question_answer 43)

In the following figures, find the area of the shaded portions. (i) (ii)

Answer:

                (i)  Area of the rectangle \[\text{ABCD}=l\times b\] \[=\text{18 cm}\times \text{1}0\text{ cm}=\text{18}0\text{ c}{{\text{m}}^{\text{2}}}\] Area of the right triangle EAF          \[\text{= }\frac{\text{1}}{\text{2}}\text{ (base  }\!\!\times\!\!\text{  height)}\] \[=\frac{1}{2}(6\times 1\text{0) c}{{\text{m}}^{\text{2}}}\text{= 30 c}{{\text{m}}^{\text{2}}}\] Area of the right triangle CBE \[\text{= }\frac{\text{1}}{\text{2}}\text{(base  }\!\!\times\!\!\text{  height)}\] \[=\frac{1}{2}(8\times 10\text{) c}{{\text{m}}^{\text{2}}}\text{= 40 c}{{\text{m}}^{\text{2}}}\] \[\therefore \]                  Area of shaded portion CDFE       = Area of the rectangle ABCD  ? [Area of the right triangle EAF ? Area of the right triangle CBE] \[=\text{18}0\text{ c}{{\text{m}}^{\text{2}}}-\left[ \text{3}0\text{ c}{{\text{m}}^{\text{2}}}+\text{4}0\text{ c}{{\text{m}}^{\text{2}}} \right]\] \[=\text{ 18}0\text{ c}{{\text{m}}^{\text{2}}}-\text{7}0\text{ c}{{\text{m}}^{\text{2}}}=\text{11}0\text{ c}{{\text{m}}^{\text{2}}}.\] (ii) Area of the square PQRS           \[=\text{side}\times \text{side}=\text{2}0\times \text{2}0\text{ c}{{\text{m}}^{\text{2}}}=\text{4}00\text{ c}{{\text{m}}^{\text{2}}}\] Area of the right triangle TSU          \[\text{= }\frac{\text{1}}{\text{2}}\text{(base  }\!\!\times\!\!\text{  height)}\] \[\text{= }\frac{\text{1}}{\text{2}}\text{(10  }\!\!\times\!\!\text{  10) c}{{\text{m}}^{\text{2}}}\text{= 50c}{{\text{m}}^{\text{2}}}\] Area of the right triangle URQ          \[\text{= }\frac{\text{1}}{\text{2}}\text{(base  }\!\!\times\!\!\text{  height)}\] \[\text{= }\frac{\text{1}}{\text{2}}\text{(10  }\!\!\times\!\!\text{  20) c}{{\text{m}}^{\text{2}}}\text{= 100 c}{{\text{m}}^{\text{2}}}\] Area of the right triangle TPQ          \[\text{= }\frac{\text{1}}{\text{2}}\text{(base  }\!\!\times\!\!\text{  height)}=\frac{\text{1}}{\text{2}}\text{(P T  }\!\!\times\!\!\text{  PQ)}\] \[\text{PT = PS -TS = QR-TS}\] \[=\text{2}0\text{ cm}-\text{1}0\text{ cm}=\text{1}0\text{ cm}\] \[\text{= }\frac{\text{1}}{\text{2}}\text{(10  }\!\!\times\!\!\text{  20) c}{{\text{m}}^{\text{2}}}\text{= 100 c}{{\text{m}}^{\text{2}}}\] \[\therefore \] Area of the shaded portion UQT      = Area of the square PQRS ? [Area of the right triangle TSU + Area of the right triangle URQ + Area of the right triangle TPQ] \[=\text{4}00\text{ c}{{\text{m}}^{\text{2}}}-[\text{5}0\text{ c}{{\text{m}}^{\text{2}}}+\text{1}00\text{ c}{{\text{m}}^{\text{2}}}+\text{1}00\text{ c}{{\text{m}}^{\text{2}}}]\] \[=150\,c{{m}^{2}}\].