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

Let's solve the problem step by step:

(a) Form an equation and simplify to 2x2+5x52=02x^2 + 5x - 52 = 0

The formula for the area AA of a trapezium is:

A=12×(PQ+RS)×heightA = \frac{1}{2} \times (PQ + RS) \times \text{height}

From the diagram, we know:

  • PQ=xcmPQ = x \, \text{cm}
  • RS=3x2cmRS = 3x - 2 \, \text{cm}
  • The height is PS=x+3cmPS = x + 3 \, \text{cm}

The area is given as 49 cm². Therefore, we substitute these values into the area formula:

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

Simplify the expression inside the parentheses:

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

Now move all terms to the left-hand side:

4x2+10x698=04x^2 + 10x - 6 - 98 = 0

4x2+10x104=04x^2 + 10x - 104 = 0

Finally, divide the entire equation by 2:

2x2+5x52=02x^2 + 5x - 52 = 0

This is the required equation.


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

To solve this quadratic equation, we can 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, the coefficients are:

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

Now, 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

Since the discriminant is positive, we have two real solutions. Now apply the quadratic formula:

x=5±4412(2)x = \frac{-5 \pm \sqrt{441}}{2(2)}

x=5±214x = \frac{-5 \pm 21}{4}

This gives two possible values for xx:

x=5+214=164=4x = \frac{-5 + 21}{4} = \frac{16}{4} = 4

or

x=5214=264=6.5x = \frac{-5 - 21}{4} = \frac{-26}{4} = -6.5

Since xx represents a length, we discard the negative solution, so x=4x = 4.


(c) Find the length of RS

Now that we know x=4x = 4, we can substitute this into the expression for RSRS:

RS=3x2=3(4)2=122=10cmRS = 3x - 2 = 3(4) - 2 = 12 - 2 = 10 \, \text{cm}

Thus, the length of RSRS is 10 cm.


Would you like further details or have any questions?


Related Questions:

  1. How would the area formula change if PQPQ and RSRS were not parallel?
  2. What is the significance of the discriminant in the quadratic formula?
  3. Can you solve the equation using factoring instead of the quadratic formula?
  4. How can we verify that x=4x = 4 is the correct solution using a different method?
  5. What happens to the dimensions of the trapezium if the area is doubled?

Tip:

When solving quadratic equations, always check whether the discriminant is positive, zero, or negative to determine the nature of the roots (real or complex).

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Geometry
Area of Trapezium

Formulas

Area of trapezium: A = 1/2 × (PQ + RS) × height
Quadratic formula: x = [-b ± sqrt(b² - 4ac)] / 2a

Theorems

Quadratic Formula

Suitable Grade Level

Grades 9-10