In a trapezium ABCD, AB is parallel to DC. The diagonals AC and BD intersect at P. If AP ∶ PC = 4 ∶ (4x - 4) and BP ∶ PD = (2x - 1) ∶ (2x + 4), then what is the value of x?
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Solving for x in a Trapezium Diagonals Ratio Problem
In this problem, we are given a trapezium ABCD where side AB is parallel to side DC. The diagonals AC and BD intersect at point P. We are provided with ratios of the segments of the diagonals, specifically AP : PC and BP : PD, expressed in terms of 'x'. Our goal is to find the value of x that satisfies these conditions.
Understanding Trapezium Properties and Diagonals
A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides. In trapezium ABCD, AB || DC. When the diagonals AC and BD intersect at a point P, important geometric relationships are formed, particularly involving similar triangles.
Similar Triangles formed by Trapezium Diagonals
Consider the triangles formed by the intersecting diagonals and the parallel sides: \(\triangle APB\) and \(\triangle CPD\).
- \(\angle PAB\) and \(\angle PCD\) are alternate interior angles because AB is parallel to DC and AC is a transversal line. Therefore, \(\angle PAB = \angle PCD\).
- \(\angle PBA\) and \(\angle PDC\) are alternate interior angles because AB is parallel to DC and BD is a transversal line. Therefore, \(\angle PBA = \angle PDC\).
- \(\angle APB\) and \(\angle CPD\) are vertically opposite angles. Therefore, \(\angle APB = \angle CPD\).
Since two angles of \(\triangle APB\) are equal to two corresponding angles of \(\triangle CPD\), the third angles must also be equal. Thus, \(\triangle APB\) is similar to \(\triangle CPD\) by AAA similarity criterion.
Because \(\triangle APB \sim \triangle CPD\), the ratios of their corresponding sides are equal. The corresponding sides are AP to PC, BP to PD, and AB to DC.
So, we have the relationship: \[\frac{AP}{PC} = \frac{BP}{PD} = \frac{AB}{DC}\]
Setting up the Equation using Given Ratios
We are given the ratios:
- AP : PC = 4 : (4x - 4)
- BP : PD = (2x - 1) : (2x + 4)
Using the similarity property \(\frac{AP}{PC} = \frac{BP}{PD}\), we can set up the following equation:
\[ \frac{4}{4x - 4} = \frac{2x - 1}{2x + 4} \]Solving the Equation for x
Now, we need to solve this algebraic equation for x. We can cross-multiply to eliminate the denominators:
\[ 4 \times (2x + 4) = (4x - 4) \times (2x - 1) \]Expand both sides of the equation:
\[ 8x + 16 = (4x)(2x) + (4x)(-1) + (-4)(2x) + (-4)(-1) \] \[ 8x + 16 = 8x^2 - 4x - 8x + 4 \]Combine like terms on the right side:
\[ 8x + 16 = 8x^2 - 12x + 4 \]Move all terms to one side to form a standard quadratic equation (\(ax^2 + bx + c = 0\)):
\[ 0 = 8x^2 - 12x - 8x + 4 - 16 \] \[ 0 = 8x^2 - 20x - 12 \]We can simplify the equation by dividing all terms by the greatest common divisor, which is 4:
\[ \frac{0}{4} = \frac{8x^2}{4} - \frac{20x}{4} - \frac{12}{4} \] \[ 0 = 2x^2 - 5x - 3 \]Now we solve the quadratic equation \(2x^2 - 5x - 3 = 0\). We can solve this by factoring. We look for two numbers that multiply to \((2)(-3) = -6\) and add up to -5. These numbers are -6 and 1.
Rewrite the middle term (-5x) using -6x and 1x:
\[ 2x^2 - 6x + x - 3 = 0 \]Group terms and factor:
\[ (2x^2 - 6x) + (x - 3) = 0 \] \[ 2x(x - 3) + 1(x - 3) = 0 \]Factor out the common term \((x - 3)\):
\[ (x - 3)(2x + 1) = 0 \]Set each factor equal to zero to find the possible values of x:
\[ x - 3 = 0 \quad \text{or} \quad 2x + 1 = 0 \] \[ x = 3 \quad \text{or} \quad 2x = -1 \] \[ x = 3 \quad \text{or} \quad x = -\frac{1}{2} \]Validating the Solution for x
The ratios represent lengths of segments, which must be positive. We need to check if the values of x make the terms in the ratios positive.
The terms are \((4x - 4)\), \((2x - 1)\), and \((2x + 4)\).
Check \(x = -\frac{1}{2}\) :
- \(4x - 4 = 4(-\frac{1}{2}) - 4 = -2 - 4 = -6\). This is negative.
Since a segment length or a part of a ratio representing length cannot be negative, \(x = -\frac{1}{2}\) is not a valid solution in this geometric context.
Check \(x = 3\) :
- \(4x - 4 = 4(3) - 4 = 12 - 4 = 8\). This is positive.
- \(2x - 1 = 2(3) - 1 = 6 - 1 = 5\). This is positive.
- \(2x + 4 = 2(3) + 4 = 6 + 4 = 10\). This is positive.
Since all terms in the ratios are positive when \(x=3\), this is a valid solution.
For \(x=3\), the ratios are AP : PC = 4 : 8 = 1 : 2 and BP : PD = 5 : 10 = 1 : 2. The equality of the ratios is satisfied.
Final Answer for x
The only valid value for x is 3.
| Step | Description | Calculation/Reason |
|---|---|---|
| 1 | Identify similar triangles | \(\triangle APB \sim \triangle CPD\) (AAA similarity) |
| 2 | Write ratio equality | \(\frac{AP}{PC} = \frac{BP}{PD}\) |
| 3 | Substitute given ratios | \(\frac{4}{4x - 4} = \frac{2x - 1}{2x + 4}\) |
| 4 | Solve the equation | \(2x^2 - 5x - 3 = 0\) |
| 5 | Find possible x values | \(x = 3\), \(x = -\frac{1}{2}\) |
| 6 | Validate x values | Lengths must be positive; \(x = -\frac{1}{2}\) is invalid. |
| 7 | State valid x value | \(x = 3\) |
Revision Table: Key Concepts for Trapezium Diagonal Problems
| Concept | Description | Relevance to the Problem |
|---|---|---|
| Trapezium Properties | A quadrilateral with at least one pair of parallel sides. | AB || DC condition is crucial. |
| Intersecting Diagonals | Diagonals AC and BD intersect at P. | Forms triangles \(\triangle APB\) and \(\triangle CPD\). |
| Similar Triangles | Triangles with corresponding angles equal and corresponding sides proportional. | \(\triangle APB \sim \triangle CPD\) because of alternate interior angles and vertically opposite angles. |
| Ratio of Corresponding Sides | For similar triangles, the ratio of lengths of corresponding sides is constant. | \(\frac{AP}{PC} = \frac{BP}{PD} = \frac{AB}{DC}\) is the key relationship used. |
| Solving Quadratic Equations | Finding the values of the variable that satisfy an equation of degree 2. | The problem reduces to solving \(2x^2 - 5x - 3 = 0\). |
| Validating Solutions | Checking if the mathematical solution makes sense in the real-world or geometric context. | Ratio terms must be positive lengths, eliminating \(x = -\frac{1}{2}\). |
Additional Information: Similar Triangles in Trapeziums
The property that the triangles formed by the intersection of the diagonals and the parallel sides of a trapezium are similar (\(\triangle APB \sim \triangle CPD\)) is a fundamental result in trapezium geometry. This similarity holds true regardless of whether the trapezium is isosceles or right-angled, as long as it has at least one pair of parallel sides (AB || DC in this case). The ratio of similarity between \(\triangle APB\) and \(\triangle CPD\) is equal to the ratio of the lengths of the parallel sides, i.e., \(\frac{AB}{DC}\). This relationship is very useful for solving problems involving lengths and ratios of segments within a trapezium.
For example, if you knew the lengths of the parallel sides AB and DC, you could find the ratio \(\frac{AB}{DC}\) directly, and this ratio would be equal to \(\frac{AP}{PC}\) and \(\frac{BP}{PD}\). In this problem, we used the equality of the diagonal segment ratios (\(\frac{AP}{PC} = \frac{BP}{PD}\)) to find x, which implicitly relies on the similarity of \(\triangle APB\) and \(\triangle CPD\).
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