Question

A modernistic painting consists of triangles rectangles and pentagons all drawn so as to not overlap or share sides within each rectangle are drawn $5$ red roses and each pentagon contains $4$ carnations.

How many triangles rectangles and pentagons appear in the painting iſ the painting contains a total of $38$ geometric figures $148$ sides of geometric figures, and $122$ flowers?

How many pentagons appear in the painting?​

Answer 1

Find the number of triangles rectangles and pentagons that appear in the painting:

Given that,

$5$ red roses within each rectangle.

$4$ carnations within each pentagon.

The painting contains a total of $38$ geometric figures $148$ sides of geometric figures and $122$ flowers.

Step-1: Find the system of equations.

Let the number of triangles be $\mathrm{x}$, rectangles be $\mathrm{y}$, and pentagons be $\mathrm{z}$ in the modernistic painting. It is given that there are a total of $38$ geometric figures in the painting.

This can be represented by the equation:

$\Rightarrow \mathrm{x}+\mathrm{y}+\mathrm{z}=38$

We know that, triangles have $3$ sides each, rectangles have $4$ sides each, and pentagons have $5$ sides each and total of $148$ geometric sides in the painting.

This can be represented by the equation:

$\Rightarrow 3\mathrm{x}+4\mathrm{y}+5\mathrm{z}=148$

There are $5$ red roses within the rectangle and $4$ carnations within the pentagon. There are a total of $122$ flowers. This can be represented by the equation:

$\Rightarrow 5\mathrm{y}+4\mathrm{z}=122$

Step-2: Solve the system of equations.

Solve the system of above $3$ equations:

$\begin{array}{rcl}& \Rightarrow & \mathrm{x}+\mathrm{y}+\mathrm{z}=38\to \left(1\right)\\ & \Rightarrow & 3\mathrm{x}+4\mathrm{y}+5\mathrm{z}=148\to \left(2\right)\\ & \Rightarrow & 5\mathrm{y}+4\mathrm{z}=122\to \left(3\right)\end{array}$

Select the equations $\left(1\right)$ and $\left(2\right)$, and eliminate the variable $\mathrm{x}$:

$\begin{array}{l}\Rightarrow 3x+3y+3z=114\left[\mathrm{Multiply}\mathrm{it}\mathrm{with}\left(3\right)\right]\\ \frac{3x+4y+5z=148}{-y-2z=-34}\to \left(4\right)\end{array}$

Select the equations $\left(3\right)$ and $\left(4\right)$, and eliminate the variable $\mathrm{y}$:

$$\begin{gathered} \begin{array}{l}\Rightarrow 5\mathrm{y}+4\mathrm{z}=122\\ \frac{-5\mathrm{y}-10\mathrm{z}=-170}{-6\mathrm{z}=-48}\to \left(5\right)\end{array}\left[\mathrm{Multiply}\mathrm{it}\mathrm{with}\left(5\right)\right] \\ \mathrm{z}=\frac{-48}{-6} \\ \mathrm{z}=8 \end{gathered}$$

As pentagons be $\mathrm{z}$, so the number of pentagons that appear in the painting are $8$.

Step-3: Find the number of triangles and rectangles.

To find the number of rectangles substitute $\mathrm{z}=8$ in the equation $\left(3\right)$:

$\begin{array}{rcl}& \Rightarrow & 5\mathrm{y}+4\left(8\right)=122\\ 5\mathrm{y}+32& =& 122\\ 5\mathrm{y}& =& 122-32\\ 5\mathrm{y}& =& 90\\ \mathrm{y}& =& 18\end{array}$

As rectangles be $\mathrm{y}$, so the number of rectangles that appear in the painting are$18$.

To find the number of triangles substitute $\mathrm{y}=18$ and $\mathrm{z}=8$ in the equation $\left(1\right)$:

$\begin{array}{rcl}& \Rightarrow & \mathrm{x}+18+8=38\\ \mathrm{x}+26& =& 38\\ \mathrm{x}& =& 38-26\\ \mathrm{x}& =& 12\end{array}$

As triangles be $\mathrm{x}$, so the number of triangles that appear in the painting are $12$.

Hence, the number of triangles rectangles and pentagons that appear in the painting are $\left(12,18,8\right)$ respectively.