1. If the sides of a triangle are long, determine whether the triangle is a right-angled triangle. Vinay 3 years ago Solution: We have, Sides of triangle as On finding their squares, we get So, by converse of Pythagoras theorem the given sides cannot be the sides of a right triangle. .
![\[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> AB\text{ }=\text{ }3\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> BC\text{ }=\text{ }4\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> AC\text{ }=\text{ }6\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bc43211d54d71c9d64870d10408fc87f_l3.png "Rendered by QuickLaTeX.com")
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![\[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> A{{B}^{2}}~=\text{ }{{3}^{2}}~=\text{ }9 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> B{{C}^{2~}}=\text{ }{{4}^{2}}~=\text{ }16 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> A{{C}^{2}}~=\text{ }{{6}^{2}}~=\text{ }36 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> Since,\text{ }A{{B}^{2}}~+\text{ }B{{C}^{2}}~\ne \text{ }A{{C}^{2}} \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c232b44bc7d5682a420b00d5e0c517d4_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{3}\text{ }\mathbf{cm},\text{ }\mathbf{4}\text{ }\mathbf{cm},\text{ }\mathbf{and}\text{ }\mathbf{6}\text{ }\mathbf{cm}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4e7f08d95ac5afd7a7bf17c308701538_l3.png "Rendered by QuickLaTeX.com")
