Consider a isosceles right triangle ABC,

Consider AB = x is the length of a leg of a isosceles right triangle an AC = y is a hypotenuse. Given that, \(\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={2}\) m/s

a) We know that area of a right triangle \(\displaystyle=\frac{{1}}{{2}}\times\) base \(\displaystyle\times\) height

\(\displaystyle{A}=\frac{{1}}{{2}}{\left({x}\right)}{\left({x}\right)}\)

\(\displaystyle{A}=\frac{{1}}{{2}}{x}^{{2}}\) (1)

Differentiate A with respect to t,

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\frac{{d}}{{{\left.{d}{t}\right.}}}{\left(\frac{{1}}{{2}}{x}^{{2}}\right)}\)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\frac{{1}}{{2}}{\left({2}{x}\right)}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}\)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={x}\frac{{{\left.{d}{x}\right.}}}{{\left.{d}{t}\right.}}{\left({2}\right)}\)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={\left({2}\right)}{\left({2}\right)}\) [Given]

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={4}\)

Hence area of the triangle changing at \(\displaystyle{4}\frac{{m}^{{2}}}{{s}}\).

b) In isosceles right triangle ABC,

\(\displaystyle{A}{C}^{{2}}={A}{B}^{{2}}+{B}{C}^{{2}}\) [Using pythoras theorem]

\(\displaystyle{y}^{{2}}={x}^{{2}}+{x}^{{2}}\)

\(\displaystyle{y}^{{2}}={2}{x}^{{2}}\)

\(\displaystyle{1}={2}{x}^{{2}}\) [Given that hypotenuse y=1m]

\(\displaystyle{x}^{{2}}=\frac{{1}}{{2}}\)

\(\displaystyle{x}=\frac{{1}}{\sqrt{{2}}}\)

Substtute \(\displaystyle{x}=\frac{{1}}{\sqrt{{2}}}{\quad\text{and}\quad}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={2}\) in (2)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={\left(\frac{{1}}{\sqrt{{2}}}\right)}{\left({2}\right)}\)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}\)

Hence area of the triangle changing at sqrt2 m^2/sZSK.

c) In isosceles right triangle ABC,

\(\displaystyle{A}{C}^{{2}}={A}{B}^{{2}}+{B}{C}^{{2}}\) [Using pythoras theorem]

\(\displaystyle{y}^{{2}}={x}^{{2}}+{x}^{{2}}\)

\(\displaystyle{y}^{{2}}={2}{x}^{{2}}\)

\(\displaystyle{y}=\sqrt{{2}}{x}\)

Differentiate y with respect to t,

\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\frac{{d}}{{{\left.{d}{t}\right.}}}{\left(\sqrt{{2}}{x}\right)}\)

\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}\)

\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}\) (2)

\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}={2}\sqrt{{2}}\)

Hence hypotenuse change at \(\displaystyle{2}\sqrt{{2}}\) m/s

Consider AB = x is the length of a leg of a isosceles right triangle an AC = y is a hypotenuse. Given that, \(\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={2}\) m/s

a) We know that area of a right triangle \(\displaystyle=\frac{{1}}{{2}}\times\) base \(\displaystyle\times\) height

\(\displaystyle{A}=\frac{{1}}{{2}}{\left({x}\right)}{\left({x}\right)}\)

\(\displaystyle{A}=\frac{{1}}{{2}}{x}^{{2}}\) (1)

Differentiate A with respect to t,

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\frac{{d}}{{{\left.{d}{t}\right.}}}{\left(\frac{{1}}{{2}}{x}^{{2}}\right)}\)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\frac{{1}}{{2}}{\left({2}{x}\right)}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}\)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={x}\frac{{{\left.{d}{x}\right.}}}{{\left.{d}{t}\right.}}{\left({2}\right)}\)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={\left({2}\right)}{\left({2}\right)}\) [Given]

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={4}\)

Hence area of the triangle changing at \(\displaystyle{4}\frac{{m}^{{2}}}{{s}}\).

b) In isosceles right triangle ABC,

\(\displaystyle{A}{C}^{{2}}={A}{B}^{{2}}+{B}{C}^{{2}}\) [Using pythoras theorem]

\(\displaystyle{y}^{{2}}={x}^{{2}}+{x}^{{2}}\)

\(\displaystyle{y}^{{2}}={2}{x}^{{2}}\)

\(\displaystyle{1}={2}{x}^{{2}}\) [Given that hypotenuse y=1m]

\(\displaystyle{x}^{{2}}=\frac{{1}}{{2}}\)

\(\displaystyle{x}=\frac{{1}}{\sqrt{{2}}}\)

Substtute \(\displaystyle{x}=\frac{{1}}{\sqrt{{2}}}{\quad\text{and}\quad}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={2}\) in (2)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={\left(\frac{{1}}{\sqrt{{2}}}\right)}{\left({2}\right)}\)

\(\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}\)

Hence area of the triangle changing at sqrt2 m^2/sZSK.

c) In isosceles right triangle ABC,

\(\displaystyle{A}{C}^{{2}}={A}{B}^{{2}}+{B}{C}^{{2}}\) [Using pythoras theorem]

\(\displaystyle{y}^{{2}}={x}^{{2}}+{x}^{{2}}\)

\(\displaystyle{y}^{{2}}={2}{x}^{{2}}\)

\(\displaystyle{y}=\sqrt{{2}}{x}\)

Differentiate y with respect to t,

\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\frac{{d}}{{{\left.{d}{t}\right.}}}{\left(\sqrt{{2}}{x}\right)}\)

\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}\)

\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}\) (2)

\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}={2}\sqrt{{2}}\)

Hence hypotenuse change at \(\displaystyle{2}\sqrt{{2}}\) m/s