homework 6.4 centers of triangles

Mathematical Knowledge for Secondary Teachers

10.6 centers of triangles.

When we think about a “center” of a triangle, there are many different properties that we may want. We will now explore how some of these properties generate various “centers” of a triangle.

10.6.1 Centroid

The first “center” that we will consider is the point that is the arithmetic mean of all of the points contained inside of the triangle. From a physics perspective, this point is the center of mass, or the balance point, for the triangle. To generate this point, we need to first create some definitions.

Definition 10.7 The median of a side of a triangle is the segment connecting the midpoint of the side to the opposite vertex.

So for \(\triangle ABC\) , if \(D\) is the midpoint of \(AB\) , then \(CD\) is the median of \(AB\) .

Theorem 10.17 Medians of a triangle are concurrent at a point.

We will give an outline of the proof in a series of lemmas with some key details left to be filled in. We recommend that as you read through the proof you should use a dynamic geometry application to construct the diagrams and to study properties of the constructions.

Lemma 10.1 Let \(\triangle ABC\) be given. Let \(A'\) be the midpoint of \(BC\) , \(B'\) be the midpoint of \(AC\) , and \(C'\) be the midpoint of \(AB\) .

Then \(\triangle A'B'C\sim \triangle BAC\) with \[\frac{A'B'}{BA}= \frac{A'C}{BC} = \frac{B'C}{AC}=2\] and triangles \(\triangle AB'C'\) , \(\triangle B'A'C\) , \(\triangle C'BA'\) , and \(\triangle A'B'C'\) are all congruent and so are also all similar to \(\triangle ABC\) with ratio of \(1:2\) .

Proof . Let \(\triangle ABC\) be given. Let \(A'\) be the midpoint of \(BC\) , \(B'\) be the midpoint of \(AC\) , and \(C'\) be the midpoint of \(AB\) .

Since \(A'\) is the midpoint of \(BC\) , \(B'\) is the midpoint of \(AC\) , and angles \(\angle BCA\) and \(\angle A'CB'\) are equal, by Theorem 10.16 we know that \(\triangle A'B'C\sim \triangle BAC\) with \[\frac{A'B'}{BA}= \frac{A'C}{BC} = \frac{B'C}{AC}=\frac{1}{2}.\]

Using similar arguments, we have that \(\triangle AB'C' \sim \triangle ACB\) and \(\triangle BC'A' \sim \triangle BAC\) with a ratio of \(1:2\) .

We now know that \[\frac{A'B'}{AB}=\frac{A'C'}{AC} = \frac{B'C'}{BC} = \frac{1}{2}\] and so by Theorem 10.15 \(\triangle A'B'C' \sim \triangle ABC\) with a ratio of \(1:2\) .

So the segments connecting the midpoints of the sides of a triangle create four congruent triangles that are similar to the original triangle and whose side lengths are half the length of the original.

Related Content Standards

• (HSG.CO.10) Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to \(180^\circ\) ; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.

Lemma 10.2 The segments joining the midpoints of two sides of a triangle is parallel to the third side and half the length.

Proof . The proof the previous lemma included that the segments joining the midpoints of two sides of a triangle is half the length of the third side. So we only need to prove that it is parallel to the third side.

Since the four triangles inside of the original triangle are congruent to one another we know that the angles \(\angle BC'A'\) and \(\angle B'A'C'\) are equal. Since these are alternate interior angles of the transversal \(A'C'\) crossing \(AB\) and \(A'B'\) we see that \(A'B'\) is parallel to \(AB\) . We use the same argument to show that the other two segments are parallel to the corresponding sides of the triangle.

Lemma 10.3 Let \(\triangle ABC\) be given. Let \(A'\) be the midpoint of \(BC\) , \(B'\) be the midpoint of \(AC\) , and \(C'\) be the midpoint of \(AB\) and let \(G\) be the point of intersection of \(AA'\) and \(BB'\) .

Then \(\triangle AGB \sim \triangle A'GB'\) and \[\frac{AG}{A'G} = \frac{BG}{B'G} = \frac{AB}{A'B'} = 2.\]

Proof . Let \(\triangle ABC\) be given. Let \(A'\) be the midpoint of \(BC\) , \(B'\) be the midpoint of \(AC\) , and \(C'\) be the midpoint of \(AB\) and let \(G\) be the point of intersection of \(AA'\) and \(BB'\) .

Since \(AB\) is parallel to \(A'B'\) we know that \(\angle GA'B' = \angle GAB\) and \(\angle GB'A' = \angle GBA\) since they are alternate interior angles. Also, \(\angle A'GB' = \angle AGB\) because they are vertical angles. So \(\triangle AGB \sim \triangle A'GB'\) by the definition of similarity. Since \[\frac{AB}{A'B'} = 2\] we have the remainder of the ratios by Theorem 10.15 .

We can use the same arguments to prove the following lemma about the intersection point of \(AA'\) and \(CC'\) .

Lemma 10.4 Let \(\triangle ABC\) be given. Let \(A'\) be the midpoint of \(BC\) , \(B'\) be the midpoint of \(AC\) , and \(C'\) be the midpoint of \(AB\) and let \(G'\) be the point of intersection of \(AA'\) and \(CC'\) .

Then \(\triangle AG'C \sim \triangle A'G'C'\) and \[\frac{AG'}{A'G'} = \frac{CG'}{C'G'} = \frac{AC}{A'C'} = 2.\]

Question: Why do we give the point of intersection of \(AA'\) and \(CC'\) a separate name rather than just calling it \(G\) ?

Since \(G\) and \(G'\) are both on \(AA'\) and \(\frac{AG}{A'G} = \frac{AG'}{A'G'} = 2\) we see that \(G=G'\) .

Question: Why is proving that \(G\) and \(G'\) are at the same point along a median segment sufficient for proving that \(G=G'\) ? Be as specific as possible.

So the medians \(AA'\) , \(BB'\) , and \(CC'\) are concurrent.

Definition 10.8 The centroid of a triangle is the point of concurrence of the medians of the sides of the triangle.

10.6.2 Circumcenter

The next “center” we consider is the point whose distance is the same from the three vertices of the triangle and so is the center of the circle that circumscribes the triangle. A unique property of this point is that it is not always inside of the triangle. One can prove that the circumcenter is inside of the triangle if and only if the triangle is acute.

Figure 10.5: Sample Venn diagrams

Theorem 10.18 The perpendicular bisectors of the three edges of a triangle are concurrent at a point.

Proof . Let \(\triangle ABC\) be given and let \(O\) be the point of intersection of the perpendicular bisectors of \(AB\) and \(AC\) . Because \(O\) is on the perpendicular bisector of \(AB\) , \(AO=BO\) . Since \(O\) is on the perpendicular bisector of \(AC\) , \(AO=CO\) . By transitivity, \(BO=CO\) and so \(O\) is on the perpendicular bisector of \(BC\) . So the perpendicular bisectors are concurrent.

Definition 10.9 The circumcenter of a triangle is the point of concurrence of the perpendicular bisectors of the sides of the triangle.

10.6.3 Orthocenter

A third point related to the triangle involves the point of intersection of the altitudes of the triangle.

Definition 10.10 The altitude of a vertex of a triangle is the segment connecting the vertex of a triangle to a point on the extension of the opposite side of the triangle so that the segment is perpendicular to the opposite side.

In each of the two triangles below, \(AA'\) is the altitude of \(A\) . In the first triangle, the altitude is outside of the triangle, while it is inside of the triangle in the second example.

In order to prove that the altitudes are concurrent we will first prove a useful theorem regarding concurrency proved by Giovanni Ceva (1648-1734).

Theorem 10.19 (Ceva's Theorem) In a triangle \(\triangle ABC\) , three lines \(AD\) , \(BE\) , and \(CF\) intersect at a single point \(K\) if and only if \[\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.\]

Proof . We first assume that the three lines \(AD\) , \(BE\) , and \(CF\) are concurrent. We can then extend the lines \(BE\) and \(CF\) beyond the triangle until they meet \(GH\) , the line through \(A\) parallel to \(BC\) .

By alternate interior angles, \(\angle AHF\) and \(\angle BCF\) is equal and by vertical angles, \(\angle AFH\) and \(\angle CFB\) are equal. So by Theorem 10.14 , \(\triangle AHF \sim \triangle BCF\) . Using the same type of arguments, \(\triangle AGE \sim \triangle CBE\) .

We can also use similar arguments to show that \(\triangle AGK \sim \triangle BDK\) and \(\triangle CDK \sim \triangle AHK\) .

These similar triangles imply the following proportions:

\[ \frac{AF}{FB} = \frac{AH}{BC}, \quad \frac{CE}{EA} = \frac{BC}{AG}, \quad \frac{AG}{BD} = \frac{AK}{DK}, \quad \frac{AH}{DC}=\frac{AK}{DK}\]

From the last two proportions we conclude that \(\frac{AG}{BD} = \frac{AH}{DC}\) and so \(\frac{BD}{DC} = \frac{AG}{AH}\) .

Using these proportions we have that \[\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = \frac{AH}{BC} \cdot \frac{AG}{AH} \cdot \frac{BC}{AG} = 1.\]

In order to prove the converse, we will assume \[\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1\] and prove that this implies \(AD\) , \(BE\) , and \(CF\) are concurrent.

Assume that \(K\) is the point of intersection of \(BE\) and \(CF\) and draw the line \(AK\) until its intersection with \(BC\) at a point \(D'\) . Then, from the just proven part of the theorem it follows that \[\frac{AF}{FB} \cdot \frac{BD'}{D'C} \cdot \frac{CE}{EA} = 1.\]

On the other hand, it’s given that \[\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1,\] so that \(\frac{BD'}{D'C} = \frac{BD}{DC}\) implying that \(D\) and \(D'\) are the same point. So the segments are concurrent.

Theorem 10.20 The lines extending altitudes of the vertices of a triangle are concurrent.

Proof . Let \(\triangle ABC\) be given. Let \(A'\) be point on the line \(BC\) so that \(AA'\) is the altitude of \(A\) , \(B'\) be the point on the line \(AC\) so that \(BB'\) is the altitude of \(B\) , and \(C'\) be the point on the line \(AB\) so that \(CC'\) is the altitude of \(C\) .

We have that triangles \(\triangle ACA'\) and \(\triangle BCB'\) are similar because they both have a right angle and share an angle at \(C\) . So \(\frac{CB'}{A'C}=\frac{BB'}{AA'}\) . Similarly, \[\frac{AC'}{B'A} = \frac{CC'}{BB'} \quad \mbox{ and } \quad \frac{BA'}{C'B}=\frac{AA'}{CC'}.\]

Therefore, \[\frac{AC'}{C'B} \cdot \frac{BA'}{A'C} \cdot \frac{CB'}{B'A} = 1\] and Ceva’s Theorem implies that the lines are concurrent.

Definition 10.11 The point of concurrency of the extensions of the altitudes of the vertices of a triangle is called the orthocenter .

10.6.4 Euler Line

Leonard Euler showed in 1765 that the centroid, circumcenter, and orthocenter of a triangle are colinear and the line through these three points is called the Euler line of the triangle.

Theorem 10.21 (Euler Line) Given a triangle \(\triangle ABC\) with centroid \(G\) , circumcenter \(O\) , and orthocenter \(H\) , the points \(G\) , \(H\) , and \(O\) are colinear with \(G\) between \(O\) and \(H\) and is twice as far from the orthocenter as from the circumcenter, \(GH=2GO\) .

10.6.5 Exercises

Reflection questions about the centroid.

• At what points in the proof, if any, were axioms or common definitions used to illuminate something?
• At what points in the proof, if any, were definitions used to illuminate something?
• When, if at all, were propositions or theorems used to illuminate something?
• How were theorems used in the proof? (i.e., focus on what work went into using the theorem).
• In geometry, students often struggle with the fact that a picture of the thing they are proving “shows” the result, so it seems pointless to write everything out. Why do you think we insist on doing it this way anyway?

Prove that the angle bisectors of a triangle are concurrent. This point of concurrency is called the incenter of the triangle.

Working with Trigonometric Ratios

6.1: This Time with Strategies (5 minutes)

CCSS Standards

• HSG-SRT.C.6

This warm-up uses a problem students didn’t know how to approach at the beginning of the unit. At this point they can try again using the right triangle table. Then when students learn to look up trigonometric ratios in the calculator, that procedure will be solidly grounded in conceptual understanding.

Student Facing

Estimate the value of \(z\) .

Student Response

For access, consult one of our IM Certified Partners .

Activity Synthesis

Point out to students this is the triangle they didn’t know how to approach at the beginning of the unit. Now they have a strategy for finding side lengths of right triangles for acute angles included in the right triangle table. The purpose of this lesson is to teach a method that works for any angle, not just the ones in the table.

6.2: New Names, Same Ratios (15 minutes)

Routines and Materials

Instructional Routines

• MLR8: Discussion Supports

The purpose of this activity is to connect the work students have done in previous lessons and the warm-up to trigonometric ratios. Students learn the names of the trigonometric ratios and how to look them up in the calculator. To continue solidifying their conceptual understanding, students compare the calculator’s value to their work with the right triangle table.

Tell students, “The right triangle table is useful, but what if the angle is not a multiple of 10 degrees? There must be a way to move from estimating to calculating. The name of the column in the right triangle table, 'adjacent leg divided by the hypotenuse,' is long, so mathematicians call this ratio cosine . The column called 'opposite leg divided by the hypotenuse' is named sine . The column called 'opposite leg divided by the adjacent leg' is named tangent . These three names describe trigonometric ratios . Label the columns of the right triangle table with the corresponding trigonometric function names.”

“The word trigonometric comes from the ancient roots of trigon (think about what pentagon or hexagon mean) and metric (like measure). So the word trigonometric comes from ancient words meaning triangle measurement.”

“Cosine, sine, and tangent work like functions where the input is the measure of an angle and the output is a ratio. Scientific calculators can display the ratio for any angle, even ones not included in the table. First, identify that we have an angle measure and the length of the adjacent leg. Then identify that we are looking for the length of the hypotenuse. The ratio of adjacent leg divided by hypotenuse is called cosine. So we can write \(\cos(50)=\frac{3}{z}\) .” Demonstrate how to use the technology available to solve this type of problem. Typing \(\cos(50)\) into the calculator and then substituting gives the new equation \(0.643=\frac{3}{z}\) . (Note: if students don’t get 0.643, check the mode of the calculator. There is no need to discuss the meaning of radians now, but students will see radians later in this course.)

• Use your calculator to determine the values of \(\cos(50)\) ,  \(\sin(50)\) , and \(\tan(50)\) .
• Use your calculator to determine the values of \(\cos(40)\) ,  \(\sin(40)\) , and \(\tan(40)\) .
• How do these values compare to your chart? 
• Find the value of \(z\) .

Invite students to share their new value for \(z\) . (Solving the equation we wrote in the launch gives \(z=4.7\) units.)

Ask students:

• “How does this answer compare to using the table?” (It’s the same.)
• “Which one is more trustworthy?” (All opinions are valid.)
• “Which one is more accurate?” (The calculator is much more accurate than the table for numbers between decade numbers. The calculator gives more decimal places, and it’s possible to solve without rounding to get a more precise answer.)

Inform students it is also okay to leave answers in the form \(z=\frac{3}{\cos(50)}\)  which would be exact.

Ask students to add these definitions to their reference charts as you add them to the class reference chart:

The cosine of an acute angle in a right triangle is the ratio (quotient) of the length of the adjacent leg to the length of the hypotenuse.

\(\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\)

The sine of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the hypotenuse.

\(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\)

The tangent  of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the adjacent leg.

\(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\)

6.3: Solve These Triangles (15 minutes)

Building On

• Think Pair Share

Required Materials

• Scientific calculators

Students apply their new knowledge of trigonometric ratios to solve these problems. Since these problems ask for multiple side and angle measures, there is an increased opportunity for creativity in solving.

Monitor for students who solve multiple trigonometric equations versus those that apply the Pythagorean Theorem.

Arrange students in groups of 2. Ask students to compare their strategy with their partner’s and decide if they are both correct, even if they are different.

Students need not complete all the problems before the discussion.

Solve for \(x\) .

Solve for \(y\) .

Find all the missing sides and angle measures.

The measure of angle \(X\) is 90 degrees and angle \(Y\) is 12 degrees. Side \(XZ\) has length 2 cm.

The measure of angle \(K\) is 90 degrees and angle \(L\) is 71 degrees. Side \(LM\) has length 20 cm.

Are you ready for more?

Complete the table.

Based on this information, what do you think are the cosine, sine, and tangent of 90 degrees? Explain or show your reasoning.

Anticipated Misconceptions

If students struggle to get started, prompt them to set up ratios to find the missing sides. If students struggle to set up ratios, prompt them to identify what is known and what they are looking to find by annotating the diagram.

Invite students to contrast solving multiple trigonometric equations versus applying the Pythagorean Theorem to determine the final side length. Is one method easier? More accurate? (The Pythagorean Theorem is more familiar so I prefer it. Both methods are accurate so long as you don't round too much.)

Call attention to the triangle with no marked right angle. Ask students, “How do you know you can use trigonometric functions with this triangle?” (By the Triangle Angle Sum Theorem the third angle must be 90 degrees.)

Lesson Synthesis

Display this triangle and give students quiet work time to write as many equations as they can to show the relationships in triangle \(END\) .

Invite students to share one equation at a time until you have recorded two of each trigonometric ratio ( cosine, sine, and tangent ):

\(\cos(35)=\frac{d}{n}\)

\(\sin(35)=\frac{e}{n}\)

\(\tan(35)=\frac{e}{d}\)

\(\cos(55)=\frac{e}{n}\)

\(\sin(55)=\frac{d}{n}\)

\(\tan(55)=\frac{d}{e}\)

Tell students the value of \(d\) is 6 units. First, ask them to consider which equations would be helpful in solving for the other sides. Then give them time to do the calculations ( \(e=4.2 \) units, \(n=7.3\)  units). If students get slightly different answers, take the opportunity to discuss precision and rounding error.

6.4: Cool-down - Solve That Triangle (5 minutes)

Student lesson summary.

We have a column in the right triangle table for "adjacent leg  \(\div\)  hypotenuse." We use this ratio so frequently it has a name. It is called the cosine of the angle. We write \(\cos(25)\) to say the cosine of 25 degrees. A scientific calculator can display the cosine of any angle. This means we can more precisely calculate unknown side lengths rather than estimating using the table. The right triangle table is sometimes called a trigonometry table since cosine, sine , and tangent are trigonometric ratios . Here is what the table looks like with the ratios labeled with their special names:

If the length  \(b\)  is 7, we can find \(c\) by solving the equation \(\cos(25)=\frac{7}{c}\) . So \(c\)  is about 7.7 units. To solve for \(a\) we have 3 choices: the Pythagorean Theorem, sine, and tangent. Let’s use tangent by solving the equation \(\tan(25)=\frac{a}{7}\) . So \(a\)  is about 3.3 units. We can check our answers using the Pythagorean Theorem. It should be true that \(3.3^2+7^2=7.7^2\) . The two expressions are almost equal, which makes sense because we expect some error due to rounding.

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13.6.4: Triangles, Rectangles, and the Pythagorean Theorem

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Learning Objectives

By the end of this section, you will be able to:

• Solve applications using properties of triangles

Use the Pythagorean Theorem

• Solve applications using rectangle properties

Be prepared

Before you get started, take this readiness quiz.

• Simplify: \(12(6h)\). If you missed this problem, review Exercise 1.10.1 .
• The length of a rectangle is three less than the width. Let w represent the width. Write an expression for the length of the rectangle. If you missed this problem, review Exercise 1.3.43 .
• Solve: \(A=\frac{1}{2}bh\) for b when A=260 and h=52. If you missed this problem, review Exercise 2.6.10 .
• Simplify: \(\sqrt{144}\). If you missed this problem, review Exercise 1.9.10 .

Solve Applications Using Properties of Triangles

In this section we will use some common geometry formulas. We will adapt our problem-solving strategy so that we can solve geometry applications. The geometry formula will name the variables and give us the equation to solve. In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.

SOLVE GEOMETRY APPLICATIONS

• Read the problem and make sure all the words and ideas are understood. Draw the figure and label it with the given information.
• Identify what we are looking for.
• Label what we are looking for by choosing a variable to represent it.
• Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Solve the equation using good algebra techniques.
• Check the answer by substituting it back into the equation solved in step 5 and by making sure it makes sense in the context of the problem.
• Answer the question with a complete sentence.

We will start geometry applications by looking at the properties of triangles. Let’s review some basic facts about triangles. Triangles have three sides and three interior angles. Usually each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.

The plural of the word vertex is vertices . All triangles have three vertices . Triangles are named by their vertices: The triangle in Figure \(\PageIndex{1}\) is called \(\triangle{ABC}\).

The three angles of a triangle are related in a special way. The sum of their measures is \(180^{\circ}\). Note that we read \(m\angle{A}\) as “the measure of angle A.” So in \(\triangle{ABC}\) in Figure \(\PageIndex{1}\).

\[m \angle A+m \angle B+m \angle C=180^{\circ} \nonumber\]

Because the perimeter of a figure is the length of its boundary, the perimeter of \(\triangle{ABC}\) is the sum of the lengths of its three sides.

\[P = a + b + c \nonumber\]

To find the area of a triangle, we need to know its base and height. The height is a line that connects the base to the opposite vertex and makes a \(90^\circ\) angle with the base. We will draw \(\triangle{ABC}\) again, and now show the height, \(h\). See Figure \(\PageIndex{2}\).

TRIANGLE PROPERTIES

For \(\triangle{ABC}\)

Angle measures:

\[m \angle A+m \angle B+m \angle C=180^{\circ}\]

• The sum of the measures of the angles of a triangle is 180°.

\[P = a + b + c\]

• The perimeter is the sum of the lengths of the sides of the triangle.

\(A = \frac{1}{2}bh, b = \text{ base }, h = \text{ height }\)

• The area of a triangle is one-half the base times the height.

Example \(\PageIndex{1}\)

The measures of two angles of a triangle are 55 and 82 degrees. Find the measure of the third angle.

Try It \(\PageIndex{1}\)

The measures of two angles of a triangle are 31 and 128 degrees. Find the measure of the third angle.

Try It \(\PageIndex{2}\)

The measures of two angles of a triangle are 49 and 75 degrees. Find the measure of the third angle.

Example \(\PageIndex{2}\)

The perimeter of a triangular garden is 24 feet. The lengths of two sides are four feet and nine feet. How long is the third side?

Try It \(\PageIndex{3}\)

The perimeter of a triangular garden is 48 feet. The lengths of two sides are 18 feet and 22 feet. How long is the third side?

Try It \(\PageIndex{4}\)

The lengths of two sides of a triangular window are seven feet and five feet. The perimeter is 18 feet. How long is the third side?

Example \(\PageIndex{3}\)

The area of a triangular church window is 90 square meters. The base of the window is 15 meters. What is the window’s height?

Try It \(\PageIndex{5}\)

The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

Try It \(\PageIndex{6}\)

A triangular tent door has area 15 square feet. The height is five feet. What is the base?

The triangle properties we used so far apply to all triangles. Now we will look at one specific type of triangle—a right triangle. A right triangle has one 90° angle, which we usually mark with a small square in the corner.

Definition: RIGHT TRIANGLE

A right triangle has one 90° angle, which is often marked with a square at the vertex.

Example \(\PageIndex{4}\)

One angle of a right triangle measures 28°. What is the measure of the third angle?

Try It \(\PageIndex{7}\)

One angle of a right triangle measures 56°. What is the measure of the other small angle?

Try It \(\PageIndex{8}\)

One angle of a right triangle measures 45°. What is the measure of the other small angle?

In the examples we have seen so far, we could draw a figure and label it directly after reading the problem. In the next example, we will have to define one angle in terms of another. We will wait to draw the figure until we write expressions for all the angles we are looking for.

Example \(\PageIndex{5}\)

The measure of one angle of a right triangle is 20 degrees more than the measure of the smallest angle. Find the measures of all three angles.

Try It \(\PageIndex{9}\)

The measure of one angle of a right triangle is 50° more than the measure of the smallest angle. Find the measures of all three angles.

20°,70°,90°

Try It \(\PageIndex{10}\)

The measure of one angle of a right triangle is 30° more than the measure of the smallest angle. Find the measures of all three angles.

30°,60°,90°

We have learned how the measures of the angles of a triangle relate to each other. Now, we will learn how the lengths of the sides relate to each other. An important property that describes the relationship among the lengths of the three sides of a right triangle is called the Pythagorean Theorem . This theorem has been used around the world since ancient times. It is named after the Greek philosopher and mathematician, Pythagoras, who lived around 500 BC.

Before we state the Pythagorean Theorem, we need to introduce some terms for the sides of a triangle. Remember that a right triangle has a 90° angle, marked with a small square in the corner. The side of the triangle opposite the 90°90° angle is called the hypotenuse and each of the other sides are called legs .

The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. In symbols we say: in any right triangle, \(a^{2}+b^{2}=c^{2}\), where a and b are the lengths of the legs and cc is the length of the hypotenuse.

Writing the formula in every exercise and saying it aloud as you write it, may help you remember the Pythagorean Theorem.

THE PYTHAGOREAN THEOREM

In any right triangle, where \(a\) and \(b\) are the lengths of the legs, \(c\) is the length of the hypotenuse.

\[a^{2}+b^{2}=c^{2} \label{Ptheorem}\]

To solve exercises that use the Pythagorean Theorem (Equation \ref{Ptheorem}), we will need to find square roots. We have used the notation \(\sqrt{m}\) and the definition:

If \(m = n^{2}\), then \(\sqrt{m} = n\), for \(n\geq 0\).

For example, we found that \(\sqrt{25}\) is 5 because \(25=5^{2}\).

Because the Pythagorean Theorem contains variables that are squared, to solve for the length of a side in a right triangle, we will have to use square roots.

Example \(\PageIndex{6}\)

Use the Pythagorean Theorem to find the length of the hypotenuse shown below.

Try It \(\PageIndex{11}\)

Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.

Try It \(\PageIndex{12}\)

Example \(\PageIndex{7}\)

Use the Pythagorean Theorem to find the length of the leg shown below.

Try It \(\PageIndex{13}\)

Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.

Try It \(\PageIndex{14}\)

Example \(\PageIndex{8}\)

Kelvin is building a gazebo and wants to brace each corner by placing a 10″ piece of wood diagonally as shown above.

If he fastens the wood so that the ends of the brace are the same distance from the corner, what is the length of the legs of the right triangle formed? Approximate to the nearest tenth of an inch.

\(\begin{array} {ll} {\textbf{Step 1. }\text{Read the problem.}} &{} \\\\ {\textbf{Step 2. }\text{Identify what we are looking for.}} &{\text{the distance from the corner that the}} \\ {} &{\text{bracket should be attached}} \\ \\{\textbf{Step 3. }\text{Name. Choose a variable to represent it.}} &{\text{Let x = distance from the corner.}} \\ {\textbf{Step 4.} \text{Translate}} &{} \\ {\text{Write the appropriate formula and substitute.}} &{a^{2} + b^{2} = c^{2}} \\ {} &{x^{2} + x^{2} = 10^{2}} \\ \\ {\textbf{Step 5. Solve the equation.}} &{} \\ {} &{2x^{2} = 100} \\ {\text{Isolate the variable.}} &{x^{2} = 50} \\ {\text{Simplify. Approximate to the nearest tenth.}} &{x \approx 7.1}\\\\ {\textbf{Step 6. }\text{Check.}} &{}\\ {a^{2} + b^{2} = c^{2}} &{} \\ {(7.1)^{2} + (7.1)^{2} \approx 10^{2} \text{ Yes.}} &{} \\\\ {\textbf{Step 7. Answer the question.}} &{\text{Kelven should fasten each piece of}} \\ {} &{\text{wood approximately 7.1'' from the corner.}} \end{array}\)

Try It \(\PageIndex{15}\)

John puts the base of a 13-foot ladder five feet from the wall of his house as shown below. How far up the wall does the ladder reach?

Try It \(\PageIndex{16}\)

Randy wants to attach a 17 foot string of lights to the top of the 15 foot mast of his sailboat, as shown below. How far from the base of the mast should he attach the end of the light string?

Solve Applications Using Rectangle Properties

You may already be familiar with the properties of rectangles. Rectangles have four sides and four right (90°) angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, \(L\), and its adjacent side as the width, \(W\).

The distance around this rectangle is \(L+W+L+W\), or \(2L+2W\). This is the perimeter , \(P\), of the rectangle.

\[P=2L+2W\]

What about the area of a rectangle? Imagine a rectangular rug that is 2-feet long by 3-feet wide. Its area is 6 square feet. There are six squares in the figure.

\[\begin{array} {l} {A=6} \\ {A=2\cdot3} \\ {A=L\cdot W} \end{array}\]

The area is the length times the width. The formula for the area of a rectangle is

PROPERTIES OF RECTANGLES

• Rectangles have four sides and four right (90°) angles.
• The lengths of opposite sides are equal.

The perimeter of a rectangle is the sum of twice the length and twice the width.

The area of a rectangle is the product of the length and the width.

\[A=L·W\]

Example \(\PageIndex{9}\)

The length of a rectangle is 32 meters and the width is 20 meters. What is the perimeter?

Try It \(\PageIndex{17}\)

The length of a rectangle is 120 yards and the width is 50 yards. What is the perimeter?

Try It \(\PageIndex{18}\)

The length of a rectangle is 62 feet and the width is 48 feet. What is the perimeter?

Example \(\PageIndex{10}\)

The area of a rectangular room is 168 square feet. The length is 14 feet. What is the width?

Try It \(\PageIndex{19}\)

The area of a rectangle is 598 square feet. The length is 23 feet. What is the width?

Try It \(\PageIndex{20}\)

The width of a rectangle is 21 meters. The area is 609 square meters. What is the length?

Example \(\PageIndex{11}\)

Find the length of a rectangle with perimeter 50 inches and width 10 inches.

Try It \(\PageIndex{21}\)

Find the length of a rectangle with: perimeter 80 and width 25.

Try It \(\PageIndex{22}\)

Find the length of a rectangle with: perimeter 30 and width 6.

We have solved problems where either the length or width was given, along with the perimeter or area; now we will learn how to solve problems in which the width is defined in terms of the length. We will wait to draw the figure until we write an expression for the width so that we can label one side with that expression.

Example \(\PageIndex{12}\)

The width of a rectangle is two feet less than the length. The perimeter is 52 feet. Find the length and width.

Try It \(\PageIndex{23}\)

The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.

18 meters, 11 meters

Try It \(\PageIndex{24}\)

The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width.

19 feet, 11 feet

Example \(\PageIndex{13}\)

The length of a rectangle is four centimeters more than twice the width. The perimeter is 32 centimeters. Find the length and width.

Try It \(\PageIndex{25}\)

The length of a rectangle is eight more than twice the width. The perimeter is 64. Find the length and width.

Try It \(\PageIndex{26}\)

The width of a rectangle is six less than twice the length. The perimeter is 18. Find the length and width.

Example \(\PageIndex{14}\)

The perimeter of a rectangular swimming pool is 150 feet. The length is 15 feet more than the width. Find the length and width.

Try It \(\PageIndex{27}\)

The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the length and width.

70 feet, 30 feet

Try It \(\PageIndex{28}\)

The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width.

90 yards, 60 yards

Key Concepts

• Read the problem and make all the words and ideas are understood. Draw the figure and label it with the given information.
• Name what we are looking for by choosing a variable to represent it.
• Check the answer in the problem and make sure it makes sense.
• \(m\angle{A}+m\angle{B}+m\angle{C}=180\)
• \(P=a+b+c\)
• \(A=\frac{1}{2}bh\), b=base,h=height
• The Pythagorean Theorem In any right triangle, \(a^{2} + b^{2} = c^{2}\) where \(c\) is the length of the hypotenuse and \(a\) and \(b\) are the lengths of the legs.
• The perimeter of a rectangle is the sum of twice the length and twice the width: \(P=2L+2W\).
• The area of a rectangle is the length times the width: \(A=LW\).

• NCERT Solutions
• NCERT Solutions for Class 10
• NCERT Solutions for Class 10 Maths
• Chapter 6: Triangles
• Exercise 6.4

NCERT Solutions for Class 10 Maths Chapter 6 - Triangles Exercise 6.4

Out of the 6 exercises present in the CBSE Class 10 Chapter 6 Triangles, Exercise 6.4 deals with the calculation of areas of triangles that are similar. The solutions to the questions present in Exercise 6.4 are given below in both PDF and scrollable image format. These NCERT Class 10 solutions are prepared with proper reference, giving step-by-step explanations, by the Maths experts at BYJU’S.

Understanding the proper methods to solve the NCERT Solutions for Class 10 Maths will help the students in solving the different types of questions that are likely to be asked in the board examination. Once the students get proficient in these NCERT Class 10 Maths solutions , their problem-solving speed will increase, boosting their self-confidence.

NCERT Solutions for Class 10 Maths Chapter 6- Triangles Exercise 6.4

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Access other exercise solutions of Class 10 Maths Chapter 6 – Triangles

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)

Exercise 6.3 Solutions 16 Questions (1 Main Question with 6 Sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)

Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)

Access Answers of Maths NCERT Class 10 Chapter 6 – Triangles Exercise 6.4

1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm 2  and 121 cm 2 . If EF = 15.4 cm, find BC.

Solution: Given, ΔABC ~ ΔDEF,

Area of ΔABC = 64 cm 2

Area of ΔDEF = 121 cm 2

EF = 15.4 cm

As we know, if two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

= AC 2 /DF 2  = BC 2 /EF 2

∴ 64/121 = BC 2 /EF 2

⇒ (8/11) 2  = (BC/15.4) 2

⇒ 8/11 = BC/15.4

⇒ BC = 8×15.4/11

⇒ BC = 8 × 1.4

⇒ BC = 11.2 cm

2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In ΔAOB and ΔCOD, we have

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angle)

∴ ΔAOB ~ ΔCOD [AAA similarity criterion]

As we know, if two triangles are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding sides. Therefore,

Area of (ΔAOB)/Area of (ΔCOD) = AB 2 /CD 2

= (2CD) 2 /CD 2  [∴ AB = 2CD]

∴ Area of (ΔAOB)/Area of (ΔCOD)

= 4CD 2 /CD 2 = 4/1

Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.

Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.

We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO

Let us draw two perpendiculars, AP and DM, on line BC.

We know that area of a triangle = 1/2 × Base × Height

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ~ ΔDMO (AA similarity criterion)

∴ AP/DM = AO/DO

⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO

4. If the areas of two similar triangles are equal, prove that they are congruent.

Say, ΔABC and ΔPQR are two similar triangles and equal in area.

Now, let us prove ΔABC ≅ ΔPQR

ΔABC ~ ΔPQR

∴ Area of (ΔABC)/Area of (ΔPQR) = BC 2 /QR 2

⇒ BC 2 /QR 2  =1 [Since, Area(ΔABC) = (ΔPQR)

⇒ BC 2 /QR 2

Similarly, we can prove that

AB = PQ and AC = PR

Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

5. D, E and F are, respectively, the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Given: AM and DN are the medians of triangles ABC and DEF, respectively and ΔABC ~ ΔDEF.

We have to prove: Area(ΔABC)/Area(ΔDEF) = AM 2 /DN 2

Since, ΔABC ~ ΔDEF (Given)

∴ Area(ΔABC)/Area(ΔDEF) = (AB 2 /DE 2 ) …………………………… (i)

and, AB/DE = BC/EF = CA/FD ……………………………………… (ii)

In ΔABM and ΔDEN,

Since ΔABC ~ ΔDEF

AB/DE = BM/EN [Already proved in equation (i) ]

∴ ΔABC ~ ΔDEF [SAS similarity criterion]

⇒ AB/DE = AM/DN ………………………………………………….. (iii)

∴ ΔABM ~ ΔDEN

The areas of two similar triangles are proportional to the squares of the corresponding sides.

∴ area(ΔABC)/area(ΔDEF) = AB 2 /DE 2  = AM 2 /DN 2

Hence, proved.

7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Given, ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Area(ΔBQC) = ½ Area(ΔAPC)

ΔAPC and ΔBQC are both equilateral triangles,

∴ ΔAPC ~ ΔBQC [AAA similarity criterion]

∴ area(ΔAPC)/area(ΔBQC) = (AC 2 /BC 2 ) = AC 2 /BC 2

Since, Diagonal = √2 side = √2 BC = AC

⇒ area(ΔAPC) = 2 × area(ΔBQC)

⇒ area(ΔBQC) = 1/2area(ΔAPC)

Tick the correct answer and justify.

8. ABC and BDE are two equilateral triangles, such that D is the mid-point of BC. The ratio of the area of triangles ABC and BDE is (A) 2:1 (B) 1:2 (C) 4:1 (D) 1:4

Given , ΔABC and ΔBDE are two equilateral triangles. D is the midpoint of BC.

∴ BD = DC = 1/2BC

Let each side of the triangle be 2 a .

ΔABC ~ ΔBDE

∴ Area(ΔABC)/Area(ΔBDE) = AB 2 /BD 2  = (2 a ) 2 /( a ) 2  = 4 a 2 / a 2  = 4/1 = 4:1

Hence, the correct answer is (C).

9. Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio (A) 2:3 (B) 4:9 (C) 81:16 (D) 16:81

Given, the sides of two similar triangles are in the ratio 4:9.

Let ABC and DEF be two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

The ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides.

∴ Area(ΔABC)/Area(ΔDEF) = AB 2 /DE 2 

∴ Area(ΔABC)/Area(ΔDEF) = (4/9) 2  = 16/81 = 16:81

Hence, the correct answer is (D).

The fourth exercise of Class 10 NCERT Maths Chapter 6 Triangles deals with the topic Area of Similar Triangles. There are 9 questions in this exercise, out of which 2 are short answers with reasoning type of questions, 5 are short answer type questions, and the remaining 2 are long answer type of questions. Exercise 6.4 deals with the Area of Similar Triangles and one theorem, Theorem 6.6. The theorem that forms the base of Exercise 6.4 is given below:

• Theorem 6.6 : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

To understand the concepts that are taught in Class 10 well, there is no other effective method than solving NCERT Solutions . Solving these solutions not only helps you understand the concepts but also helps in getting acquainted with different types of questions that could be asked in the CBSE Class 10 board exam.

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Chapter 4 – Solving Triangles

Chapter 4 uses the Law of Sines and Law of Cosines to extend the ability to solve triangles beyond just right triangles. Formulas for the area of a triangle using trig functions can also be found.

Trigonometry Copyright © 2022 by Mike Weimerskirch and the University of Minnesota Board of Regents is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License , except where otherwise noted.

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