Question

In figure, $S$ and $T$ trisect the side $QR$ of a right triangle $PQR$. then $8P{R}^2=3P{R}^2+9P{S}^2$.

State whether the above statement is true or false

• A. True
• B. False

Answer 1

The correct option is B False

Since S & T trisect the side QR

$QS=TS=RT$

$QR=3x$ & $QT=2x$

In right triangle $PQR\Rightarrow P{R}^2=P{Q}^2+Q{R}^2$

$P{R}^2=P{Q}^2+(3x{)}^2=P{Q}^2+9x^2$

In right triangle $PQS\Rightarrow P{S}^2=P{Q}^2+Q{S}^2$

$\Rightarrow P{S}^2=P{Q}^2+x^2$

In right triangle $PQT\Rightarrow P{T}^2=P{Q}^2+O{T}^2$

$P{T}^2=P{Q}^2+(2x{)}^2$

we get $8P{T}^2=3P{R}^2+5P{S}^2$