Answer:
Hence, the perimeter of the figure is:
(√29+ 2√5 +√13+ √17+√73) units=26.1 units.
Step-by-step explanation:
The perimeter of the figure is the length of all the line segments.
i.e. Line segment QR,RS,ST,TU and QU.
We know that the distance between two points (a,b) and (c,d) is given as:
[tex]\sqrt{(c-a)^2+(d-b)^2}[/tex].
- Length of segment Q(2,0)R(4,5) is equal to the distance between these the coordinates of these vertex.
i.e. distance between (2,0) and (4,5) is:
[tex]\sqrt{(4-2)^2+(5-0)^2}\\\\=\sqrt{2^2+5^2}\\\\=\sqrt{4+25}\\\\=\sqrt{29}[/tex]
Hence length of line segment QR is: √29 units.
- Length of segment R(4,5)S(8,7) is equal to the distance between these the coordinates of these vertex.
i.e. the distance between (4,5) and (8,7) is:
[tex]\sqrt{(8-4)^2+(7-5)^2}\\\\=\sqrt{4^2+2^2}\\\\=\sqrt{16+4}\\\\=\sqrt{20}=2\sqrt{5}[/tex]
Hence length of line segment RS is: 2√5 units.
- Length of segment S(8,7)T(6,4) is equal to the distance between these the coordinates of these vertex.
i.e. the distance between (8,7) and (6,4) is:
[tex]\sqrt{(8-6)^2+(7-4)^2}\\\\=\sqrt{2^2+3^2}\\\\=\sqrt{4+9}\\\\=\sqrt{13}[/tex]
Hence length of line segment RS is: √13 units.
- Length of segment T(6,4)U(10,3) is equal to the distance between these the coordinates of these vertex.
i.e. the distance between (6,4) and (10,3) is:
[tex]\sqrt{(10-6)^2+(3-4)^2}\\\\=\sqrt{4^2+(-1)^2}\\\\=\sqrt{16+1}\\\\=\sqrt{17}[/tex]
Hence length of line segment RS is: √17 units.
- Length of segment U(10,3)Q(2,0) is equal to the distance between these the coordinates of these vertex.
i.e. the distance between (10,3) and (2,0) is:
[tex]\sqrt{(10-2)^2+(3-0)^2}\\\\=\sqrt{8^2+3^2}\\\\=\sqrt{64+9}\\\\=\sqrt{73}[/tex]
Hence length of line segment RS is: √73 units.
Hence, the perimeter of the figure is:
(√29+ 2√5 +√13+ √17+√73) units=26.1 units
