![\[\begin{array}{*{35}{l}}</strong> <!-- /wp:paragraph --> <!-- wp:paragraph --> <strong> \left( \mathbf{i} \right)\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{7}\text{ }\mathbf{cm},\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{24}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{c}\text{ }=\text{ }\mathbf{25}\text{ }\mathbf{cm} \\</strong> <!-- /wp:paragraph --> <!-- wp:paragraph --> <strong> \left( \mathbf{ii} \right)\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{9}\text{ }\mathbf{cm},\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{16}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{c}\text{ }=\text{ }\mathbf{18}\text{ }\mathbf{cm} \\</strong> <!-- /wp:paragraph --> <!-- wp:paragraph --> <strong> \left( \mathbf{iii} \right)\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{1}.\mathbf{6}\text{ }\mathbf{cm},\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{3}.\mathbf{8}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{c}\text{ }=\text{ }\mathbf{4}\text{ }\mathbf{cm} \\</strong> <!-- /wp:paragraph --> <!-- wp:paragraph --> <strong> \left( \mathbf{iv} \right)\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{8}\text{ }\mathbf{cm},\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{c}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm} \\</strong> <!-- /wp:paragraph --> <!-- wp:paragraph --> <strong>\end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-35f2fe1d6e2f2bef57a02b3e77d7ac8b_l3.png "Rendered by QuickLaTeX.com")
![\[\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1833cc7760eba792ec467db2c623cfe8_l3.png "Rendered by QuickLaTeX.com")
Solutions:
(i) Given,
![\[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> a\text{ }=\text{ }7\text{ }cm,\text{ }b\text{ }=\text{ }24\text{ }cm\text{ }and\text{ }c\text{ }=\text{ }25\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \therefore {{a}^{2}}~=\text{ }49,\text{ }{{b}^{2}}~=\text{ }576\text{ }and\text{ }{{c}^{2}}~=\text{ }625 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> Since,\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~=\text{ }49\text{ }+\text{ }576\text{ }=\text{ }625\text{ }=\text{ }{{c}^{2}} \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a43a866c93a0fb6dd63a2f705ef171e0_l3.png "Rendered by QuickLaTeX.com")
Then, by converse of Pythagoras theorem
The given sides are of a right triangle.
(ii) Given,
![\[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> a\text{ }=\text{ }9\text{ }cm,\text{ }b\text{ }=\text{ }16\text{ }cm\text{ }and\text{ }c\text{ }=\text{ }18\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \therefore {{a}^{2}}~=\text{ }81,\text{ }{{b}^{2}}~=\text{ }256\text{ }and\text{ }{{c}^{2}}~=\text{ }324 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> Since,\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~=\text{ }81\text{ }+\text{ }256\text{ }=\text{ }337\text{ }\ne \text{ }{{c}^{2}} \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-08e37c5c1f9e0e6d9e8e5304bb2db473_l3.png "Rendered by QuickLaTeX.com")
Then, by converse of Pythagoras theorem
The given sides cannot be of a right triangle.
(iii) Given,
![\[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> a\text{ }=\text{ }1.6\text{ }cm,\text{ }b\text{ }=\text{ }3.8\text{ }cm\text{ }and\text{ }C\text{ }=\text{ }4\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \therefore {{a}^{2}}~=\text{ }2.56,\text{ }{{b}^{2}}~=\text{ }14.44\text{ }and\text{ }{{c}^{2}}~=\text{ }16 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-80c932121a25ce1cabbdaee70c973aa7_l3.png "Rendered by QuickLaTeX.com")
Since,
![\[{{a}^{2}}~+\text{ }{{b}^{2}}~=\text{ }2.56\text{ }+\text{ }14.44\text{ }=\text{ }17\text{ }\ne \text{ }{{c}^{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-343ad649396f49b03209fc595236dd40_l3.png "Rendered by QuickLaTeX.com")
Then, by converse of Pythagoras theorem
The given sides cannot be of a right triangle.
(iv) Given,
![\[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> a\text{ }=\text{ }8\text{ }cm,\text{ }b\text{ }=\text{ }10\text{ }cm\text{ }and\text{ }C\text{ }=\text{ }6\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \therefore {{a}^{2}}~=\text{ }64,\text{ }{{b}^{2}}~=\text{ }100\text{ }and\text{ }{{c}^{2}}~=\text{ }36 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4d2d0e17c717dd0213693d99ddb41508_l3.png "Rendered by QuickLaTeX.com")
Since,
![\[{{a}^{2}}~+\text{ }{{c}^{2}}~=\text{ }64\text{ }+\text{ }36\text{ }=\text{ }100\text{ }=\text{ }{{b}^{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-daf2af62ef50ee4a52abc0ad1b3e5ff9_l3.png "Rendered by QuickLaTeX.com")
Then, by converse of Pythagoras theorem
The given sides are of a right triangle