Example Question - geometric similarity
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Geometric Similarity and Ratio Calculation in Triangles
<p>(a) In triangle ABM and CDM:</p> <p>\[\angle MAB = \angle MCD\] (Vertically opposite angles are equal)</p> <p>\[\angle AMB = \angle CMD\] (Each equals \(90^\circ\))</p> <p>Therefore, by AA (Angle-Angle) similarity criterion, \(\triangle ABM \sim \triangle CDM\).</p> <p>(b) The ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides:</p> <p>\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle CDM}} = \left(\frac{AM}{CM}\right)^2\]</p> <p>(c) \[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \frac{\frac{1}{2} AM \times BM}{\frac{1}{2} AC \times BC}\]</p> <p>Given that \(BM = MC\) and \(AM = MC\),</p> <p>\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \left(\frac{MC}{AC}\right) \times \left(\frac{AM}{BC}\right)\]</p> <p>Since \(AB = AC\), \(BC = AB - AM\),</p> <p>\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \frac{MC}{AC} \times \frac{AM}{AB - AM}\]</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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