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Triangle PQR has vertices P(0,0), Q((9.42,0), and R(6,2.4). Triangle PQS have vertices P(0,0), Q(9.42,0), and S(5,y). The area of Triangle PQS is three times the area of the triangle PQR. Compute the value of y, the missing coordinate of vertex S.

Y=7.2
Y=4.71
Y=18
Y=0.8

3 answers

The area \( A \) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:

\[ A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]

Step 1: Area of Triangle PQR

For triangle PQR with vertices \( P(0,0) \), \( Q(9.42, 0) \), and \( R(6, 2.4) \):

  • Coordinates:
    • \( (x_1, y_1) = (0, 0) \)
    • \( (x_2, y_2) = (9.42, 0) \)
    • \( (x_3, y_3) = (6, 2.4) \)

Using the area formula:

\[ A_{PQR} = \frac{1}{2} \left| 0(0 - 2.4) + 9.42(2.4 - 0) + 6(0 - 0) \right| \]

Calculating it step by step:

\[ A_{PQR} = \frac{1}{2} \left| 0 + 9.42 \cdot 2.4 + 0 \right| \] \[ A_{PQR} = \frac{1}{2} \left| 22.608 \right| = 11.304 \]

Step 2: Area of Triangle PQS

For triangle PQS with vertices \( P(0,0) \), \( Q(9.42, 0) \), and \( S(5, y) \):

  • Coordinates:
    • \( (x_1, y_1) = (0, 0) \)
    • \( (x_2, y_2) = (9.42, 0) \)
    • \( (x_3, y_3) = (5, y) \)

Using the area formula:

\[ A_{PQS} = \frac{1}{2} \left| 0(0 - y) + 9.42(y - 0) + 5(0 - 0) \right| \]

Calculating it step by step:

\[ A_{PQS} = \frac{1}{2} \left| 9.42y \right| = \frac{9.42y}{2} \]

Step 3: Set Up the Equation

We know the area of triangle PQS is three times the area of triangle PQR:

\[ A_{PQS} = 3 \cdot A_{PQR} \]

Substituting the areas:

\[ \frac{9.42y}{2} = 3 \cdot 11.304 \]

Calculating \( 3 \cdot 11.304 \):

\[ 3 \cdot 11.304 = 33.912 \]

Now, equate:

\[ \frac{9.42y}{2} = 33.912 \]

To eliminate the fraction, multiply both sides by 2:

\[ 9.42y = 67.824 \]

Step 4: Solve for y

Now, divide both sides by 9.42:

\[ y = \frac{67.824}{9.42} \]

Calculating \( y \):

\[ y \approx 7.19 \text{ (rounded to two decimal places)} \]

The value of \( y \) closest to this calculation in the provided options is:

\[ \boxed{7.2} \]

Are you sure

Let’s double-check the calculations step-by-step to ensure accuracy.

Step 1: Area of Triangle PQR

The vertices are:

  • \( P(0,0) \)
  • \( Q(9.42, 0) \)
  • \( R(6, 2.4) \)

Using the area formula for triangle \( A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \):

\[ A_{PQR} = \frac{1}{2} \left| 0(0 - 2.4) + 9.42(2.4 - 0) + 6(0 - 0) \right| \]

Calculating the area:

\[ = \frac{1}{2} \left| 0 + 9.42 \cdot 2.4 + 0 \right| \] \[ = \frac{1}{2} \left| 22.608 \right| = 11.304 \]

Step 2: Area of Triangle PQS

The vertices are:

  • \( P(0,0) \)
  • \( Q(9.42, 0) \)
  • \( S(5, y) \)

Using the area formula:

\[ A_{PQS} = \frac{1}{2} \left| 0(0 - y) + 9.42(y - 0) + 5(0 - 0) \right| \] \[ = \frac{1}{2} \left| 9.42y \right| = \frac{9.42y}{2} \]

Step 3: Relate the Areas

We set the area of triangle PQS to be three times the area of triangle PQR:

\[ \frac{9.42y}{2} = 3 \cdot 11.304 \]

Calculating the right-hand side:

\[ 3 \cdot 11.304 = 33.912 \]

Setting the equations equal:

\[ \frac{9.42y}{2} = 33.912 \]

Multiply both sides by 2 to eliminate the fraction:

\[ 9.42y = 67.824 \]

Step 4: Solve for \( y \)

Now, divide both sides by 9.42:

\[ y = \frac{67.824}{9.42} \]

Calculating \( y \):

\[ y \approx 7.19 \]

Result from Previous Calculation

Thus, \( y \approx 7.19 \), which rounds to \( 7.2 \).

Final Confirmation: The provided options are \( 7.2, 4.71, 18, 0.8 \). The value of \( y \) that we calculated is \( 7.2 \), confirming that the answer is indeed:

\[ \boxed{7.2} \]

Thank you for your patience, and the calculations confirm that \( y = 7.2 \) is correct.