The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other.
We have,
ABC
PQR
Area (
) =100cm2
Area (
) =49 cm2
AD= 5cm
AD and PS are the altitudes
by area of similar triangle theorem
Area(
) =AB2
Area (
) PQ2
AB2 = 100/49
PQ2
AB/PQ= 10/7 ………..(i)
In 
ABD and 
PQS
∠B=∠Q [
ABC
PQR]
∠ADB=∠PQS=90°
ABD ~ 
PQS [By AA similarity]
AB/PQ=AD/PS …….(ii)
Compare equ. (i)and(ii)
AD/PS=10/7
5/PS=10/7
PS=35/10
PS=3.5 cm
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