Answer:
(a) In the right triangle ABC, \[\because \] \[\angle B={{90}^{\text{o}}}\] |Given \[\therefore \] By Pythagoras theorem \[A{{C}^{2}}=A{{B}^{2}}=B{{C}^{2}}\] \[\Rightarrow \] \[A{{C}^{2}}={{6}^{2}}+{{8}^{2}}\] \[\Rightarrow \] \[A{{C}^{2}}=36+64\] \[\Rightarrow \] \[A{{C}^{2}}=100\] \[\Rightarrow \] \[AC=\sqrt{100}\]
Therefore, . Hence, AC is equal to 10 cm. (b) In the right triangle ABC, \[\because \,\,\,\,\,\angle B={{90}^{\text{o}}}\] |Given \[\therefore \] By Pythagoras theorem \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] \[\Rightarrow \] \[{{13}^{2}}=A{{B}^{2}}+{{5}^{2}}\] \[\Rightarrow \] \[169=A{{B}^{2}}+25\] \[\Rightarrow \] \[A{{B}^{2}}=169-25\] \[\Rightarrow \] \[A{{B}^{2}}=144\] \[\Rightarrow \] \[AB=\sqrt{144}\] 100 1 ?1 20 00 ? 0 0
Therefore, \[\sqrt{144}\,=12\]. Hence, AB is equal to 12 cm. 12 1 \[\overline{1}\] \[\overline{44}\] ?1 22 44 ?44 0
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