According to the question:,
P’ is the image of P
![\[(3,4)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f169a581fb87d7f2926d5169a10bbd8c_l3.png "Rendered by QuickLaTeX.com")
reflected in x- axis and O’ is the image of O the origin in the line P’P.
(iii) Perimeter of POP’O’ is (
![\[4\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3c0fa656ae249b0b1e03c8cd2e7be583_l3.png "Rendered by QuickLaTeX.com")
x OP) units.
Let Q be the point of intersection of diagonals OO’ and PP’.
So, OQ =
![\[3\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-df8bcc09db12acd0f6bda88b4b8135fb_l3.png "Rendered by QuickLaTeX.com")
units and OP =
units
Hence,
![\[OP\text{ }=\text{ }\surd [{{\left( OQ \right)}^{2}}~+\text{ }{{\left( PQ \right)}^{2}}]\text{ }=\text{ }\surd \text{ }({{3}^{2}}~+\text{ }{{4}^{2}})\text{ }=\text{ }\surd \left( 9\text{ }+\text{ }16 \right)\text{ }=\text{ }\surd 25\text{ }=\text{ }5\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-79d6940bfd6eb23d0685e4a9ab506b01_l3.png "Rendered by QuickLaTeX.com")
units
Thus, the perimeter of POP’O’ =
![\[4\text{ }\times \text{ }5\text{ }=\text{ }20\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-29e571d163a3471875e7ee100627bc10_l3.png "Rendered by QuickLaTeX.com")
units
