Find the length of A
ill give brainliest.
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Answer:
a = 5
Step-by-step explanation:
Topic: Pythagorean Theorem
In this problem, we need to solve for a missing side in a right triangle using the Pythagorean theorem.
Formula: [tex]a^2 + b^2 = c^2[/tex]
Parts:
Step 1: Plug in information
We want to use the info we have and plug in the values for the variables
"a" is unknown, so we can leave it like that
"b" is the second-longest and it is a leg. In this case, it is 12
"c" is the hypotenuse and the longest side. This has a value of 13
We can use this to rewrite the equation
a² + 12² = 13²
Step 2: Solve for a
We want all terms with "a" on one side, and we want "a" alone, no exponents, nothing.
a² + 12² = 13²
a² = 13² - 12²
a = √(13² - 12²)
Step 3: Solve for a
The rest is the same, just solving the equations. Remember, square rooting and squaring are not distributive to adding and subtracting. This means that √(13² - 12²) is not 13 - 12.
a = √(13² - 12²)
a = √(169 - 144)
a = √(25)
a = 5
-Chetan K
Answer:
The length of a is 5.
Step-by-step explanation:
Solution :
Here, we have given that the two sides of triangle are 12 and 13.
Finding the third side of triangle by pythagorean theorem formula :
[tex]{\longrightarrow{\pmb{\sf{{(c)}^{2} = {(a)}^{2} + {(b)}^{2}}}}}[/tex]
Substituting all the given values in the formula to find the third side of triangle :
[tex]\begin{gathered}\quad{\longrightarrow{\sf{{(c)}^{2} = {(a)}^{2} + {(b)}^{2}}}}\\\\\quad{\longrightarrow{\sf{{(13)}^{2} = {(a)}^{2} + {(12)}^{2}}}}\\\\\quad{\longrightarrow{\sf{{(13 \times 13)} = {(a)}^{2} + {(12 \times 12)}}}}\\\\\quad{\longrightarrow{\sf{(169)= {(a)}^{2} + (144)}}}\\\\\quad{\longrightarrow{\sf{169 = {(a)}^{2} + 144}}}\\\\\quad{\longrightarrow{\sf{{(a)}^{2} = 169 - 144}}}\\\\\quad{\longrightarrow{\sf{{(a)}^{2} = 25 }}}\\\\\quad{\longrightarrow{\sf{a= \sqrt{25}}}}\\\\\quad{\longrightarrow{\sf{a= \sqrt{5 \times 5}}}}\\\\\quad{\longrightarrow{\sf{a= 5}}}\\\\ \qquad\star{\underline{\boxed{\frak{\red{a= 5}}}}}\end{gathered}[/tex]
Hence, the length of a is 5.
[tex]\rule{300}{2.5}[/tex]