Q14 of 168 Page 5

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.

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We have,


ABCPQR


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 [ABCPQR]


∠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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