S O L U T I O N :
[tex] \text{Let us assume the breadth be } \purple{\frak{a}}[/tex]
The length is 5m more than its breadth
[tex] \text{The length be} \: \green{\frak{a + 5}}[/tex]
Here we need to get both breadth and length by finding out a
[tex] \hookrightarrow \frak{{ \sf P}erimeter_{(Rectangle)} =2 (Length + Breadth)} \\ \\ \hookrightarrow \frak{150 =2 (a + 5 + a)} \\ \\ \hookrightarrow \frak{150=2 (2a + 5)} \\ \\ \hookrightarrow \frak{150=4a + 10} \\ \\ \hookrightarrow \frak{4a = 150 - 10} \\ \\ \hookrightarrow \frak{4a = 140} \\ \\ \hookrightarrow \frak{a = \cancel \frac{140}{4} } \\ \\ \star \quad \underline{\boxed{\frak{ \blue{a = 35}}}}[/tex]
[tex] \bigstar \underline{ \frak{Finding \: them}} : [/tex]
[tex] \qquad\odot \: \: \frak{Length = a + 5 = 35 + 5 = \pink{40m}} \\ \\
\qquad\odot \: \: \frak{Breadth = a= \pink{35m}}[/tex]
