Question

the area if an isosceles triangle is 240 cm and the length of each one of its equal side is 26 cm . find the base of the triangle.

Answer 1

given,

area of the ∆ABC =240cm²

let AB=AC=26cm

draw a AD is perpendicular to BC

let CD be x ,AD be h

area of ∆ADC=1/2*CD*AD

=> ar(ABC)/2=1/2*x*h

=>240/2=1/2*x*h

=>xh=240

=>h=240/x

by Pythagoras thereon,

AC²=AD²+CD

=>26²=h²+x²

=>676=(240/x)²+x²

=>676=57600/x²+x²

=>676=57600+x⁴/x²

=>x⁴-676x²+57600=0

let 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 10

BC=BD+CD

=CD+CD. ( the AD divides the BC into two equal parts.)

=24+24,10+10

=48 or 20

the possible values of base is 48 and 20

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Answer 2

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