the area of an isosceles triangle is 60 cm square and the length of each of its sides is 13 cm . Find its base .
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Draw altitude AD.
Ar(ΔABC)=60cm2
12AD×BC=60
bh=60 where h is height and b is CD
By Pythagoras theorem
AC2=b2+h2
b2+h2=132=169
(b+h)2=b2+h2+2bh=169+2×60=289
b+h=289−−−√=17cm …equation 1
(b−h)2=b2+h2−2bh=169−120=49
b−h=49−−√=±7cm …equation 2
equation 1 + equation2
2b=±7+17=24cm,10cm
As b=CD so 2b=BC=base
Thus, base = 24cm or 10 cm
Ar(ΔABC)=60cm2
12AD×BC=60
bh=60 where h is height and b is CD
By Pythagoras theorem
AC2=b2+h2
b2+h2=132=169
(b+h)2=b2+h2+2bh=169+2×60=289
b+h=289−−−√=17cm …equation 1
(b−h)2=b2+h2−2bh=169−120=49
b−h=49−−√=±7cm …equation 2
equation 1 + equation2
2b=±7+17=24cm,10cm
As b=CD so 2b=BC=base
Thus, base = 24cm or 10 cm
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