question_answer 1)

\[\text{ }\!\!\Delta\!\!\text{ ABC}\]is isosceles with \[\text{AB}=\text{AC}=\text{7}.\text{5 cm}\], and \[\text{BC}=\text{9}~\text{cm}\]. The height AD from A to BC, is 6 cm. Find the area of \[\Delta \Alpha \Beta C\]. What will be the height from C to AB i.e., CE?

Answer:

                Area of \[\text{ }\!\!\Delta\!\!\text{ ABC}\]        \[\text{= }\frac{\text{1}}{\text{2}}\text{(bese  }\!\!\times\!\!\text{  height)}\]                                                 \[\text{= }\frac{\text{1}}{\text{2}}\text{(BC  }\!\!\times\!\!\text{  AD)}\]                                                 \[\text{=}\frac{\text{1}}{\text{2}}\text{(9 }\!\!\times\!\!\text{ 6) = 27 c}{{\text{m}}^{\text{2}}}\]                 Hence, Area of \[\text{ }\!\!\Delta\!\!\text{ ABC}\]is \[\text{27 c}{{\text{m}}^{2}}\]                 Again, Area of \[\Delta \Alpha \Beta C\] \[\text{= }\frac{\text{1}}{\text{2}}\text{(Base  }\!\!\times\!\!\text{  height)}\]                                                 \[\text{= }\frac{\text{1}}{\text{2}}\text{(AB  }\!\!\times\!\!\text{  CE)}\]                 \[\Rightarrow \]                               \[\text{27 = }\frac{\text{1}}{\text{2}}\text{(7}\text{.5  }\!\!\times\!\!\text{  CE)}\]                 \[\Rightarrow \]                               \[\text{CE = }\frac{\text{27 }\!\!\times\!\!\text{ 2}}{\text{7}\text{.5}}\text{=7}\text{.2 cm}\] Hence, the height from C to AB i.e., CE is 7.2 cm