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| 1. |
Find area of isosceles triangleabc in which ab=AC=6cm is inscribed ina a circle of radius 9cm |
| Answer» Let O be the centre of the circle and let P be the mid-point of BC. Then ,{tex}O P \\perp B C{/tex}Since\xa0∆ ABC is isosceles and P is the mid-point of BC. Therefore, {tex}A P \\perp B C{/tex}\xa0as median from the vertex in an isosceles triangle is perpendicular to the base.Let A P = x and PB = CP = y.Applying Pythagoras theorem in\xa0∆ APB and ∆ OPB, we have AB2 = BP2 + AP2and, OB2 = OP2 + BP2{tex}\\Rightarrow{/tex} 36 = y2 + x2 ..........(i)and, 81 = (9 - x){tex}^2{/tex} + y{tex}^2{/tex} ...(ii){tex}\\Rightarrow{/tex}\xa081 - 36 = [(9 - x)2 + y2\xa0\xa0]- (y2 + x2 )[Subtracting (i) from (ii)]{tex}\\Rightarrow{/tex}\xa045 = 81 - 18x{tex}\\Rightarrow{/tex}\xa0x = 2 cm, therefore AP = x = 2 cm .....(iii)Putting x = 2 in (i), we get36 = y2 + 4\xa0{tex}\\Rightarrow{/tex} y2 = 32\xa0{tex}\\Rightarrow{/tex} y = {tex}4 \\sqrt { 2 }{/tex} cm{tex}{/tex}Therefore,BC = 2BP = 2y = {tex}8 \\sqrt { 2 }{/tex}\xa0.cm .....(iv)Therefore,Area of\xa0∆ ABC=\xa0{tex}\\frac 1 2{/tex}BC × AP=\xa0{tex}\\frac 1 2{/tex}× 2y × x [ from (iii) & (iv) ]= xy= 2 × 4√2 = 8√2 cm2. | |