Question

How do you use the formula $A=\dfrac{1}{2}bh$ to write a simplified expression for the area of a right triangle with leg length of $4x{{y}^{-1}}$ and $7{{x}^{2}}$ ?

Answer 1

Hint: In this problem we have to calculate the area of the right-angle triangle with the given data. In the problem, they have mentioned that we need to use the formula $A=\dfrac{1}{2}bh$. In the problem, we have the leg length and the height of the right-angle triangle. So, we will substitute those values in the given formula and simplify it by using the exponential and algebraic formulas. After simplifying the obtained equation, then we will get the required result.

Formula Use:
1.Area of the triangle is $A=\dfrac{1}{2}bh$ .
2. ${{a}^{-1}}=\dfrac{1}{a}$ .

Complete step by step procedure:
Given that, the right-angle triangle with a leg length of $4x{{y}^{-1}}$ and $7{{x}^{2}}$ .
In the question, they have mentioned the area of the triangle formula. We are going to use this formula to find the area of the triangle.
The given leg length that is the base length of the triangle is denoted by
$b=4x{{y}^{-1}}$ .
We have the exponential formula ${{a}^{-1}}=\dfrac{1}{a}$ in the above equation, then we will get
 $b=\dfrac{4x}{y}$
They have also given the height of triangle is denoted by
$h=7{{x}^{2}}$.
For calculating the area of the triangle, we are going to use the given area of the triangle formula
$A=\dfrac{1}{2}bh$.
Substituting the values of base and height of the given triangle, then we will get
$A=\dfrac{1}{2}\times \left( \dfrac{4x}{y} \right)\times 7{{x}^{2}}$
Using the exponential rule ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ , then we will get
$A=\dfrac{14{{x}^{3}}}{y}$

Hence the area of the right-angled triangle is $\dfrac{14{{x}^{3}}}{y}$.

Note:
In this problem, they have mentioned the formula for the area of the triangle. In some cases, they don’t mention the area formula but they asked to calculate the area. In that case, also we need to use the formula $A=\dfrac{1}{2}bh$ and simplify the equation to get the result.