[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]
Since, the Area of smaller rectangle is 35,
Let's calculate it's length ~
[tex]\qquad \sf \dashrightarrow \:a = l \cdot w[/tex]
[tex]\qquad \sf \dashrightarrow \:35 = l \cdot5[/tex]
[tex]\qquad \sf \dashrightarrow \:l = 35 \div 5[/tex]
[tex]\qquad \sf \dashrightarrow \:l = 7 \: ft[/tex]
And the two rectangles are similar ~ so the ratio of their sides are equal as well ~
Assume length of greater rectangle be x
[tex]\qquad \sf \dashrightarrow \: \dfrac{25}{5} = \dfrac{x}{7} [/tex]
[tex]\qquad \sf \dashrightarrow \:x = 5 \times 7[/tex]
[tex]\qquad \sf \dashrightarrow \:x = 35[/tex]
Now, since the length and width of the greater rectangle is known, let's find it's Area ~
[tex]\qquad \sf \dashrightarrow \:a = l \cdot w[/tex]
[tex]\qquad \sf \dashrightarrow \:a = 35 \sdot25[/tex]
[tex]\qquad \sf \dashrightarrow \:a = 875 \: \: ft {}^{2} [/tex]
So, Area of the greater rectangle is 875 ft²
