(11) Give possible lengths of the legs of a right triangle with a hypotenuse of .

Ask by Simpson Bradley. in the United States

Mar 25,2025

Question

(11) Give possible lengths of the legs of a right triangle with a hypotenuse of $$ \sqrt{265} $$. $$ a^{2}+b^{2}=c^{2} $$ $$ 6^{2}+ $$

UpStudy AI · Accepted Answer

We start with the Pythagorean theorem:

$$
a^2 + b^2 = c^2
$$

Here the hypotenuse is given as

$$
c = \sqrt{265}
$$

so that

$$
a^2 + b^2 = 265.
$$

There are infinitely many choices for the numbers $$a$$ and $$b$$ that satisfy this equation. However, if we are looking for integer lengths, we want integers $$a$$ and $$b$$ such that

$$
a^2 + b^2 = 265.
$$

Let’s search for possible pairs:

1. Try $$a = 3$$:

$$
   3^2 + b^2 = 265 \quad \Rightarrow \quad 9 + b^2 = 265 \quad \Rightarrow \quad b^2 = 256.
   $$
  
   Thus,

$$
   b = 16.
   $$

So one possible pair is $$a = 3$$ and $$b = 16$$.

2. Try $$a = 11$$:

$$
   11^2 + b^2 = 265 \quad \Rightarrow \quad 121 + b^2 = 265 \quad \Rightarrow \quad b^2 = 144.
   $$
  
   This gives

$$
   b = 12.
   $$

So another possible pair is $$a = 11$$ and $$b = 12$$.

We can check both:

- For $$a = 3$$ and $$b = 16$$:

$$
  3^2 + 16^2 = 9 + 256 = 265.
  $$

- For $$a = 11$$ and $$b = 12$$:

$$
  11^2 + 12^2 = 121 + 144 = 265.
  $$

Thus, the right triangle with a hypotenuse of $$\sqrt{265}$$ can have legs of lengths $$3$$ and $$16$$ or $$11$$ and $$12$$.

It is important to note that if the leg lengths are not required to be integers then many other pairs $$(a,b)$$ satisfying

$$
a^2 + b^2 = 265
$$

are possible. For example, if one leg is chosen to be $$6$$, then the other leg would be

$$
b = \sqrt{265 - 6^2} = \sqrt{265 - 36} = \sqrt{229}.
$$

In summary, one set of integer possibilities is:

$$
\boxed{(a,b)=(3,16)\ \text{or}\ (11,12)}.
$$
Possible lengths for the legs are 3 and 16, or 11 and 12.

(11) Give possible lengths of the legs of a right triangle with a hypotenuse of .

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