In figure, altitude RH = 15 and ¯¯¯¯¯¯¯ST is drawn parallel to ¯¯¯¯¯¯¯¯QP, What must be the length of ¯¯¯¯¯¯¯¯RJ so that the area of ΔRST=13 the area of ΔRQP ?
5√3
5√2
7
5
cannot be determined from the information given
A
5
B
5√3
C
5√2
D
7
E
cannot be determined from the information given
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Solution
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The triangle RST and RPQ are similar
So we get RJ/RH=ST/QP
Given area of triangle RST is equal to 1/3rd of the area of triangle RQP
Which implies 1/2×RJ×ST=1/3×1/2×RH×PQ
Which gives RJ×RJ×QP/RH=1/3×RH×PQ
There fore we get RJ=√75=5√3
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