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31.) $\\triangle fgh$ is an equilateral triangle with $fg = x + 8$, $gh…

Question

31.) $\triangle fgh$ is an equilateral triangle with $fg = x + 8$, $gh = 2x - 7$, and $fh = 3x - 22$.

$x = \underline{\quad\quad\quad}$ $fg = \underline{\quad\quad\quad}$ $gh = \underline{\quad\quad\quad}$ $fh = \underline{\quad\quad\quad}$

32.) $\triangle lmn$ is an isosceles triangle, with legs $lm$ and $ln$, $lm = 3x - 2$, $ln = 2x + 1$, and $mn = 5x - 2$.

$x = \underline{\quad\quad\quad}$ $lm = \underline{\quad\quad\quad}$ $ln = \underline{\quad\quad\quad}$ $mn = \underline{\quad\quad\quad}$

33.) $\triangle abc$ is an isosceles triangle, with legs $ab$ and $ac$, $ab = 4x - 3$, $bc = x + 9$, and $ac = 6x - 7$.

$x = \underline{\quad\quad\quad}$ $ab = \underline{\quad\quad\quad}$ $ac = \underline{\quad\quad\quad}$ $bc = \underline{\quad\quad\quad}$

Explanation:

Problem 31:

Step1: Set FG = GH (equilateral triangle)

In an equilateral triangle, all sides are equal. So, set \( FG = GH \):
\( x + 8 = 2x - 7 \)

Step2: Solve for x

Subtract \( x \) from both sides: \( 8 = x - 7 \)
Add 7 to both sides: \( x = 15 \)

Step3: Find FG, GH, FH

  • \( FG = x + 8 = 15 + 8 = 23 \)
  • \( GH = 2x - 7 = 2(15) - 7 = 23 \)
  • \( FH = 3x - 22 = 3(15) - 22 = 23 \)
Problem 32:

Step1: Set LM = LN (isosceles legs)

In an isosceles triangle, legs \( LM \) and \( LN \) are equal. So, set \( LM = LN \):
\( 3x - 2 = 2x + 1 \)

Step2: Solve for x

Subtract \( 2x \) from both sides: \( x - 2 = 1 \)
Add 2 to both sides: \( x = 3 \)

Step3: Find LM, LN, MN

  • \( LM = 3x - 2 = 3(3) - 2 = 7 \)
  • \( LN = 2x + 1 = 2(3) + 1 = 7 \)
  • \( MN = 5x - 2 = 5(3) - 2 = 13 \)
Problem 33:

Step1: Set AB = AC (isosceles legs)

In an isosceles triangle, legs \( AB \) and \( AC \) are equal. So, set \( AB = AC \):
\( 4x - 3 = 6x - 7 \)

Step2: Solve for x

Subtract \( 4x \) from both sides: \( -3 = 2x - 7 \)
Add 7 to both sides: \( 4 = 2x \)
Divide by 2: \( x = 2 \)

Step3: Find AB, AC, BC

  • \( AB = 4x - 3 = 4(2) - 3 = 5 \)
  • \( AC = 6x - 7 = 6(2) - 7 = 5 \)
  • \( BC = x + 9 = 2 + 9 = 11 \)

Answer:

s:

Problem 31:

\( x = \boldsymbol{15} \), \( FG = \boldsymbol{23} \), \( GH = \boldsymbol{23} \), \( FH = \boldsymbol{23} \)

Problem 32:

\( x = \boldsymbol{3} \), \( LM = \boldsymbol{7} \), \( LN = \boldsymbol{7} \), \( MN = \boldsymbol{13} \)

Problem 33:

\( x = \boldsymbol{2} \), \( AB = \boldsymbol{5} \), \( AC = \boldsymbol{5} \), \( BC = \boldsymbol{11} \)