Given: The sum of the catheters in a triangle is 27 cm
To Determine: The area of the triangle
Solution
Please note the below
Let the first cathetus be x, then the second cathetus would be
![\begin{gathered} c_1=x \\ c_2=27-x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8kc7ax385275dqjmc2ojy7nsftn6lnib70.png)
For the second right triangle
![\begin{gathered} c_1=2 \\ c_2=7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/omufh613tk7vcwxmwhhowic4twqbumonhc.png)
Since the two right triangles are corresponding to each other, then the ratio of their cathethers are equal
Therefore
![\begin{gathered} (x)/(27-x)=(2)/(7) \\ 7x=2(27-x) \\ 7x=54-2x \\ 7x+2x=54 \\ 9x=54 \\ x=(54)/(9) \\ x=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3ivgunrn6nbbhyx2sbsg17ekv3dvk6pdvy.png)
So, the cathethers for the first right triangle would be
![\begin{gathered} c_1=x:c_2=27-x \\ c_1=6 \\ c_2=27-6 \\ c_2=21 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/18nfv63fdnmr1nxzq33ob5t5ewv7c32agj.png)
Note that the catheters formed the base and the height of the first triangle. The area of a triangle can be calculated using the formula below
![\begin{gathered} Area(triangle)=(1)/(2)* base* height \\ Area(triangle)=(1)/(2)*6cm*21cm \\ Area(triangle)=3cm*21cm \\ Area(triangle)=63cm^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7hls9woa0oe6ycwrvp2hs2j7s8h8dtktxg.png)
Hence, the area of the first triangle is 63cm²