Straight Lines
Geometric Configuration with Lines
Grade 11
Question:
<p>Find the area of the square ABCD where <i>A(a,0)</i>, <i>B(0,a)</i>, <i>C(a,a)</i>, and the configuration involves point <i>F</i> at <i>\left(-\frac{a}{3}, a\right)</i> with <i>BF = \frac{a}{4}(a + BF)</i>.</p>
<p>(A) 36</p>
<p>(B) 42</p>
<p>(C) 49</p>
<p>(D) 64</p>
Step-by-Step Solution
Key Concept: Use similarity of triangles and the given constraint on BF to find the side length a, then calculate the area.
<p><strong>Step 1:</strong> From the given condition <i>BF = \frac{a}{4}(a + BF)</i>, we solve for <i>BF</i>:</p><p>\[4BF = a + BF \implies 3BF = a \implies BF = \frac{a}{3}\]</p><p><strong>Step 2:</strong> Using the similarity relation <i>\triangle BFE \sim \triangle CFA</i> and geometric constraints with <i>E\left(0, \frac{3a}{4}\right)</i>:</p><p>\[AE = \sqrt{a^2 + \left(\frac{3a}{4}\right)^2} = \sqrt{a^2 + \frac{9a^2}{16}} = \frac{5a}{4}\]</p><p><strong>Step 3:</strong> From the constraint that <i>l = \frac{15}{4}</i> and <i>l = \frac{3a}{4}(5)</i>:</p><p>\[\frac{15}{4} = \frac{3a}{4} \cdot 5 \implies a = 7\]</p><p><strong>Step 4:</strong> Area of square = <i>a²</i> = 49</p><p>∴ Answer is <i>(C) 49</i>.</p>
Correct Answer: C