PQRS is a rectangle in which PQ=9cm and PS=6cm. T is a point in PQ such that PT=7cm and RV is the perpendicular from R to ST. Calculate ST and RV.
Hello
The length of ST you can calculate using Pythogoras.
If V is a point on ST such that RV is perpendicular to ST you could use coordinate geometry to find V.
If the coordinates of the vertices are, for instance, P(0,6), Q(9,6), R(9,0), S(0,0). You could find the slope and or the equation of the line through ST and the equation of the line perpendicular to ST and passing through R. The intersection of the two lines will give you V. There is a formula for finding the perpendicualr distance of a point from a line, you could use that too.
Hope that helps.
Here is one way.
In right triangle SPT, by Pythagorean theorem,
(ST)^2 = (PS)^2 +(PT)^2
ST = sqrt(6^2 +7^2) = sqrt(85) = 9.2195...cm -------------answer.
Area of right triangle SPT = (1/2)(6)(7) = 21 sq.cm
Area of right triangle TQR = (1/2)(9-7)(6) = 6 sq.cm
So, area of triangle STR = 6*9 -21 -6 = 27 sq.cm
Area of triangle STR = (1/2)(RV)(ST) also.
Hence,
27 = (1/2)(RV)(sqrt(85))
RV = (27*2) /sqrt(85) = 5.8571 cm -------------answer.
Hello, sakibniaz!
$\displaystyle PQRS$ is a rectangle in which $\displaystyle PQ$ = 9cm and $\displaystyle PS$ = 6cm.
$\displaystyle T$ is a point in $\displaystyle PQ$ such that $\displaystyle PT$ = 7 cm and $\displaystyle RV \perp ST.$
Calculate $\displaystyle ST$ and $\displaystyle RV.$Code:S 9 R o - - - - - - - - - - - o | * θ * | | * * | 6 | * V * | | o | | θ * | o - - - - - - - - o - - o P 7 T 2 Q
In right triangle $\displaystyle SPT\!:\;\;ST^2 \:=\:6^2 + 7^2 \:=\:85 \quad\Rightarrow\quad \boxed{ST \:=\:\sqrt{85}}$
Since $\displaystyle \angle STP = \angle RST = \theta,\;\;\Delta SPT \sim \Delta RVS$
Hence: .$\displaystyle \frac{RV}{SR} = \frac{SP}{ST} \quad\Rightarrow\quad \frac{RV}{9} \:=\:\frac{6}{\sqrt{85}}$
. . Therefore: .$\displaystyle RV \:=\:\frac{2}{3\sqrt{85}} \quad\Rightarrow\quad\boxed{ RV\:=\:\frac{2\sqrt{85}}{255}} $