Question

In figure, altitude RH = 15 and $\overline{ST}$ is drawn parallel to $\overline{QP}$, What must be the length of $\overline{RJ}$ so that the area of $\Delta RST=\frac{1}{3}$ the area of $\Delta$RQP ?

• A. 5
• B. $5\sqrt{3}$
• C. $5\sqrt{2}$
• D. 7
• E. cannot be determined from the information given

Answer 1

Answer: (B)

$5\sqrt{3}$

• The triangle $RST$ and $RPQ$ are similar
• So we get $RJ/RH=ST/QP$
• Given area of triangle $RST$ is equal to $1/3$rd of the area of triangle $RQP$
• Which implies $1/2\times RJ\times ST=1/3\times 1/2\times RH\times PQ$
• Which gives $RJ\times RJ\times QP/RH=1/3\times RH\times PQ$
• There fore we get $RJ=\sqrt{75}=5\sqrt{3}$