Q31. - Geometry

Question

a. Find the area of the right triangle in terms of a and b.

b. Find the area of the right triangle in terms of c and h.

c. Solve for h in terms of the other variables.

d. A right triangle has legs 6 and 8. Find the lengths of the altitude and the median to the hypotenuse.

Step-by-Step Solution

Answer

 a. The area of the right triangle in terms of a and b is$\frac{ab}{2}$ .

b. The area of the right triangle in terms of c and h is $\frac{ch}{2}$.

 c. h in the terms of the other variables is $\frac{ab}{c}$.

d. The length of the altitude and median to the hypotenuse is$4.8$ .

Step 1. Analyze the given information.

Here,$BC=a,CA=b,CD=h$ and $AB=c$.

Step 2. Concept Used.

Area of triangle can be found using the formula

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

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

Step 3. Find the area of triangle.

Area of $△ABC$

$\begin{array}{l}b=a\\ h=b\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(a\right)\left(b\right)\\ A=\frac{ab}{2}\end{array}$

Therefore, the area of the right triangle in terms of  and b is$\frac{ab}{2}$$\frac{ab}{2}$ .

Step 1. Analyze the given information.

Here,$BC=a,CA=b,CD=h$ and $AB=c$.

Step 2. Concept Used.

Area of triangle can be found using the formula

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

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

Step 3. Find the area of triangle.

Area of $△ABC$

$\begin{array}{l}b=c\\ h=h\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(c\right)\left(h\right)\\ A=\frac{ch}{2}\end{array}$

Therefore, the area of the right triangle in terms of c and h is$\frac{ch}{2}$ .

Step 1. Analyze the given information.

Here,$BC=a,CA=b,CD=h$ and$AB=c$ .

Step 2. Concept Used.

Area of triangle can be found using the formula

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

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

Step 3. Find the area of triangles.

Area of $△ABC$

$\begin{array}{l}b=a\\ h=b\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(a\right)\left(b\right)\\ A=\frac{ab}{2}\end{array}$

We can also say, that the Area of $△ABC$is

$\begin{array}{l}b=c\\ h=h\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(c\right)\left(h\right)\\ A=\frac{ch}{2}\end{array}$

Step 4. Equating the area of triangle.

$\begin{array}{l}A=\frac{ab}{2}\\ A=\frac{ch}{2}\\ ab=ch\\ h=\frac{ab}{c}\end{array}$

Therefore,h in the terms of the other variables is $\frac{ab}{c}$.

Step 1. Analyze the given information.

Here$BC=a=6,CA=b=8,CD=h=\frac{ab}{c}$, and$AB=c$ .

Step 2. Use Pythagoras theorem.

We can use Pythagorean Theorem. 

Pythagorean Theorem define that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of other sides.

 $\begin{array}{l}{a}^2+b^2=c^2\\ 6^2+8^2=c^2\\ c^2=100\\ c=10\end{array}$

Step 3. Find the length of altitude.

Substitute $a=6,b=8,c=10$ in $h=\frac{ab}{c}$

$\begin{array}{l}h=\frac{6\left(8\right)}{10}\\ h=4.8\end{array}$

Therefore, the length of the altitude and median to the hypotenuse is$4.8$ .