is approximately 22.6. Rounding to the nearest tenth, the length of the diagonal is 22.6 ft.
Now, for the second question. To find the area of an isosceles right triangle, you need to know the length of the hypotenuse. Let's call it "C" and the legs of the triangle "a" and "b". In an isosceles right triangle, the two legs have the same length.
We know the length of the hypotenuse is 30 meters, so we have C = 30.
Using the Pythagorean theorem, we have a^2 + a^2 = C^2. Simplifying, we get 2a^2 = C^2.
Plugging in the values, we have 2a^2 = 30^2. Simplifying further, we get 2a^2 = 900.
To solve for "a", divide both sides of the equation by 2. We get a^2 = 450.
Taking the square root of both sides, we have a = √450.
Using a calculator, we find that √450 is approximately 21.2.
Since the area of a triangle is given by the formula A = (base * height)/2, and in this case, both the base and height are equal to "a", we can substitute the value of "a" to find the area.
So, the area is A = (21.2 * 21.2)/2, which is approximately 224.7 square meters when rounded to the nearest tenth.