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Question

In given figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. (Use $$\pi \, = \, 22/7$$)
973843_8129b0620af148dc896fea005f80bf38.png


Solution


In right $$\triangle AED$$, using Pythagoras theorem,

$${ AD }^{ 2 }={ AE }^{ 2 }+{ DE }^{ 2 }={ 9 }^{ 2 }+{ 12 }^{ 2 }=81+144=225cm$$

$$\Rightarrow \quad AD=\sqrt { 225 } =15cm=diameter$$

Required area = ar. ABCD + ar. semi-circle - ar $$\triangle AED=\left( 20\times 15 \right) +\dfrac { 1 }{ 2 } \times \dfrac { 22 }{ 7 } \times { \left( \dfrac { 15 }{ 2 }  \right)  }^{ 2 }-\dfrac { 1 }{ 2 } \times 9\times 12$$

$$=300+88.3928-54=334.3928{ cm }^{ 2 }$$

952917_973843_ans_a8e769c9b0544897b007489743a836f1.png

Mathematics

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