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Time slot 2-48 Calculation for problem 1(ques no 12) a^2 +b^2 =169 ab =60 Now (a+b)^2 =a^2 +b^2 +2ab = 169+2*60=169+120=289 =17^2 Hence a+b = 17( a, b are length and hence we reject -ve value) --(1) (a-b) ^2=a^2+b^2 -2ab = 169-2*60=169-120=49=7^2 Hence a - b = 7( argument as above) --(2) From (1) & (2) we may get a =(17+7)/2=12 units b =(17-7)/2=5 units ------ A Method 13^2 = 169 =144+25=12^2+5^2 Hence a=12 ,b=5 Or a=5, b =12 ----- (13)^2=(12+1)^2 =12^2 +(1^2+2*12*1) =12^2+(1+24) =12^2+25 =12^2+5^2 Hence a=12,b=5 Or a=5, b =12 ----

Figure drawn is not correct. The triangle is obtuse. Hence the orthocentre will be outside of the triangle. At point C the obtuse angle will be formed.

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