Question

devoir 1.2 question 1: déterminez laire et le périmètre de la figure suivante. arrondis au dixième près si nécessaire. 3x - 4 21 cm

Explanation:

Step1: Find the value of \(x\)

We know that the bottom - most horizontal side is \(21\) cm. So, \(5x+3x - 4+5x=21\). Combining like terms gives \(13x-4 = 21\). Adding \(4\) to both sides: \(13x=21 + 4=25\), then \(x=\frac{25}{13}\approx1.923\).

Step2: Calculate the perimeter

The perimeter \(P\) of the figure is the sum of all the side - lengths.
The vertical side - lengths: \(2(x + 5)\) and the horizontal side - lengths: \(2(5x)+(3x - 4)\).
\(P=2(x + 5)+2(5x)+(3x - 4)\)
Substitute \(x=\frac{25}{13}\) into the formula:
\[

$$\begin{align*} P&=2(\frac{25}{13}+ 5)+2\times5\times\frac{25}{13}+(3\times\frac{25}{13}-4)\\ &=2(\frac{25 + 65}{13})+\frac{250}{13}+\frac{75}{13}-4\\ &=2\times\frac{90}{13}+\frac{250 + 75}{13}-4\\ &=\frac{180}{13}+\frac{325}{13}-4\\ &=\frac{180+325}{13}-4\\ &=\frac{505}{13}-4\\ &=\frac{505}{13}-\frac{52}{13}\\ &=\frac{505 - 52}{13}=\frac{453}{13}\approx34.8 \end{align*}$$

\]

Step3: Calculate the area

We can consider the figure as a large rectangle with two small rectangles removed.
The large rectangle has dimensions \(l = 5x+3x - 4+5x=13x - 4\) and \(w=x + 5\).
The area of the small rectangles is \(2\times5\times x = 10x\).
The area of the large rectangle \(A_{1}=(13x - 4)(x + 5)=13x^{2}+65x-4x-20=13x^{2}+61x - 20\).
The area of the two small rectangles \(A_{2}=10x\).
The area of the figure \(A=A_{1}-A_{2}=13x^{2}+61x - 20-10x=13x^{2}+51x - 20\).
Substitute \(x=\frac{25}{13}\) into the area formula:
\[

$$\begin{align*} A&=13\times(\frac{25}{13})^{2}+51\times\frac{25}{13}-20\\ &=13\times\frac{625}{169}+\frac{1275}{13}-20\\ &=\frac{625}{13}+\frac{1275}{13}-20\\ &=\frac{625 + 1275}{13}-20\\ &=\frac{1900}{13}-20\\ &=\frac{1900}{13}-\frac{260}{13}\\ &=\frac{1900 - 260}{13}=\frac{1640}{13}\approx126.2 \end{align*}$$

\]

Answer:

Perimeter \(\approx34.8\) cm, Area \(\approx126.2\) \(cm^{2}\)