SOLUTION:
Step 1 :
Set up the right triangle with the following:
side a = x
side b = 13 + 3x
side c = 14 + 3x
Using pythagorean theorem :
![a^2+b^2=c^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/fdnnfwrccw5g60jmi691r5gcz9ekxf8waa.png)
![x^2+(13+3x)^2=(14+3x)^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/o8aepib0zsgl97spvdwhfk4kdoylloemjh.png)
Step 2 :
Expand this equation by using foil to get:
![x^2+(169+78x+9x^2)=196+84x+9x^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/ca4i90z4cjgar0yvsus02chut2g6vv5gjj.png)
Simplifying this equation will give you:
![x^2\text{ - 6 x - 27 = 0}](https://img.qammunity.org/2023/formulas/mathematics/high-school/y2non5b4ezrwh958zw5vztk4yhc3iol2v5.png)
Step 3 :
Factor this to get:
(x-9)(x +3) = 0
Solving this will give you x = 9 or x = -3.
The side length cannot be negative so discard the -3.
Step 4 :
Your side lengths will be:
![\begin{gathered} a\text{ = x = 9 m} \\ b\text{ = 13 + 3 x = 13 + 3 ( 9 ) = 13 + 27 = 40 m} \\ \text{c = 14 + 3 x = 14 + 3 ( 9 ) = 14 + 27 = 41 m} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/n8e4eo7vv0i8mfzmtmzrvtc7ydalp1pgkk.png)
CONCLUSION:
The side lengths of the triangle are:
9 m , 40 m and 41 m