Question

The altitude of a triangle is five-thirds the length of the corresponding base. If the altitude is increased by 4cm and the base is decreased by 2cm, the area of the triangle remains the same. Find the altitude and the base of the triangle.

Answer 1

Hint: Assume that the altitude of the triangle is x and the length of the corresponding base is y. Use the statement in the question to form two linear equations in two variables x and y. Solve for x and y using any one of the known methods like substitution, elimination, graphically or using matrices. The value of x and y will give the value of the length of altitude and the length of the corresponding base.

Complete step-by-step answer:

Let x be the length of altitude of the triangle and y the length of the corresponding base.
Since altitude is one-fifth the length of the base, we have
$x=\dfrac{5y}{3}\text{ (i)}$
Also, according to the question increasing altitude by 4 cm and decreasing base by 2 cm does not change the area of the triangle.
Since area of the triangle is equal to $\dfrac{1}{2}bh$, we have
$\begin{align}
  & \dfrac{1}{2}xy=\dfrac{1}{2}\left( x+4 \right)\left( y-2 \right) \\
 & \Rightarrow xy=\left( x+4 \right)\left( y-2 \right) \\
\end{align}$
Expanding the expression on RHS, we get
$\begin{align}
  & xy=xy-2x+4y-8 \\
 & \Rightarrow 2x=4y-8\text{ } \\
 & \Rightarrow x=2y-4\text{ (ii)} \\
\end{align}$
Substituting the value of x from equation (i) in equation (ii), we get
$\dfrac{5y}{3}=2y-4$
Adding 4 on both sides, we get
$\dfrac{5y}{3}+4=2y$
Subtracting $\dfrac{5y}{3}$ from both sides, we get
$\begin{align}
  & \dfrac{5y}{3}+4-\dfrac{5y}{3}=2y-\dfrac{5y}{3} \\
 & \Rightarrow \dfrac{6y-5y}{3}=4 \\
 & \Rightarrow \dfrac{y}{3}=4 \\
\end{align}$
Multiplying both sides by 3, we get
$\begin{align}
  & \dfrac{3y}{3}=12 \\
 & \Rightarrow y=12 \\
\end{align}$
Substituting the value of y in equation (ii), we get
x = 2(12)-4 = 24-4 = 20.
Hence the altitude of the triangle is 20cm, and the length of the base is 12 cm.

Note: Alternatively we can solve the equations graphically also.
We plot the two lines on the same graph and then find the coordinates of the point of intersection of lines. The plot of the given system is shown below.

As is evident from the graph, the given lines intersect at point A(20,12).
Hence x = 20 and y = 12, which is the same as obtained above.