Big Ideas Math Geometry, 2014
BI
Big Ideas Math Geometry, 2014 View details
6. Solving Right Triangles
Continue to next subchapter

Exercise 32 Page 506

Using the information about the perimeter of the rectangle and the ratio of the width of the rectangle to its length, write a system of equations.

Side Lengths: AB=CD=2cm, DA=BC=6cm
Angle Measures: m∠ BAD=m∠ DCB=90^(∘), m∠ ABD=m∠ CDB≈ 71.6^(∘), m∠ ADB=m∠ CBD≈ 18.4^(∘)

Practice makes perfect

We are given a rectangle ABCD whose perimeter is 16cm. We also know that Segment BD divides the rectangle into two congruent triangles.

The congruent triangles are △ ABD and △ CDB. Since all angles of a rectangle measure 90^(∘), both triangles are right triangles. We are asked to find the side lengths and angle measures of these triangles. Let's do it one at a time.

Side Lengths of the Triangles

We will start by finding the side lengths of the triangles. Recall that the perimeter of the rectangle is 16cm and the ratio of its width to its length is 1:3. Using this information, we can write a system of equations. Let w represent the width of the rectangle and l its length. 2w+2l=16 & (I) w:l=1:3 & (II) We will solve the system for w and l using the Substitution Method. First we will solve Equation II for l.
2w+2l=16 w:l=1:3
(II):Solve for l
2w+2l=16 wl= 13
2w+2l=16 3w=l
2w+2l=16 l=3w
Next, let's substitute 3w for l in Equation I and solve it for w.
2w+2l=16 l=3w
2w+2( 3w)=16 l=3w
(I):Solve for w
2w+6w=16 l=3w
8w=16 l=3w
w=2 l=3w
Finally we will substitute 2 for w in Equation II and calculate the value of l.
w=2 l=3w
w=2 l=3( 2)
w=2 l=6
We have that the width of the rectangle is 2cm and its length is 6cm.
The length and the width of the rectangle correspond to the lengths of the legs of △ ABD and △ CDB. To determine the length of the hypotenuse of the congruent triangles, we will use the Pythagorean Theorem. Let's focus on the triangle △ ABD. AB^2+DA^2=BD^2 ⇔ 2^2+6^2=BD^2 We will solve the above equation for BD.
2^2+6^2=BD^2
Solve for BD
4+36=BD^2
40=BD^2
BD^2=40
BD=sqrt(40)
BD=sqrt(4* 10)
BD=sqrt(4)*sqrt(10)
BD=2sqrt(10)
We have that the hypotenuse of the congruent triangles measures 2sqrt(10)cm.

Angle Measures of the Triangles

Now we will find the angle measures of the triangles. We already know that the triangles are right triangles, m∠ BAD=m∠ DCB=90^(∘).

Since the triangles are congruent, we only need to find the angle measures of one of the triangles. Thus, let's focus on △ ABD. We will start by finding m∠ ABD. The tangent ratio of ∠ ABD is the ratio between DA and AB. tan∠ ABD=DA/AB ⇔ tan∠ ABD=6/2 To find m∠ ABD, we will use the inverse tangent ratio. Note that ∠ ABD is an acute angle. tan∠ ABD=6/2 ⇔ m∠ ABD=tan^(- 1)6/2 Let's approximate the inverse tangent of 62.
m∠ ABD=tan^(- 1)6/2
m∠ ABD=tan^(- 1)3
m∠ ABD≈ 71.6^(∘)
Angle ∠ ABD measures about 71.6^(∘). Recall that by the Triangle Sum Theorem the sum of the measures of the interior angles of a triangle is 180^(∘). m∠ BAD+m∠ ABD+m∠ ADB=180^(∘) Let's substitute 90^(∘) for m∠ BAD and 71.6^(∘) for m∠ ABD into the above equation, and solve it for m∠ ADB.
m∠ BAD+m∠ ABD+m∠ ADB=180^(∘)
90^(∘)+ 71.6^(∘)+m∠ ADB=180^(∘)
Solve for m∠ ADB
161.6^(∘)+m∠ ADB=180^(∘)
m∠ ADB=18.4^(∘)
Angle ∠ ADB measures about 18.4^(∘). We have found the measures of all angles of △ ABD and △ CDB. Note that ∠ ABD corresponds to ∠ CDB and ∠ ADB corresponds to ∠ CBD. Corresponding angles of congruent figures are congruent.