The original area of the triangle is:
![A_{\text{original}}=243in^2_{}](https://img.qammunity.org/2023/formulas/mathematics/college/fbs3511pbpfbmbtoei8rmkhmv0wdct2loh.png)
Let's call the original length "l" and the original width "w".
And now we remember the formula to calculate the area of a triangle using the length (or base) and the width (or height):
![A=(l* w)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/s1a8fj1rgihqxpmtuj4y94ibdase4d0f66.png)
So for the original triangle:
![(l* w)/(2)=243in^2](https://img.qammunity.org/2023/formulas/mathematics/college/mzh2yf4reydyns9inyz3ca358mqliyysc6.png)
Now, since we are told that the length and width are reduced to 1/3 their orifinal length, the new length is:
![(l)/(3)](https://img.qammunity.org/2023/formulas/mathematics/college/mecrnfdq9rf1qcxncfwr99q6x1gzbezh3k.png)
And the new width is:
![(w)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/9n8a2pmwd1avqw1jmjg8ngoj302wiy98h8.png)
And using this length and width, the area of the new triangle will be calculated as follows:
![A_{\text{new}}=((l)/(3)*(w)/(3))/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/93s7ws6lml6x1n65vk0f0wpc7ocikjkuhm.png)
Solving the operations in the numerator:
![\begin{gathered} A_{\text{new}}=((l* w)/(3*3))/(2) \\ \\ A_{\text{new}}=((l* w)/(9))/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mk8a77rpadccmoif2tmh0k1m6iwhtlwmb2.png)
We can re-write this expression as follows:
![A_{\text{new}}=(1)/(9)((l* w)/(2))](https://img.qammunity.org/2023/formulas/mathematics/college/xv6fl3l8b1e4zsiz5gy613604iqef00jjg.png)
And we know that for this triangle the expression in parentheses is equal to:
![(l* w)/(2)=243in^2](https://img.qammunity.org/2023/formulas/mathematics/college/mzh2yf4reydyns9inyz3ca358mqliyysc6.png)
Substituting this into the expression to find the new area:
![\begin{gathered} A_{\text{new}}=(1)/(9)(243in^2) \\ \\ A_{\text{new}}=27in^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4sg6i12phixxuada9w492bzttb52869apg.png)
Answer:
the new area is 27 in^2.