Math Problem Statement

The diagram shows trapezium PQRS. PQ is parallel to RS and angle PSR = 90°. PQ = x cm, PS = (x + 3) cm and RS = (3x - 2) cm. (a) Given that the area of the trapezium is 49 cm?, form an equation in x and show that it simplifies to 2x* + 5x - 52 = 0. (b) Solve the equation 2x2 + 5x - 52 = 0. (c) Hence, find the length of RS.

The diagram shows trapezium PQRS. PQ is parallel to RS and angle PSR = 90°. PQ = x cm, PS = (x + 3) cm and RS = (3x - 2) cm. (a) Given that the area of the trapezium is 49 cm², form an equation in x and show that it simplifies to 2x² + 5x - 52 = 0. (b) Solve the equation 2x² + 5x - 52 = 0. (c) Hence, find the length of RS.

Solution

Step-by-step Solution

The trapezium PQRSPQRS has the following dimensions:

  • PQ=xcmPQ = x \, \text{cm}
  • PS=x+3cmPS = x + 3 \, \text{cm}
  • RS=3x2cmRS = 3x - 2 \, \text{cm}
  • PSR=90\angle PSR = 90^\circ, making PSRSPS \perp RS.

Part (a): Form an equation for the area of the trapezium

The area AA of a trapezium is given by the formula:

A=12×(a+b)×hA = \frac{1}{2} \times (a + b) \times h

Where:

  • aa and bb are the lengths of the parallel sides (PQPQ and RSRS),
  • hh is the perpendicular height (which is PSPS in this case).

Substituting the given values: PQ=xcm,RS=(3x2)cm,PS=(x+3)cmPQ = x \, \text{cm}, \quad RS = (3x - 2) \, \text{cm}, \quad PS = (x + 3) \, \text{cm}

The area of the trapezium is given as 49cm249 \, \text{cm}^2. Therefore, substituting into the area formula:

49=12×(x+(3x2))×(x+3)49 = \frac{1}{2} \times (x + (3x - 2)) \times (x + 3)

Simplifying: 49=12×(4x2)×(x+3)49 = \frac{1}{2} \times (4x - 2) \times (x + 3)

Multiply both sides by 2 to eliminate the fraction: 98=(4x2)(x+3)98 = (4x - 2)(x + 3)

Now, expand the right-hand side: 98=4x(x+3)2(x+3)98 = 4x(x + 3) - 2(x + 3)

98=4x2+12x2x698 = 4x^2 + 12x - 2x - 6

98=4x2+10x698 = 4x^2 + 10x - 6

Subtract 98 from both sides: 0=4x2+10x6980 = 4x^2 + 10x - 6 - 98

0=4x2+10x1040 = 4x^2 + 10x - 104

Now divide the entire equation by 2 to simplify: 0=2x2+5x520 = 2x^2 + 5x - 52

This is the required equation.

Part (b): Solve the quadratic equation 2x2+5x52=02x^2 + 5x - 52 = 0

We will use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For the equation 2x2+5x52=02x^2 + 5x - 52 = 0, we have:

  • a=2a = 2,
  • b=5b = 5,
  • c=52c = -52.

First, calculate the discriminant: Δ=b24ac=524(2)(52)\Delta = b^2 - 4ac = 5^2 - 4(2)(-52) Δ=25+416=441\Delta = 25 + 416 = 441

Now, substitute into the quadratic formula: x=5±4412(2)x = \frac{-5 \pm \sqrt{441}}{2(2)} x=5±214x = \frac{-5 \pm 21}{4}

Thus, the two possible values for xx are: x1=5+214=164=4x_1 = \frac{-5 + 21}{4} = \frac{16}{4} = 4 x2=5214=264=6.5x_2 = \frac{-5 - 21}{4} = \frac{-26}{4} = -6.5

Since xx represents a length, it must be positive. Therefore, x=4x = 4.

Part (c): Find the length of RSRS

Now that we know x=4x = 4, we can find the length of RSRS: RS=3x2=3(4)2=122=10cmRS = 3x - 2 = 3(4) - 2 = 12 - 2 = 10 \, \text{cm}

Final Answers:

  • (a) The equation simplifies to 2x2+5x52=02x^2 + 5x - 52 = 0.
  • (b) The solution to the equation is x=4x = 4.
  • (c) The length of RSRS is 10cm10 \, \text{cm}.

Let me know if you need any additional details or have questions!


Questions for deeper understanding:

  1. How would the equation change if the area of the trapezium were different?
  2. What are the properties of trapeziums that help in solving such problems?
  3. Why did we discard the negative value of xx in the solution?
  4. How would you calculate the perimeter of trapezium PQRSPQRS using the values obtained?
  5. What would be the next steps if the angle PSR\angle PSR was different?

Tip: When using the quadratic formula, always check whether both roots are applicable based on the context of the problem (e.g., lengths must be positive).

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Quadratic Equations
Area of Trapezium

Formulas

Area of a trapezium A = (1/2) * (a + b) * h
Quadratic equation formula ax² + bx + c = 0
Quadratic formula x = (-b ± √(b² - 4ac)) / (2a)

Theorems

Properties of trapeziums
Quadratic formula

Suitable Grade Level

Grades 8-10