The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.
We have,
ABCPQR
Area () =121cm2
Area () =64cm2
AD= 12.1cm
AD and PS are the medians
By area of similar triangle theorem
Area() =AB2
Area () PQ2
AB2 =121
PQ2 64
AB =11 ………… (i)
PQ 8
ABCPQR
AB/PQ=BC/QR [Corresponding parts of similar triangles are proportional] AB/PQ=2BD/2QS [AD and BD are medians]
AB/PQ=BD/QS ………… (ii)
In ABD and PQS
∠B=∠Q [ABCPQS]
AB/PQ=BD/QS [from (ii)]
ABD ~ PQS [By AA similarity]
AB/PQ=AD/PS Compare equ. (i)and(ii)
AD/PS=11/8
12.1/PS=11/8
PS=12.1x8/8
PS= 8.8 cm