Heres an ez question for yall

Mathematics

1 answer:

SSSSS [86.1K]2 years ago

7 0

Answer: Choice B

$\left\{ \ 3, 5, \sqrt{34} \ \right\}$

=========================================================

Explanation:

We'll be using the converse of the pythagorean theorem.

If we had a triangle with sides a,b,c and $a^2+b^2 = c^2$ was true, then we have a right triangle. Note: c is always the longest side

For the first answer choice we have: a = 5, b = 8, c = 12

Then,

$a^2+b^2 = c^2\\\\5^2+8^2 = 12^2\\\\25+64 = 144\\\\89 = 144\\\\$

which is false. This shows that we do not have a right triangle with sides 5,8,12. This rules out choice A. Choice C is a similar story.

Choice B on the other hand works because

$a^2+b^2 = c^2\\\\3^2+5^2 = (\sqrt{34})^2\\\\9+25 = 34\\\\34 = 34\\\\$

We have a true statement at the end, which confirms <u>choice B</u> is a right triangle.

You might be interested in

Find the area of the circle whose equation is x2+y2=6x-8y

Semenov [28]

Answer:

Given that the equation of a circle is :

$\green{ \boxed{\boxed{\begin{array}{cc} {x}^{2} + {y}^{2} = 6x - 8y \\ = > {x}^{2} + {y}^{2} - 6x + 8y = 0 \\ = > {x}^{2} + {y}^{2} + 2 \times ( - 3) \times x + 2 \times 4 \times y = 0 \\ \\ \sf \: standard \: equation \: o f \: circle \: is : \\ {x}^{2} + {x}^{2} + 2gx + 2fy + c = 0 \\ \\ \sf \: by \: comparing \\ \\ g = - 3 \\ f = 4 \\ c = 0 \\ \\ \sf \: radius \: \: r = \sqrt{ {g}^{2} + {f}^{2} - c } \\ = \sqrt{ {( - 3)}^{2} + {4}^{2} - 0 } \\ = \sqrt{9 + 16} \\ = \sqrt{25} \\ = 5 \: unit \\ \\ \bf \: area \: = \pi {r}^{2} \\ = \pi \times {5}^{2} \\ =\pink{ 25\pi \: { unit }^{2} }\end{array}}}}$