Q29. - Geometry

Question

a. Find the ratio of the areas of $△QRT$ and $△QTS$.

b. If the area of $△QRS$ is240 . Find the length of the altitude from s to .$\overleftrightarrow{QR}$

Step-by-Step Solution

Answer

 a.The ratio of the areas of $△QRT$ and $△QTS$ is$2:3$ .

 b. The altitude from s to$QR$  is 240.

Step 1. Given information.

$QR=24,RT=12$and $TS=18$

Step 2. Concept Used.

Area of triangle can be found using the formula

$A=\frac{1}{2}bh$

Whereb=  base of triangle andh= height of triangle.

Step 3. Find the area of triangles.

Area of triangle $△QRT$ is

$\begin{array}{l}b=RT=12\\ h=h\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(12\right)\left(h\right)\\ A\left(QRT\right)=6h\end{array}$

Area of triangle $△QTS$ is

$\begin{array}{l}b=TS=18\\ h=h\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(18\right)\left(h\right)\\ A\left(QTS\right)=9h\end{array}$

Step 4. Find the ratio of areas.

So, the ratio of the areas of $△QRT$ and $△QTS$ will be

$\begin{array}{c}\frac{A\left(\Delta QRT\right)}{A\left(\Delta QTS\right)}=\frac{6h}{9h}\\ =\frac{2}{3}\\ =2:3\end{array}$

Therefore, the ratio of the areas of $△QRT$ and $△QTS$ is $2:3$.

Step 1. Given information.

Area of $△QRS=240$

Let the altitude be x

Step 2. Concept Used.

Area of triangle can be found using the formula

$A=\frac{1}{2}bh$

Where b= base of triangle and h=height of triangle.

Step 3. Use the formula of the area of triangle.

$\begin{array}{l}b=QR=24\\ h=x\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(24\right)\left(x\right)\\ A\left(QRS\right)=12x\\ 240=12x\\ x=20\end{array}$

Therefore, the altitude from s to $QR$ is 20.