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Here is a rectangle and a right-angled triangle - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 1

Question 23

Here is a rectangle and a right-angled triangle. All measurements are in centimetres. The area of the rectangle is greater than the area of the triangle. Find the ... show full transcript

Worked Solution & Example Answer:Here is a rectangle and a right-angled triangle - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 1

Step 1

Step 1: Express the area of the rectangle

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Answer

The dimensions of the rectangle are given as:

Length: $3x - 2$ cm
Width: $x - 1$ cm

The area of the rectangle $A_{rectangle}$ can be expressed as: $A_{rectangle} = (3x - 2)(x - 1)$ Expanding this gives: $A_{rectangle} = 3x^2 - 3x - 2$

Step 2

Step 2: Express the area of the triangle

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Answer

The base and height of the right-angled triangle are given as:

Base: $x$ cm
Height: $2x$ cm

The area of the triangle $A_{triangle}$ can be expressed as: $A_{triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times x \times 2x = x^2$

Step 3

Step 3: Set up the inequality

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To find the set of possible values for $x$ , we need to set up the inequality based on the problem statement: $A_{rectangle} > A_{triangle}$ This translates to: $3x^2 - 3x - 2 > x^2$

Step 4

Step 4: Rearrange and simplify the inequality

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Rearranging the inequality gives: $3x^2 - 3x - 2 - x^2 > 0$ Which simplifies to: $2x^2 - 3x - 2 > 0$

Step 5

Step 5: Solve the quadratic inequality

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To solve the inequality $2x^2 - 3x - 2 > 0$ , we first find the roots using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 2$ , $b = -3$ , and $c = -2$ : $x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}$ This gives: $x = 2 \quad \text{(approximately)} \quad \text{or} \quad x = -\frac{1}{2}$ Hence, the critical points are $x = 2$ and $x = -\frac{1}{2}$ .

Step 6

Step 6: Determine the intervals

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To analyze the intervals:

Test points in the intervals $(-\infty, -\frac{1}{2})$ , $(-\frac{1}{2}, 2)$ , and $(2, +\infty)$ to find where the inequality holds true.

After testing these intervals, we find that the inequality holds for: $x > 2$ .

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Here is a rectangle and a right-angled triangle - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 1

Worked Solution & Example Answer:Here is a rectangle and a right-angled triangle - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 1

Step 1: Express the area of the rectangle

Step 2: Express the area of the triangle

Step 3: Set up the inequality

Step 4: Rearrange and simplify the inequality

Step 5: Solve the quadratic inequality

Step 6: Determine the intervals

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