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In the figure, EF || AC, BC = 10 cm, AB ...

In the figure, EF || AC, BC = 10 cm, AB = 13 cm and EC = 2 cm, find AF

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To solve the problem step by step, we will use the information given and apply Thales' theorem. ### Step 1: Understand the Given Information We have a figure where EF is parallel to AC. The lengths given are: - BC = 10 cm - AB = 13 cm - EC = 2 cm We need to find the length of AF. ### Step 2: Calculate BE Since EF is parallel to AC, we can find the length of BE using the relation: \[ BE = BC - EC \] Substituting the values: \[ BE = 10 \, \text{cm} - 2 \, \text{cm} = 8 \, \text{cm} \] ### Step 3: Set Up the Variables Let: - AF = x cm - BF = AB - AF = 13 cm - x cm ### Step 4: Apply Thales' Theorem According to Thales' theorem, the ratios of the segments are equal: \[ \frac{BF}{AF} = \frac{BE}{EC} \] Substituting the known values: \[ \frac{13 - x}{x} = \frac{8}{2} \] This simplifies to: \[ \frac{13 - x}{x} = 4 \] ### Step 5: Cross-Multiply to Solve for x Cross-multiplying gives: \[ 13 - x = 4x \] Now, rearranging the equation: \[ 13 = 4x + x \] \[ 13 = 5x \] ### Step 6: Solve for x Dividing both sides by 5: \[ x = \frac{13}{5} \] Calculating this gives: \[ x = 2.6 \, \text{cm} \] ### Step 7: Conclusion Thus, the length of AF is: \[ AF = 2.6 \, \text{cm} \]
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ICSE-MEASURES OF CENTRAL TENDENCY (MEAN, MEDIAN, QUARTILES AND MODE)-EXERCISE 24 (E)
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