Example Question - congruent triangles
Here are examples of questions we've helped users solve.
Determining the True Statement About Angles in a Geometric Diagram
Let angle BCD = x Since AB \cong DE and \angle A = \angle D (given m \angle B = 43^{\circ}), triangles ABC and CDE are congruent by the ASA (Angle-Side-Angle) criterion. Therefore, m \angle BCE = m \angle BCD = x (corresponding angles of congruent triangles are equal). In \triangle BCE, m \angle BCE + m \angle BCE + m \angle CEF = 180^{\circ} (sum of angles in a triangle) x + x + 152^{\circ} = 180^{\circ} 2x = 180^{\circ} - 152^{\circ} 2x = 28^{\circ} x = 14^{\circ} Since m \angle BCE = x, m \angle BCE = 14^{\circ}. But there is no statement that says m \angle BCE = 14^{\circ}. Now we need to check angle ACD: angle ACD = angle BCE (by congruent triangles ABC and CDE) angle ACD = 14^{\circ} According to the choices given: \text{If statement 3 is } \angle ACD = 71^{\circ}, \text{then it is false, as we calculated it to be } 14^{\circ}. The only statement we did not refute directly is statement 1: \text{If statement 1 is } m \angle D = 28^{\circ}, \text{ angles D and B would sum to } 43^{\circ} + 28^{\circ} = 71^{\circ}, \text{ which is not the straight angle sum of } 180^{\circ}. \text{ Therefore, statement 1 is false.} Assuming the diagram and markings are accurate (as we are bound by the image provided), we have shown that statement 1 and statement 3 are both incorrect. Statement 2 asserts m \angle A = 43^{\circ}, which we can infer from the congruence of \triangle ABC and \triangle CDE since AB \cong DE and \angle B \cong \angle D. So by process of elimination and confirming with the congruent triangles, the true statement is: \text{Statement 2: } m \angle A = 43^{\circ}.
Congruent Triangles and Side Lengths
The image shows a pair of congruent triangles, ΔDEG and ΔEFG, with DE congruent to EF, DG equal to 3a, and FG equal to a + 42. In congruent triangles, corresponding sides are equal in length. Therefore: DE = EF Since FG is the sum of DG and EG, and EG is equal to DE (because DE = EF and EF = EG by congruency), you can express FG as: FG = DG + EG Given: DG = 3a EG = DE = EF (Because of the congruency between ΔDEG and ΔEFG) Since DE is congruent to EF, that implies EG = EF. So using the information that FG = DG + EG, we can substitute the given values into the equation: FG = 3a + EF We were also given that FG = a + 42. This allows us to set up the following equation since they both represent FG: 3a + EF = a + 42 However, to find FG, we do not actually need to solve for a or EF individually since FG equals a + 42 by the given information. Therefore: FG = a + 42 This is the expression for FG, and without additional information or numerical values provided for a, this is as simplified as it gets.
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