Example Question - area ratio
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Similarity and Ratio in a Parallelogram
<p>\text{(a) Similarity Reasoning:}</p> <p>Triangle \triangle AXM \:\text{is similar to}\: \triangle CXD \:\text{because}:</p> <p>\text{1. } \angle AXM = \angle CXD \:\text{opposite angles are equal in a parallelogram}.</p> <p>\text{2. } \angle A = \angle C \:\text{alternate angles are equal as AD} \parallel \text{BC in a parallelogram}.</p> <p>\text{Therefore, by AA similarity criterion, } \triangle AXM \sim \triangle CXD</p> <p>\text{(b) Area Ratio Calculation:}</p> <p>\text{Since } \triangle AXM \sim \triangle CXD,\: \text{the ratio of their areas is the square of the ratio of their corresponding sides.}</p> <p>\text{The corresponding sides are } AM \:\text{and} \: CD.\:</p> <p>AM = \frac{1}{2}AD \:\text{since M is the midpoint of AD}.</p> <p>CD = AD \:\text{as opposite sides of a parallelogram are equal}.</p> <p>\text{Therefore, the ratio of } AM \:\text{to} \: CD \:\text{is} \:\frac{AM}{CD} = \frac{1/2 \cdot AD}{AD} = \frac{1}{2}.</p> <p>\text{The ratio of areas is } (\frac{AM}{CD})^2 = (\frac{1}{2})^2 = \frac{1}{4}.</p> <p>\text{Hence,} \:\frac{\text{area of } \triangle AXM}{\text{area of } \triangle CXD} = \frac{1}{4}.</p>
Analysis of Geometric Similarity and Area Ratio in a Quadrilateral
(a) <p>\angle AXM = \angle CXD \quad (\text{vertically opposite angles are equal})</p> <p>\angle AMX = \angle CDX \quad (\text{corresponding angles of parallel lines are equal})</p> <p>\angle A = \angle C \quad (\text{given})</p> <p>\text{By AA similarity criterion,} \triangle AXM \sim \triangle CXD.</p> (b) <p>\frac{\text{area of }\triangle AMX}{\text{area of }\triangle CXD} = \left(\frac{AM}{CD}\right)^2</p> <p>\text{Since }\triangle AXM \sim \triangle CXD\text{, their sides are proportional. Therefore, } \frac{AM}{CD} = \frac{AX}{CX}</p> <p>\text{Therefore, the area ratio is } \left(\frac{AX}{CX}\right)^2.</p>
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