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Consider two triangles $\triangle abc$ and $\triangle def$ such that $ab=de$ and $ac=df$.Also area of $\triangle abc$ is equal to area of $\triangle def$.Now draw $cm$ perpendicular to $ab$ and $fn$ perpendicular to $de$.$ab$ and $de$ are equal and area of triangles is also equal so $cm$ should be equal to $fn$.Now $\triangle amc$ and$\triangle dnf$ are congruent by right angle triangle congruence(since hypoteneous $ac$ and $df$ are equal).therefore $\angle bac$ is equal to $\angle edf$.Now in $\triangle abc$ and $\triangle def$ by SAS both $\triangle abc$ and $\triangle def$ are congruent so $bc=ef$.I don't know where i am wrong. Show edit:Please read my proof and point out what's wrong $\endgroup$ 2
Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion . In other words, similar triangles are the same shape, but not necessarily the same size. The triangles are congruent if, in addition to this, their corresponding sides are of equal length. The side lengths of two similar triangles are proportional. That is, if Δ U V W is similar to Δ X Y Z , then the following equation holds: U V X Y = U W X Z = V W Y Z This common ratio is called the scale factor . The symbol ∼ is used to indicate similarity.
Example: Δ U V W ∼ Δ X Y Z . If U V = 3 , V W = 4 , U W = 5 and X Y = 12 , find X Z and Y Z . Draw a figure to help yourself visualize.
 Write out the proportion. Make sure you have the corresponding sides right. 3 12 = 5 X Z = 4 Y Z The scale factor here is 3 12 = 1 4 . Solving these equations gives X Z = 20 and Y Z = 16 . The concepts of similarity and scale factor can be extended to other figures besides triangles.
In this explainer, we will learn how to identify triangles that have the same area when their bases are equal in length and the vertices opposite to these bases are on a parallel line to them. To see why this result holds true, let’s consider the following scenario. We have two parallel lines, ⃖⃗𝐴𝐵 and ⃖⃗𝐶𝐷, and two triangles that share a base, △𝐴𝐵𝐶 and △𝐴𝐵𝐷. We can prove that the areas of these triangles are equal by first recalling that the area of a triangle is given by half the length of its base multiplied by its perpendicular height. So, let’s add perpendicular lines from points 𝐶 and 𝐷 to find the perpendicular heights of each triangle. We will call the points of intersection between these perpendicular lines and the parallel lines ⃖⃗𝐴𝐵 and ⃖⃗𝐶𝐷 𝐸 and 𝐹 as shown. We can note that all the angles in 𝐸𝐶𝐷𝐹 are right angles, since ∠𝐸 and ∠𝐹 are right angles and ⃖⃗𝐶𝐷⫽⃖⃗𝐴𝐵. Thus, 𝐸𝐶𝐷𝐹 is a rectangle. Therefore, the lengths of 𝐶𝐸 and 𝐷𝐹 are equal. We can now show that the areas of the triangles are equal by finding expressions for their areas: areaarea△𝐴𝐵𝐶=12(𝐴𝐵)(𝐶𝐸),△𝐴𝐵𝐷=12(𝐴𝐵)(𝐷𝐹). Since 𝐶𝐸=𝐷𝐹, we have areaarea△𝐴𝐵𝐶=△𝐴𝐵𝐷. We have proven the following result. If two triangles share a base and the vertices opposite this base lie on a straight line parallel to the base, then they have equal areas. Let’s now see an example of applying this theorem to find triangles with equal areas. Which of the following has the same area as △𝐷𝐸𝐵? We can answer this question by recalling that two triangles that share a base and have the vertex opposite the base lie on a straight line parallel to the base will have equal areas. We can then note that ⃖⃗𝐷𝐸⫽⃖⃗𝐵𝐶, so any triangle with a base of 𝐷𝐸 and a final vertex on ⃖⃗𝐵𝐶 will have an equal area to △𝐷𝐸𝐵. In particular, △𝐸𝐷𝐶 shares the base 𝐷𝐸 with △𝐷𝐸𝐵, and its vertex 𝐶 lies on 𝐵𝐶, so it has the same area as △𝐷𝐸𝐵, which is option A. In our next example, we will need to apply this theorem to determine the area of a triangle. Given that the area of △𝑌𝐴𝐵=568cm, find the area of △𝑋𝐶𝐷. Let’s start by marking the given triangle and the triangle whose area we wish to find on the given diagram. We can split each of these triangles into two smaller triangles along the line 𝑋𝑌 to get the following. Let’s consider the top two triangles first, as shown. We note that these triangles share a base, 𝑋𝑌, and their opposite vertices 𝐴 and 𝐷 both lie on a line parallel to the base. Hence, we know that these triangles have equal area. Let’s now consider the bottom two triangles, as shown. We can once again note that these two triangles share a base, 𝑋𝑌, and their opposite vertices 𝐵 and 𝐶 both lie on a line parallel to the base. Hence, we know that these triangles have equal areas. Since these triangles have the same areas and they combine to make the larger triangles, 𝑌𝐴𝐵 and 𝑋𝐶𝐷, these must also have the same area. Hence, the area of triangle 𝑋𝐶𝐷 is 568 cm2. In our next example, we will show that if two triangles lie on two parallel lines and they have bases of the same length, then they have the same area. Given that ⃖⃗𝑀𝑋⫽⃖⃗𝐾𝐷, which of the following has the same area as △𝑁𝑀𝐾? We are given a pair of parallel lines, so we can use the fact that if two triangles share a base and the vertices opposite this base lie on a straight line parallel to the base, then they have equal areas. If we choose 𝑀𝑁 as the base of the triangle, then we can choose any point on ⃖⃗𝐷𝐾 as the final vertex of the triangle to find a triangle of equal area to △𝑁𝑀𝐾. Hence, △𝑁𝑀𝐾, △𝑁𝑀𝐶, △𝑁𝑀𝑍, △𝑁𝑀𝑂, and △𝑁𝑀𝐷 all have the same area. However, none of these are options for this question. Instead, let’s use the fact that the area of a triangle is half the length of its base multiplied by its perpendicular height. We choose 𝑀𝑁 as the base of the triangle and can add the perpendicular height ℎ to the diagram as shown. Hence, area△𝑁𝑀𝐾=12(𝑀𝑁)×ℎ. We can use the same method to determine the areas of △𝐶𝑍𝐻 and △𝑂𝐷𝑋. We add the perpendicular lines from the bases to the vertex and note that all of the green lines are parallel. We note that since each of these is a transversal of parallel lines, they also meet ⃖⃗𝑀𝑋 at right angles. Thus, they all form rectangles with sections of ⃖⃗𝑀𝑋 and ⃖⃗𝐷𝐾, so each perpendicular line has the same length of ℎ. Therefore, areaarea△𝐶𝑍𝐻=12(𝐶𝑍)×ℎ,△𝑂𝐷𝑋=12(𝑂𝐷)×ℎ. Finally, since 𝑀𝑁, 𝐶𝑍, and 𝑂𝐷 all have the same length, we can conclude that triangles △𝑁𝑀𝐾, △𝐶𝑍𝐻, and △𝑂𝐷𝑋 all have the same area. Hence, △𝐶𝑍𝐻 has the same area as △𝑁𝑀𝐾, which is option C. In the previous example, we showed the following property. If two triangles lie on two parallel lines and they have bases of the same length, then they have the same area. In our next example, we will consider how the median of a triangle splits the area of the original triangle. If the area of △𝐷𝐸𝐶=6.99cm, find the area of △𝐴𝐵𝐶. We want to determine the area of △𝐴𝐵𝐶 and to do this we are given the area of △𝐷𝐸𝐶. This means that we will want to compare the areas of some triangles to that of △𝐷𝐸𝐶. To do this, we recall that the area of a triangle is half the length of its base multiplied by its perpendicular height. If we choose 𝐶𝐸 to be the base of this triangle, we get the following. We call the point on the perpendicular 𝐹; we can then see that area△𝐷𝐸𝐶=12(𝐶𝐸)(𝐷𝐹). From the diagram, we can note that 𝐴𝐸=𝐶𝐸. In fact, this tells us that 𝐷𝐸 is a median of triangle △𝐴𝐶𝐷. Since triangles △𝐷𝐴𝐸 and △𝐷𝐸𝐶 have the same base length, we can check to see if they have the same perpendicular height. Choosing 𝐴𝐸 as the base, the perpendicular from 𝐷 to ⃖⃗𝐴𝐸 will also intersect at 𝐹, so areaarea△𝐷𝐴𝐸=12(𝐴𝐸)(𝐷𝐹)=12(𝐶𝐸)(𝐷𝐹)=△𝐷𝐸𝐶. Since these triangles combine to make △𝐴𝐶𝐷, we have areacm△𝐴𝐶𝐷=6.99+6.99=13.98. We can apply the exact same reasoning to show that △𝐴𝐶𝐷 and △𝐴𝐷𝐵 have the same area. We see that both triangles have the same base length, since 𝐶𝐷=𝐷𝐵, and these bases lie on the same straight line. Finally, they share the vertex point 𝐴, so the perpendicular distance from the base to 𝐴 will be the same for both triangles. Hence, their areas are the same and so areacm△𝐴𝐵𝐷=13.98. Since △𝐴𝐵𝐶 is the combination of these triangles, we have areaareaareacm△𝐴𝐵𝐶=△𝐴𝐵𝐷+△𝐴𝐶𝐷=13.98+13.98=27.96. In our previous example, we showed two useful results. First, we saw that the median of a triangle will split the triangle into two triangles with the same area. Second, we saw that two triangles with congruent bases on the same straight line that share the opposite vertex will have equal areas since their perpendicular heights are equal. We can write these results formally as follows. Any median of a triangle will split the triangle into two triangles with the same area. Any two triangles with congruent bases that lie on the same straight line and share a common vertex opposite the base have the same area. In our next example, we will apply this property to find triangles of equal area to a given triangle. Which triangle has the same area as △𝐿𝐵𝐶? AnswerWe note that we are given that 𝐵𝐶=𝐷𝑋; we can then recall that any two triangles with congruent bases that lie on the same straight line and share a common vertex opposite the base have the same area. Hence, △𝐿𝐵𝐶 and △𝐷𝑋𝐿 have the same area. Thus far, we have concentrated on finding triangles with equal areas to a given triangle or using these results to determine areas. However, we can also ask the same questions in reverse. For example, if two triangles of equal area share a base and their vertices opposite the base lie on the same side, what can we say about these vertices? To help us understand the situation, let’s first sketch this information. We know that △𝐴𝐵𝐶 and △𝐴𝐵𝐷 have the same area; we can find expressions for the area of each triangle by using half the length of the base multiplied by the perpendicular height. Adding the perpendiculars to the diagram gives us the following. We now have areaarea△𝐴𝐵𝐶=12(𝐴𝐵)(𝐶𝐹),△𝐴𝐵𝐷=12(𝐴𝐵)(𝐷𝐺). Since the areas of the triangles are equal, we must have that 𝐶𝐹=𝐺𝐷. Next, we note that these lines are perpendicular to ⃖⃗𝐺𝐹 and are of the same length; hence, we must have that 𝐶𝐷𝐺𝐹 is a rectangle and in particular this means that ⃖⃗𝐶𝐷⫽⃖⃗𝐺𝐹. We have proven the following result. If two triangles share a base and have equal areas and the vertices opposite the base lie on the same side of the base, then these vertices lie on a straight line parallel to the base. It is worth noting that this result also holds if the triangles have congruent bases on the same line. We can write this formally as follows. If two triangles have congruent bases on a straight line and have equal areas and the vertices opposite the base lie on the same side of the base, then these vertices lie on a straight line parallel to the base. Let’s now see an example of applying this theorem to identify a geometric property from a diagram. If the areas of △𝐿𝑁𝐴 and △𝑌𝐴𝐺 are the same, which of the following must be true?
AnswerWe start by noting that each of triangles △𝐿𝑁𝐴 and △𝑌𝐴𝐺 is composed of two smaller triangles; we can compare the area of these smaller triangles. Let’s start by comparing triangles △𝐴𝐺𝐸 and △𝐴𝐷𝑁; we can do this by adding the perpendicular distance from 𝐴 to ⃖⃗𝐺𝑁 to the diagram as shown. We recall that any two triangles with congruent bases that lie on the same straight line and share a common vertex opposite the base have the same area. Hence, areaarea△𝐴𝐺𝐸=△𝐴𝐷𝑁. Combining this result with the fact that triangles △𝐿𝑁𝐴 and △𝑌𝐴𝐺 have the same area means that triangles △𝑌𝐺𝐸 and △𝐿𝑁𝐷 must also have the same area. We then recall that if two triangles have congruent bases on a straight line and have equal areas and the vertices opposite the base lie on the same side of the base, then these vertices lie on a straight line parallel to the base. Hence, 𝑌𝐿⫽𝑁𝐺, which is option E. In our final example, we will apply these theorems and properties to determine a geometric property from a given diagram. Points 𝑍, 𝐻, and 𝐷 are collinear. If the areas of △𝑍𝐶𝐻 and △𝐶𝐻𝐷 are the same, which of the following must be true?
AnswerWe first note that ⃖⃗𝐻𝐹 is parallel to ⃖⃗𝐶𝑍 and ⃖⃗𝐻𝑍 is parallel to ⃖⃗𝐶𝐹. Thus, 𝐶𝑍𝐻𝐹 is a parallelogram. Hence, its diagonal, 𝐶𝐻, splits the parallelogram into two equal-area triangles, △𝑍𝐶𝐻 and △𝐶𝐻𝐹. It is also worth noting that saying three points are collinear means that they all lie on the same straight line. Therefore, since areaarea△𝑍𝐶𝐻=△𝐶𝐻𝐷, we can conclude that area△𝐶𝐻𝐷=△𝐶𝐻𝐹. We can also note that these triangles share the same base, 𝐶𝐻. We can then recall that if two triangles share a base and have equal areas and the vertices opposite the base lie on the same side of the base, then these vertices lie on a straight line parallel to the base. Since triangles △𝐶𝐻𝐷 and △𝐶𝐻𝐹 have the same area, share a base 𝐶𝐻, and have vertices on the same side of the base, we can conclude that the base is parallel to the line between the vertices opposite the base. That is 𝐶𝐻⫽𝐹𝐷, which is option B. In the previous example, there are actually many different ways of determining the result. For example, we could use the fact that if the areas of △𝑍𝐶𝐻 and △𝐶𝐻𝐷 are the same, then their bases lying on the same line are congruent, which leads to 𝐻𝐷=𝑍𝐻=𝐹𝐶, and as 𝐻𝐷⫽𝐹𝐶, so 𝐻𝐷𝐹𝐶 is a parallelogram. Let’s finish by recapping some of the important points from this explainer.
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Will two triangles of same area always be congruent why
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