A thin brass sheet at 10°C and a thin steel sheet at 20°C have the same surface area. The common temperature at which both would have the same area is (coefficient of linear expansion for brass and steel are 19 x 10^-6/°C and 11 x 10^-6/°C respectively)

The area of brass sheet at 10°C = a₁ b₁
Area of steel sheet at 20°C = a₂ b₂
Given that    ${\mathrm{a}}_1{\mathrm{b}}_1={\mathrm{a}}_2{\mathrm{b}}_2 ....\left(1\right)$
Let the common temperature be T. Then at temperature T, the lengths of the two sheets are respectively
${\mathrm{a}}_1\left(1+{\mathrm{\alpha }}_{\mathrm{B}}(10-\mathrm{T})\right)\text{ and }{\mathrm{a}}_2\left(1+{\mathrm{\alpha }}_{\mathrm{S}}(20-\mathrm{T})\right)$
Now,  ${\mathrm{a}}_1\left(1+{\mathrm{\alpha }}_{\mathrm{B}}(10-\mathrm{T})\right)\times {\mathrm{b}}_1\left(1+{\mathrm{\alpha }}_{\mathrm{B}}(10-\mathrm{T}\right))$
            $={\mathrm{a}}_2\left(1+{\mathrm{\alpha }}_{\mathrm{S}}(20-\mathrm{T})\right)\times {\mathrm{b}}_2\left(1+{\mathrm{\alpha }}_{\mathrm{S}}(20-\mathrm{T}\right))$
or   ${\left(1+{\mathrm{\alpha }}_{\mathrm{B}}(10-\mathrm{T})\right)}^2={\left(1+{\mathrm{\alpha }}_{\mathrm{S}}(20-\mathrm{T})\right)}^2\because \left({\mathrm{a}}_1{\mathrm{b}}_1={\mathrm{a}}_2{\mathrm{b}}_2\right)$
Substituting the values of ${\mathrm{\alpha }}_{\mathrm{B}}\text{ and }{\mathrm{\alpha }}_{\mathrm{S}}$ and solving we get
$\mathrm{T}=-3\cdot {75}^{\circ }\mathrm{C}$