Answer:
Area of \[\Delta \text{ABC}\] \[\text{=}\frac{\text{1}}{\text{2}}\text{(base }\!\!\times\!\!\text{ height)}\] \[\text{=}\frac{\text{1}}{\text{2}}\text{ (AB }\!\!\times\!\!\text{ AC)}\] \[\text{= }\frac{\text{1}}{\text{2}}\text{ (5 }\!\!\times\!\!\text{ 12) = 30 c}{{\text{m}}^{\text{2}}}\] Hence, the area of \[\text{ }\!\!\Delta\!\!\text{ ABC}\] is \[\text{3}0\text{ c}{{\text{m}}^{\text{2}}}\]. Area of \[\text{ }\!\!\Delta\!\!\text{ ABC}\] \[\text{= }\frac{\text{1}}{\text{2}}\text{(base }\!\!\times\!\!\text{ height)}\] \[\text{= }\frac{\text{1}}{\text{2}}\text{ (BC }\!\!\times\!\!\text{ AD)}\] \[\Rightarrow \] \[\text{30 =}\frac{\text{1}}{\text{2}}\text{ (13 }\!\!\times\!\!\text{ AD)}\] \[\Rightarrow \] \[\text{13 + AD = 30 }\!\!\times\!\!\text{ 2}\] \[\Rightarrow \] \[\text{AD = }\frac{\text{30 }\!\!\times\!\!\text{ 2}}{\text{13}}\text{ = }\frac{\text{60}}{\text{13}}\text{ cm}\] Hence, the length of AD is \[\frac{\text{60}}{\text{13}}\text{ cm}\]
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