Congruence and Similarity

We want to decide if triangles ABC and DEF are congruent.

To decide, let's recall an important property.

Two figures are congruent if the second figure can be obtained from the first figure by a sequence of rotations, reflections, and translations.

This means that we need to look for a sequence of transformations that will map △ ABC onto △ DEF. Let's start with a reflection. We will be reflecting △ ABC across the line x=6.

Looking at the graph, we suspect that △ DEF is a translation of △ ABC. Then, let's also translate △ ABC up 4 units. If △ ABC maps onto △ DEF, then we will know for sure that the triangles are congruent.

We found that △ ABC can be mapped onto △ DEF after a reflection and translation. This means that the triangles are congruent since a sequence of transformations exists that maps △ ABC onto △ DEF.

We want to decide if the triangles ABC and GHI are congruent.

To decide, let's recall the definition of congruent figures.

Two figures are congruent if the second figure can be obtained from the first figure by a sequence of rotations, reflections, and translations.

We need to look for a sequence of transformations that will map △ ABC onto △ GHI. Let's start with a translation. We will map the point C to its corresponding vertex in △ GHI, the point I. This means translating the triangle 3 units right and 1 unit down.

Now let's reflect the triangle so that the side BC aligns with the side HI. This means reflecting △ ABC across the line y=- x+9.

We see that two vertices of △ ABC did not map onto their corresponding vertices in △ GHI. Note that there is no way to match all of the vertices when mapping. The reason is that the corresponding lengths in the triangles are not equal. This means that our triangles are not congruent.

We want to know whether the trapezoid ABCD is similar to the trapezoid EFGH. Let's take a look at the given diagram.

Similar figures have the same shape and congruent angles. Additionally, the corresponding side lengths are proportional. In our case, the pairs of corresponding angles are ∠ A and ∠ E, ∠ B and ∠ F, ∠ C and ∠ G, and ∠ D and ∠ H. We can see from the diagram that the corresponding angles are congruent.

Now let's consider the lengths of the corresponding sides. In our case, the pairs of corresponding sides are AB and EF, BC and FG, CD and GH, and AD and EH.

In similar figures, the lengths of corresponding sides are proportional. \begin{gathered} \dfrac{AB}{EF} \stackrel{?}{=} \dfrac{BC}{FG} \stackrel{?}{=} \dfrac {CD}{GH} \stackrel{?}{=} \dfrac{AD}{EH} \end{gathered} Let's verify if that is the case with our trapezoids!

The lengths of corresponding sides of ABCD and EFGH are proportional. We have also found that the corresponding angles are congruent. This means that the quadrilaterals are similar.

We are asked whether △ MNO is similar to △ PQO. Let's take a look at the given diagram.

We know that in similar figures, the corresponding side lengths are proportional. Both triangles are right triangles, so the corresponding sides are the shorter legs, the longer legs, and the hypotenuses. In other words, the pairs of corresponding sides are the following. Shorter Legs:& MN and PQ Longer Legs:& NO and QO Hypotenuses:& OM and OP If △ MNO is similar to △ PQO, then the lengths of these sides must be proportional. MN/PQ ? = NO/QO ? = OM/OP To verify this, let's find the values of these fractions. For the first fraction, we can see from the diagram that MN = 6 and PQ = 12.

With this information, we can find the value of MNPQ.

Now we will find the value of the second fraction, NOQO. Once again we can see from the diagram that NO = 8 and QO = 10.

Let's find the value of NOQO.

The value of NOQO is 45, which is not equal to 12, the value of MNPQ. Therefore, the corresponding side lengths are not proportional in our triangles. MN/PQ &? = NO/QO ? = OM/OP &⇓ 1/2 &≠ 4/5 ? = OM/OP * This means that △ MNO is not similar to △ PQO.

We are told that ABCDE and FGHIJ are two congruent polygons. We need to identify some of their corresponding parts.

Let's start by identifying the corresponding vertices of the polygons. We can do this by analyzing the angles at different vertices closely. For example, notice that there is one acute angle in ABCDE at C and there is only one acute angle at H in FGHIJ. The remaining relations can be found in similar fashion. ccc ABCDE & ≅ & FGHIJ A & ↔ & F B & ↔ & G C & ↔ & H D & ↔ & I E & ↔ & J Now we can use this information to find the corresponding parts. First, consider ∠ B. In FGHIJ, the vertex that corresponds to B is G. This means that that ∠ B and ∠ G are two corresponding angles. ∠ B ↔ ∠ G Next, we need to find the side that corresponds to AE. We know that vertex A corresponds to F and E corresponds to J. This allows us to conclude that FJ is the corresponding side to AE. AE ↔ FJ Now, we can analyze ∠ D. Between ABCDE and FGHIJ, vertex D corresponds to I. Therefore, ∠ D and ∠ I are corresponding angles. ∠ D ↔ ∠ I Lastly, we will consider the side CD. We will use the fact that vertex C corresponds to H and vertex D corresponds to I. This implies that CD corresponds to HI. CD ↔ HI

We are told that KLMNO and PQRST are two similar polygons. We need to identify some of their corresponding parts.

Let's start by identifying the corresponding vertices of the polygons. We can do this by analyzing the angles at different vertices closely. Notice that the two most obtuse angles in both figures are ∠ M and ∠ R. Also, the adjacent more acute angle ∠ N looks similar to ∠ S. ccc KLMNO & ≅ & PQRST K & & P L & & Q M & & R N & & S O & & T Now we can use this information to find the corresponding parts. First, consider MN. In PQRST, vertex M corresponds to R and N corresponds to S. This means that the side MN corresponds to RS. MN ↔ RS Next, we need to find the side that corresponds to OK. We know that vertex O corresponds to T and K corresponds to P. This allows us to conclude that OK is the corresponding side to TP. OK ↔ TP Now, we can analyze ∠ L. Between KLMNO and PQRST, vertex L corresponds to Q. Therefore, ∠ L and ∠ Q are corresponding angles. ∠ L ↔ ∠ Q Lastly, we will consider ∠ N. We will use the fact that vertex N corresponds to S. This implies that ∠ N indeed corresponds to ∠ S. ∠ N ↔ ∠ S