question_answer 15)

\[\Delta \Alpha \Beta C\] is right-angled at A. AD is perpendicular to BC. If \[\text{AB}=\text{5 cm},\text{ }\]\[\text{BC}=\text{23 cm}\] and \[\text{AC}=\text{12 cm}\]. Find the area of \[\Delta \Alpha \Beta C\]. Also find the length of AD.

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 \[\frac{\text{60}}{\text{13}}\text{ cm}\]