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2) determine the perimeter p. (level 2)

Question

  1. determine the perimeter p. (level 2)

Explanation:

Step1: Analyze the vertical sides

The vertical sides: \(3x+(x - 1)+(x - 1)=3x+x - 1+x - 1=5x-2\)

Step2: Analyze the horizontal sides

The horizontal sides: \(20 + 14+(20-(x - 1)-(x - 1))=20 + 14+(20 - x + 1 - x + 1)=56-2x\)

Step3: Calculate the perimeter

The perimeter \(P=(5x - 2)+(56-2x)+(5x - 2)+(56-2x)\)

$$ LATEXBLOCK0 $$

Another way: Using the property of translation of line - segments.
If we translate the horizontal and vertical "indented" line - segments, the figure can be thought of as a large rectangle.
The length of the large rectangle in the horizontal direction is \(20\), and in the vertical direction is \(3x\).
The "missing" part: In the horizontal direction, the sum of the two indented horizontal segments is \(20-14 = 6\), and each indented horizontal segment is \(x - 1\), so \(2(x - 1)=6\), solving \(x-1 = 3\), then \(x=4\)
The perimeter of the original figure is equal to the perimeter of a rectangle with length \(20\) and width \(3x\) plus \(2\times(3x-(x - 1))\)
The perimeter of a rectangle with length \(l = 20\) and width \(w = 3x\) is \(2(l + w)=2(20 + 3x)\)
The extra part \(2\times(3x-(x - 1))=2(2x + 1)=4x+2\)

$$ LATEXBLOCK1 $$

Wait, there is a mistake above. Let's use the first - principle method again.
If we consider the lengths of all the sides:
Top - horizontal side: \(20\)
Bottom - horizontal side: \(20\)
Left - vertical side: \(3x\)
Right - vertical side: \(x-1\)
The two indented vertical sides: \(3x-(x - 1)\) and \(3x-(x - 1)\)
The two indented horizontal sides: \(x - 1\) and \(x - 1\)

$$ LATEXBLOCK2 $$

Oops, wrong again. Let's use the translation method correctly.
If we translate the indented horizontal segments to form a complete length of \(20\) in the horizontal direction and the indented vertical segments.
The figure's perimeter is equal to the perimeter of a rectangle with length \(20\) and width \(3x\) plus \(2\times(3x-(x - 1))\)
First, find \(x\): Since \(2(x - 1)=20 - 14\), so \(x-1 = 3\), \(x = 4\)
The perimeter of a rectangle with length \(l = 20\) and width \(w=3x\) is \(2(l + w)=2(20+12)=64\)
The extra part: \(2\times(3x-(x - 1))\), substitute \(x = 4\), \(2\times(12 - 3)=18\)
No, let's start from the basic:
The sum of the horizontal sides: \(20+20=40\)
The sum of the vertical sides: \(3x+(x - 1)+(x - 1)+(x - 1)+(x - 1)+(3x-(x - 1))+(3x-(x - 1))\)

$$ LATEXBLOCK3 $$

\(P=40+11x-2=11x + 38\)
Since \(2(x - 1)=20 - 14\), \(x-1=3\), \(x = 4\)

$$ LATEXBLOCK4 $$

Answer:

The perimeter \(P = 82\mathrm{cm}\)