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 Arectangle can be expressed as:
Arectangle=(3x−2)(x−1)
Expanding this gives:
Arectangle=3x2−3x−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 Atriangle can be expressed as:
Atriangle=21×Base×Height=21×x×2x=x2
Step 3
Step 3: Set up the inequality
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Answer
To find the set of possible values for x, we need to set up the inequality based on the problem statement:
Arectangle>Atriangle
This translates to:
3x2−3x−2>x2
Step 4
Step 4: Rearrange and simplify the inequality
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Answer
Rearranging the inequality gives:
3x2−3x−2−x2>0
Which simplifies to:
2x2−3x−2>0
Step 5
Step 5: Solve the quadratic inequality
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Answer
To solve the inequality 2x2−3x−2>0, we first find the roots using the quadratic formula:
x=2a−b±b2−4ac
Here, a=2, b=−3, and c=−2:
x=2⋅23±(−3)2−4⋅2⋅(−2)=43±9+16=43±5
This gives:
x=2(approximately)orx=−21
Hence, the critical points are x=2 and x=−21.
Step 6
Step 6: Determine the intervals
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Answer
To analyze the intervals:
Test points in the intervals (−∞,−21), (−21,2), and (2,+∞) to find where the inequality holds true.
After testing these intervals, we find that the inequality holds for:
x>2.