 Hi and welcome to the session. Let us discuss the following question, question size. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure 12.24. Find the area of the design. This is the given figure 12.24. Let us now start with the solution. We are given a circular table cover of radius 32 cm. So we can write radius of given circle that is r is equal to 32 cm. Now we know area of circle is equal to pi r square where r is the radius of the circle. Now area of this given circle is equal to pi multiplied by square of 32 cm square. Now substituting 22 upon 7 for pi we get area of given circle is equal to 22 upon 7 multiplied by 32 multiplied by 32 cm square. Now simply find we get 22528 upon 7 cm square. Now clearly we can see to find the area of the design we will subtract area of equilateral triangle ABC from area of the given circle. So now we will find out area of triangle ABC. We know area of equilateral triangle is equal to root 3 upon 4 multiplied by side square. Now we will find out side of triangle ABC. Now we know triangle ABC is the equilateral triangle. So angle BAC is equal to 60 degrees. We know all angles of equilateral triangle measure 60 degrees. Now let us do one construction. Mark center of circle as O and join radius OB and OC. We know angle BAC is equal to 60 degrees. So this implies angle BOC is equal to 120 degrees. Let us recall the property that angle subtended by an arc of the circle at the center is double the angle subtended by it on remaining portion of the circle. So we can write angle BOC is equal to twice angle BAC. Now we know angle BAC is equal to 60 degrees. This implies angle BOC is equal to 2 multiplied by 60 degrees. Now we get angle BOC is equal to 120 degrees. Now draw OM perpendicular to BC. Here OM is perpendicular to BC. Now we will consider triangle VOM and triangle COM. Now clearly we can see this angle is equal to this angle. OM is common and OB is equal to OC as they are radius of the same circle. So these two triangles are congruent by RHS congruency rule. Now we can write in triangle BOM and COM OB is equal to OC as they are radius of same circle. OM is equal to OM common. Angle OMB is equal to angle OMC as they both are right angles. Now we conclude triangle BOM is congruent to triangle COM by RHS congruency condition. Now we know congruent parts of congruent triangles are equal. So BM is equal to CM by CPCT and angle BOM is equal to angle COM by CPCT. We have already shown that angle BOC is equal to 120 degrees and we also know that angle BOC is equal to angle BOM plus angle COM. Now here we have shown that by CPCT angle BOM is equal to angle COM. So we can write angle BOC is equal to angle BOM plus angle BOM. We know angle COM is equal to angle BOM. So we can substitute angle BOM for angle COM. Now this implies angle BOC is equal to twice angle BOM. Now we know angle BOC is equal to 120 degrees. So substituting 120 degrees for angle BOC we get 120 degrees is equal to twice angle BOM. Now dividing both the sides by 2 we get 60 degrees is equal to angle BOM or we can simply write it as angle BOM is equal to 60 degrees. Now we get this angle is equal to 60 degrees in this triangle. Now clearly we can see this is a right triangle and sine 60 degrees is equal to BM upon OB. We know sine theta is equal to perpendicular upon hypotenuse. Here with respect to angle BOM, BM is the perpendicular and OB is the hypotenuse. So we can write in triangle BOM sine 60 degrees is equal to BM upon OB. Now we know sine 60 degrees is equal to root 3 upon 2 and we are given that OB that is radius of the given circle is equal to 32 centimeters. So substituting corresponding value of OB and sine 60 degrees in this expression we get this expression. Now multiplying both the sides of this expression by 32 we get 32 multiplied by root 3 upon 2 is equal to BM. Now we will cancel common factor 2 from numerator and denominator both and we get 16 root 3 is equal to BM or we can simply write it as BM is equal to 16 root 3 centimeters. Now we know that BM is equal to MC is equal to 16 root 3 centimeters. BM is equal to MC we have already shown it above. So we get BM is equal to MC is equal to 16 root 3 centimeters. Now clearly we can see BC is equal to BM plus MC. Now substituting corresponding values of BM and MC in this expression we get 16 root 3 plus 16 root 3 centimeters is equal to BC. Now this implies BC is equal to 32 root 3 centimeters. Now this is the required side of equilateral triangle BC. So now we will find out area of equilateral triangle ABC it is equal to root 3 upon 4 multiplied by square of BC. Now we know BC is equal to 32 root 3 centimeters. So we will substitute 32 root 3 for BC in this expression and we get root 3 upon 4 multiplied by 32 root 3 centimeters square. Now simply find we get area of equilateral triangle ABC is equal to 768 root 3 centimeters square. Now clearly we can see area of the design is equal to area of given circle minus area of equilateral triangle ABC. So we can write area of design is equal to area of given circle minus area of triangle ABC. Now we have already shown that area of given circle is equal to 225 28 upon 7 centimeters square and area of triangle ABC is 768 root 3 centimeters square. So we get area of design is equal to 225 28 upon 7 minus 768 root 3 centimeters square. So this is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.