An isosceles triangle with base 48cm has area 240cmsquare find length of remaining two sides
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∆ ABC is isoceles . BC=48cm,we can take AC=48cm. area of ∆ is 240cm so we can write that........
AB+BC+CA=240cm
AB+48+48=240
AB+96=240
AB=240-96=144
hence,our third side i.e. AB=144cm & AC was 48cm as the ∆ was isoceles.
AB+BC+CA=240cm
AB+48+48=240
AB+96=240
AB=240-96=144
hence,our third side i.e. AB=144cm & AC was 48cm as the ∆ was isoceles.
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