In this problem, we have a diagram of a paper airplane, and we must compute its area and perimeter.
The area of a triangle with height h and base b is:
[tex]A=\sqrt{s\cdot(s-a)\cdot(s-b)\cdot(s-c)}.[/tex]Where:
• a, b and c are the lengths of the sides of the triangle,
,• s is the semi perimeter, which is given by:
[tex]s=\frac{a+b+c}{2}.[/tex](1) The area of the airplane is given by:
[tex]A=A_b-A_s.[/tex]Where:
• Ab is the area of the bigger triangle,
,• As is the area of the smaller triangle.
Bigger triangle
The bigger triangle has sides a = b = 5 cm and c = 4 cm, is semi perimeter is given by:
[tex]s=\frac{5cm+5cm+4cm}{2}=7cm.[/tex]The area of the bigger triangle is:
[tex]A_b=\sqrt{7cm\cdot(7cm-5cm)\cdot(7cm-5cm)\cdot(7cm-4cm)}=2\sqrt{21}\text{ }cm^2.[/tex]Smaller triangle
The smaller triangle has sides a = b = 3 cm and c = 4 cm, is semi perimeter is given by:
[tex]s=\frac{3cm+3cm+4cm}{2}=5cm.[/tex]The area of the smaller triangle is:
[tex]A_s=\sqrt{5cm\cdot(5cm-3cm)\cdot(5cm-3cm)\cdot(5cm-4cm)}=2\sqrt{5}\text{ }cm^2.[/tex]Replacing these results in the formula for the area of the airplane, we get:
[tex]A=2\sqrt{21}\text{ }cm^2-2\sqrt{5}\text{ }cm^2=2(\sqrt{21}-\sqrt{5})\text{ }cm^2\cong4.69\text{ }cm^2.[/tex](2) The perimeter of the airplane is the sum of the length of the sides:
[tex]P=5cm+5cm+3cm+3cm=16cm.[/tex]AnswerThe area and perimeter of the airplane are:
[tex]\begin{gathered} A=2(\sqrt{21}-\sqrt{5})\cong4.61cm^2, \\ P=16\text{ }cm. \end{gathered}[/tex]