Answer:
Perimeter = 10√2 cm ≈ 14.14 cm (2 d.p.)
Area = 12 cm²
Step-by-step explanation:
To calculate the perimeter and area of rectangle EFGH, we first need to find its width and length.
From inspection of the given diagram:
- The width of the rectangle EFGH is HE (the hypotenuse of right triangle HAE).
- The length of rectangle EFGH is EF (the hypotenuse of right triangle EBF).
As the width and length of the rectangle are the hypotenuse of right triangles, we can use Pythagoras Theorem.
[tex]\boxed{\begin{minipage}{9 cm}\underline{Pythagoras Theorem} \\\\$a^2+b^2=c^2$\\\\where:\\ \phantom{ww}$\bullet$ $a$ and $b$ are the legs of the right triangle. \\ \phantom{ww}$\bullet$ $c$ is the hypotenuse (longest side) of the right triangle.\\\end{minipage}}[/tex]
If AE = AH = 2 cm, then triangle HAE is a right triangle with both legs measuring 2 cm. To find the hypotenuse HE, use Pythagoras Theorem.
[tex]\begin{aligned}AE^2+AH^2&=HE^2\\2^2+2^2&=HE^2\\4+4&=HE^2\\HE^2&=8\\HE&=\sqrt{8}\\HE&=2\sqrt{2}\; \sf cm\end{aligned}[/tex]
As ABCD is a square where AB = 5 cm, and given AE = AH = FC = GC = 2 cm, then:
If EB = BF = 3 cm, then triangle EBF is a right triangle with both legs measuring 3 cm in length. To find the hypotenuse EF, use Pythagoras Theorem.
[tex]\begin{aligned}EB^2+BF^2&=EF^2\\3^2+3^2&=EF^2\\9+9&=EF^2\\EF^2&=18\\EF&=\sqrt{18}\\EF&=3\sqrt{2}\; \sf cm\end{aligned}[/tex]
Therefore, the dimensions of rectangle EFGH are:
- width HE = 2√2 cm
- length EF = 3√2 cm
The perimeter of a rectangle is twice the sum of its width and length.
Therefore, the perimeter of rectangle EFGH is:
[tex]\begin{aligned}\textsf{Perimeter of rectangle $EFGH$}&=2(\sf width + length)\\&=2(2\sqrt{2}+3\sqrt{2})\\&=2(5\sqrt{2})\\&=10\sqrt{2}\\&\approx 14.14\; \sf cm\; (2\;d.p.)\end{aligned}[/tex]
The area of a rectangle is the product of its width and length.
Therefore, the area of rectangle EFGH is:
[tex]\begin{aligned}\textsf{Area of rectangle $EFGH$}&=\sf width \cdot length\\&=2\sqrt{2}\cdot 3\sqrt{2}\\&=6\sqrt{2}\sqrt{2}\\&=6 \cdot 2\\&=12\;\sf cm^2\end{aligned}[/tex]
