Answer:
guessing that the area of the inner rectangle is 10m²
and the other rectangle is 16m*10m= 160m²
we can say that
(16-x) * (10-x) = 10
for the area of the inner rectangle
let's calculate the lefthand side of the equation
160 - 16x -10x + x² = 10
160 -26x +x² = 10
subtract 10 on both sides
150 -26x +x² = 0
rearrange
x² -26x + 150 = 0
looks like something we can throw the pq formula on to solve for x
with p = -26 and q = 150
[tex]x = - \frac{ - 26}{2} \frac{ + }{ - } \sqrt{( \frac{ - 26}{2})^{2} - 150 } [/tex]
[tex]x = 13 \frac{ + }{ - } \sqrt{169 - 150 } [/tex]
[tex]x = 13 \frac{ + }{ - } \sqrt{19 } [/tex]
[tex]x1 = 8.64[/tex]
[tex]x2 =17.36 [/tex]
we need to interpret this solutions
but to be sure on this, I need the unreadable text of the photographed problem
