Question

An isosceles triangle with base 48 cm has area 240 sq cm.find remaining two sides of triangle

Answer 1

Given,area of the ∆ABC =240cm²let AB=AC=26cmdraw a AD is perpendicular to BClet CD be x ,AD be harea of ∆ADC=1/2*CD*AD=> ar(ABC)/2=1/2*x*h=>240/2=1/2*x*h=>xh=240=>h=240/xby Pythagoras thereon,AC²=AD²+CD=>26²=h²+x²=>676=(240/x)²+x²=>676=57600/x²+x²=>676=57600+x⁴/x²=>x⁴-676x²+57600=0let p=x²=>p²-676p+57600=0=>p=(-(-676))±√(-676)²-4(1)(57600)/2=>p=676±√456976-230400/2=>p=676±√226576/2=>p=676±√(476)²/2=>p=676±476/2=>p=676+476/2 or 676-476/2=>p=1152/2 or 200/2=>p=576 or 100=>p=x²=576 or 100=>x=√576 or √100=>x=24 or 10=>CD=24 or 10BC=BD+CD=CD+CD. ( the AD divides the BC into two equal parts.)=24+24,10+10=48 or 20