On Your Own

Two-Dimensional Shapes T383

Unit 6

Applications

1 a. One isosceles triangle can be built with sides 12 cm, 12 cm, and 5 cm.

b. An infinite number of parallelograms can be built.

c. An infinite number of kites can be built.

d. The kite is the only convex quadrilateral that is not a parallelogram that can be built. Therefore, from Part c, an infinite number of quadrilaterals that are not parallelograms can be built. (An infinite number of nonconvex quadrilaterals, called darts, can also be built.)

2 a. AR + BR + CR + DR

b. By the Triangle Inequality, AP + PC > AC = AR + RC. So, position R requires less piping than position P.

c. BP + PD > BD = BR + RD. So, AP + PC + BP + PD > AR + RC + BR + RD. Locating the refinery at point R will require less piping.

d. For any other point, the same argument as in Part a would show that R is better. If the point is on one of the diagonals, but not at the intersection of the diagonals, then it is not on the other diagonal. So, one application of the Triangle Inequality shows that the distance has now increased.

3 a. Quadrilateral linkage

b. Side CD is the frame, side AB is the coupler, side AD is the driver crank, and CB is the follower crank.

c. CD could be shortened to equal AB and the pedal DA extended to equal CB. A sports trainer would not recommend this adjustment. The pedal would not completely rotate since a person’s thigh cannot completely rotate. As is, a person’s thigh only needs to make a small arc to continue the pedaling.

On Your Own

Two-Dimensional Shapes T384

Unit 6

4 a. Sufficient by the SAS congruence condition

b. Not sufficient

c. Sufficient by the ASA congruence condition

d. Not sufficient

e. Sufficient by the SSS congruence condition

5 a. �DMA � �QPT by the ASA congruence condition.

b. The triangles are not necessarily congruent.

c. �BLJ � �SUE by the SAS congruence condition.

d. Not necessarily congruent

e. �ART � �MDI by the SSS congruence condition.

f. Since ∠J � ∠P, �OBJ � �ITP by the ASA congruence condition.

On Your Own

Two-Dimensional Shapes T385

Unit 6

6 a. �ADJ � �CGH by the SAS congruence condition. So, −−

DJ and −−

GH must be the same length since they are corresponding parts of congruent triangles. (∠A � ∠C since the truss is isosceles.)

b. −−

AE � −−

CF (since AD + DE = CG + GF). So, �EAJ � �FCH by the SAS congruence condition. Braces

−− EJ and

−− FH must be cut the same length

because they are corresponding parts of congruent triangles.

c. �IAE � �ICF by the SAS congruence condition. So, −−

EI and −−

FI must be the same length.

d. EBFI is a kite since −−

BE � −−

BF and −−

EI � −−

FI .

7 a. When you position the carpenter’s square so that PQ = RQ and PS = RS, you have �PQS � �RQS by the SSS congruence condition (since QS = QS).

b. Since �PQS � �RQS, corresponding angles ∠PQS and ∠RQS are congruent.

On Your Own

Two-Dimensional Shapes T386

NOTE The solution to Task 9 is on page T387.

Unit 6

8 a. See the diagram at the right.

b. Step 1 of the algorithm assures that BX = BY. Step 2 of the algorithm assures that XD = YD. Of course, BD = BD. So by the SSS congruence condition, �BXD � �BYD. Thus, ∠XBD � ∠YBD, and so

� BD bisects

∠XBY. Another name for ∠XBY is ∠ABC, so � BD bisects ∠ABC.

The algorithms here and using the carpenter’s square are essentially the same. You position the carpenter’s square so that BX = BY and XD = YD. Then connect the vertex B to the vertex of the carpenter’s square to find the angle bisector.

c. Yes, there is nothing about the algorithm that limits its use to acute angles.

d. i. Step 1: With the compass point at P, draw an arc that intersects

� PA and

� PB . Call the intersection points X and Y,

respectively.

Step 2: With the compass point at point X (and then at point Y)

and using a radius greater than 1 _ 2 XY, draw two arcs of the

same radius that intersect above (or below) line AB. Label the point of intersection D.

Step 3: Draw the line PD. � PD is perpendicular to � AB .

ii. Step 1 assures that −− XP �

−− YP . Step 2 assures that −−

XD � −−

YD . Of course,

−− PD �

−− PD . So by the SSS congruence condition,

�PXD � �PYD. Thus, the two angles at P are congruent. Their measures sum to 180° because they form a straight angle. So, each angle is a right angle.

e. i. Step 1: With the compass at point A (and then at point B) and

using a radius greater than 1 _ 2 AB, draw two large arcs that

intersect −− AB and each other on both sides of −−

AB .

Step 2: Draw a line that connects the two intersection points C and D of the arcs. � CD is the perpendicular bisector of −−

AB .

ii. Step 1 of the algorithm assures that AC = BC = AD = BD. −−

CD is a side of �ACD and �BCD, so �ACD � �BCD (SSS). ∠ACE � ∠BCE because the angles are corresponding angles of these triangles. Since

−− CE is a common side for �ACE and

�BCE, these triangles are congruent (SAS). Thus, −− AE �

−− EB

and ∠AEC � ∠BEC because they are corresponding parts of congruent triangles. Since ∠AEC and ∠BEC form a straight line, their measures sum to 180°. Since the two angles have equal measures, each angle is a right angle. Therefore, � CD is the perpendicular bisector of

−− AB .

On Your Own

Two-Dimensional Shapes T387

TOOLKIT NOTE You may wish to have students add this property of parallelograms to their lists.

INSTRUCTIONAL NOTE This is an opportunity to share with students the construction for copying an angle. With that additional construction, students could design the other algorithm for SAS and ASA.

Unit 6

9 Responses will vary. The algorithms should be based on one of the triangle congruence conditions. (You may wish to have students discuss their algorithms with each other in order to see that more than one algorithm is possible.) One way to construct the triangle would be to:

Step 1: Draw a segment with a straight edge and mark off −− AB using the

compass. Label the new segment −−

XY .

Step 2: Set the compass to the length of −−

BC . Place the compass point at Y and draw an arc above

−− XY .

Step 3: Repeat Step 2 with −−

AC by placing the compass point at X.

Step 4: Label the intersection of the arcs Z and draw −−

XZ and −−

YZ .

�ABC � �XYZ because �XYZ was formed using the same three side lengths as �ABC. Thus, we used the SSS triangle congruence condition.

10 a. Anna’s conclusion follows logically from knowing that ABCD is a parallelogram.

b. Andy’s conclusion also follows logically from knowing that ABCD is a parallelogram.

On Your Own

Two-Dimensional Shapes T388

NOTE Task 12 Part c is the converse of the property of the diagonals of a kite examined in the Check Your Understanding on page 377 in the Student Edition.

Unit 6

11 INSTRUCTIONAL NOTE This task is counterintuitive for students and many adults. They are often surprised by the mathematical conclusion. You may wish to use this task to reinforce the idea that mathematics can provide reasoning to support or give a counterexample to, initial intuition.

a. The buckling will occur so that the highest point of the buckle is in the middle of the 220-foot-long rail.

b. Responses will vary. After answering Part c, students should see that a gym bag definitely would fit under the rail.

c. Using the Pythagorean Theorem:

(1,320.6)2 = 1,3202 + h2

1,584.36 = h2

39.8 ≈ h

The height of the triangle is approximately 39.8 inches.

d. Since the railing is curved, the straight line from end to middle actually would be less than 1,320.6 inches. Our estimate of the height of the buckle is more than the actual value. Even if it were only half our estimate, a gym bag would easily fit under it!

e. Expansion points allow the rails to expand without buckling.

12 a. The diagonal strips must bisect each other, that is, they must be attached at their midpoints.

b. The diagonal strips must be equal in length and attached at their midpoints.

c. One diagonal strip must be the perpendicular bisector of the other strip. In the diagram,

−− SQ is a perpendicular bisector

of −−

PR . To show quadrilateral PQRS is a kite, note that

−− PT �

−− TR . ∠PTS � ∠RTS (since both angles

are 90°), and −−

ST � −−

ST . By the SAS congruence condition, �PTS � �RTS.

−− PS �

−− RS since they

are corresponding parts of congruent triangles. Similarly, because

−− PT �

−− TR , ∠PTQ � ∠RTQ,

−−

TQ � −−

TQ and, �PTQ � �RTQ by the SAS congruence condition, and

−− PQ �

−− RQ . Quadrilateral

PQRS has two distinct pairs of consecutive sides the same length. So, it is a kite.

d. The diagonal strips must be equal in length and perpendicular bisectors of each other.

In the diagram, since SQ = PR and T is the midpoint of

−− SQ and

−− PR , quadrilateral PQRS is a

rectangle by Part b. Since it is the case that

−− SQ ⊥

−− PR , students should reason that

�PTQ � �RTQ � �RTS � �STP and thus PQ = QR = RS = SP. So, quadrilateral PQRS is a square.

On Your Own

Two-Dimensional Shapes T389

Unit 6

13 ∠1 � ∠2 because they are vertical angles. �PH′A � �PHA by the SAS congruence condition as marked in the diagram below. So, ∠2 � ∠3 and ∠1 � ∠3 by substitution.

Connections

14 a. The linkage is not rigid since the angles can change.

b. Using 2 braces is the fewest number needed to make the linkage rigid.

c. 3 braces; 5 braces. There are many different choices for the placement of the braces; in every case, the linkage is triangulated.

d. n - 3 braces will be required and n - 2 triangles will be formed.

15 a. A = b × h. The diagrams show that the triangle in the left side of the first figure can be shifted over to the right side as shown in the second figure. This forms a rectangle with base b and height h.

b. A = 1 _ 2 (b × h). The diagram shows that if a triangle is rotated about

a midpoint of a side, the triangle and its congruent image form a parallelogram with base b and height h. Since the area of a parallelogram is b × h, the area of one triangle is half that of the parallelogram.