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 Question from Lisa, a student: Hi! An isosceles triangle has an area of 25.6 sq cm. The two equal sides are 8.4 cm long. Calculate the two possible lengths of the third side. Thanks a lot!!!

Hi Lisa,

In my triangle ADC the lengths of AD and CA are each equal to 8.4 cm. The area of the triangle ADC is

25.6 = 1/2 b h

where b is the length of the base DC and h is the height, AC.

Triangle ABC is a right triangle and hence by Pythagoras' Theorem

h2 + (b/2)2 = 8.42 or b/2 = √[8.42 - h2]

Substitute this value for b/2 into the expression for the area to get

25.6 = h √[8.42 - h2]

At this point I would square each side to eliminate the square root. This gives a quartic in h but if you substitute x = h2 then you have a quadratic in x which you can solve. Finally h is the square root of x but remember that h must be positive.

Penny

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