Question

1. détermine la valeur du périmètre si l’aire du trapèze est de 160 cm².

(diagram of a right trapezoid with top base 10 cm, left leg 3x - 4, right leg x, and slant side 14 cm. handwritten notes: 10cm + 14cm + 3x - 4 + x + 10cm = 160 cm? 10cm + 14cm + x + 3x - 4 = 160 cm? 24 cm)

Explanation:

Step1: Recall trapezoid area formula

The area \( A \) of a trapezoid is given by \( A=\frac{(a + b)h}{2} \), where \( a \) and \( b \) are the lengths of the two parallel sides, and \( h \) is the height. Here, the parallel sides are \( 10 \, \text{cm} \) and \( x \), and the height \( h = 3x - 4 \), and \( A = 160 \, \text{cm}^2 \). So we have the equation:
\[ 160=\frac{(10 + x)(3x - 4)}{2} \]

Step2: Solve the equation for \( x \)

Multiply both sides by 2:
\[ 320=(10 + x)(3x - 4) \]
Expand the right - hand side:
\[ 320 = 30x-40+3x^{2}-4x \]
Simplify the right - hand side:
\[ 3x^{2}+26x - 40 - 320=0 \]
\[ 3x^{2}+26x - 360 = 0 \]
We can use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for the quadratic equation \( ax^{2}+bx + c = 0 \). Here, \( a = 3 \), \( b = 26 \), \( c=-360 \).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(26)^{2}-4\times3\times(-360)=676 + 4320 = 5000 - 676+4320=4996 \)? Wait, no, \( 26^{2}=676 \), \( 4\times3\times360 = 4320 \), so \( \Delta=676+4320 = 4996 \)? Wait, that can't be right. Wait, maybe I made a mistake in expansion. Let's re - expand \( (10 + x)(3x - 4) \):
\( (10 + x)(3x - 4)=10\times3x-10\times4+x\times3x - x\times4=30x-40 + 3x^{2}-4x=3x^{2}+26x - 40 \). So the equation is \( 3x^{2}+26x-40 = 320 \), so \( 3x^{2}+26x - 360 = 0 \). Wait, maybe we can try to factor or use another approach. Wait, maybe the height is \( 10 \)? No, the right angle is between the side of length \( 10 \) and \( 3x - 4 \), so the two parallel sides are \( 10 \) and \( x \), and the height is \( 3x - 4 \). Wait, maybe I misread the figure. Wait, the other non - parallel side is \( 14 \). Wait, maybe the trapezoid is a right trapezoid, so the side of length \( 14 \) is the hypotenuse of a right triangle with legs \( (x - 10) \) and \( (3x - 4) \). So by Pythagoras theorem: \( 14^{2}=(x - 10)^{2}+(3x - 4)^{2} \)
\[ 196=x^{2}-20x + 100+9x^{2}-24x + 16 \]
\[ 196 = 10x^{2}-44x+116 \]
\[ 10x^{2}-44x - 80 = 0 \]
Divide both sides by 2:
\[ 5x^{2}-22x - 40 = 0 \]
Using quadratic formula \( x=\frac{22\pm\sqrt{(-22)^{2}-4\times5\times(-40)}}{2\times5}=\frac{22\pm\sqrt{484 + 800}}{10}=\frac{22\pm\sqrt{1284}}{10}=\frac{22\pm35.83}{10} \)
We take the positive root: \( x=\frac{22 + 35.83}{10}\approx\frac{57.83}{10}\approx5.78 \)? No, that doesn't make sense. Wait, maybe the first approach is wrong. Wait, the area is \( 160 \). Let's assume that the two parallel sides are \( 3x - 4 \) and \( 14 \)? No, the right angle is at the top, so the vertical side is \( 3x - 4 \), the top base is \( 10 \), the bottom base is \( x \), and the slant side is \( 14 \). So the height is \( 3x - 4 \), the difference in the bases is \( x - 10 \). Then by Pythagoras, \( 14^{2}=(x - 10)^{2}+(3x - 4)^{2} \). Wait, maybe the area formula is correct. Let's go back to the area formula.
\( A=\frac{(a + b)h}{2} \), \( a = 10 \), \( b=x \), \( h = 3x - 4 \), \( A = 160 \). So \( (10 + x)(3x - 4)=320 \)
\( 3x^{2}+26x-40 - 320 = 0 \)
\( 3x^{2}+26x - 360 = 0 \)
Dividing by common factor? 3,26,360 have no common factor. Quadratic formula: \( x=\frac{-26\pm\sqrt{26^{2}-4\times3\times(-360)}}{2\times3}=\frac{-26\pm\sqrt{676 + 4320}}{6}=\frac{-26\pm\sqrt{4996}}{6}=\frac{-26\pm70.68}{6} \)
We take the positive root: \( x=\frac{-26 + 70.68}{6}=\frac{44.68}{6}\approx7.45 \)? No, this is getting messy. Wait, maybe the height is \( 10 \), and the bases are \( 3x - 4 \) and \( x \). Wait, the right angle is between the side of length \( 10 \) and \( 3x - 4 \), so \( 10 \) is the height. Then area \( A=\frac{(a + b)h}{2}=\frac{(x+(3x - 4))\times10}{2}=160 \)
\[ \frac{(4x - 4)\times10}{2}=160 \]
\[ (4x - 4)\times5 = 160 \]
\[ 4x-4=\frac{160}{5}=32 \]
\[ 4x=36 \]
\[ x = 9 \]
Ah! That makes sense. I misidentified the height. The right angle is between the side of length \( 10 \) and \( 3x - 4 \), so the height of the trapezoid is \( 10 \), and the two parallel sides are \( 3x - 4 \) and \( x \). So the area formula is \( A=\frac{( (3x - 4)+x)\times10}{2} \)

Step3: Solve for \( x \) with correct height

Given \( A = 160 \), \( h = 10 \), \( a=3x - 4 \), \( b = x \)
\[ 160=\frac{(3x - 4+x)\times10}{2} \]
Simplify the right - hand side:
\[ 160=\frac{(4x - 4)\times10}{2}=(4x - 4)\times5 \]
Divide both sides by 5:
\[ 32=4x - 4 \]
Add 4 to both sides:
\[ 36 = 4x \]
Divide both sides by 4:
\[ x = 9 \]

Step4: Find the lengths of all sides

Now that \( x = 9 \), the length of the side \( 3x - 4=3\times9 - 4=27 - 4 = 23 \)
The sides of the trapezoid are \( 10 \, \text{cm} \), \( 23 \, \text{cm} \), \( 9 \, \text{cm} \), and \( 14 \, \text{cm} \)

Step5: Calculate the perimeter

The perimeter \( P \) of a trapezoid is the sum of all its sides. So \( P=10 + 23+9 + 14 \)
\[ P=10+23 = 33;33 + 9=42;42+14 = 56 \]

Answer:

The perimeter of the trapezoid is \( 56 \, \text{cm} \)