Geometry / Mr. Hansen
5/21/2002

Name: _________________________

Key to Suggested Review Problems for Chapters 14 and 15

 

pp.703-704

1.

ÐA, ÐB, ÐC

2.

(a) sWZ by Hinge Thm. (note: you must observe that XZ=XZ by reflexive)
(b) sAB since hypot. is always longest
(c) sQR since it’s opposite the largest Ð in DPQR

3.

(a) ÐCBD by Converse Hinge Thm.; OK to say "Hinge Thm." (but note: you must observe that BD=BD by reflexive)
(b) ÐX since it’s opposite longest side
(c) Ð1 since ext. Ð is always larger than any remote int. Ð (note: you can’t compare Ð2 and ÐP; all you can say for sure is that Ð1 is the largest of the three)

6.

(a) sAE @ sAB, sED @ sEC
(b) sBE (opp. smaller Ð)
(c) hypotenuse (since rt. D )
(d) sDE (opp. larger Ð )
(e) Observe that sAE is shortest in D ADE (opp. smallest Ð ), and sCE is tied for shortest in D CED (same reason). But EB < AB (opp. smaller Ð in D ABE), and EB < CE (opp. smaller Ð in D CBE). Therefore EB is the smallest length of all. Answer: sEB. (Note: On the test, you may have a diagram that is not to scale. This was a "friendly" diagram since sEB not only looked shortest but actually was the shortest.)

7.

(a) since 20 + 20 does not exceed the remaining side

8.

Use backward chaining: We could solve this by Hinge Thm. if we had the setup conditions and knew that Ð 1 > Ð 2. But Ð 1 is an ext. Ð to DPQR, and the Hinge Thm. conditions are satisfied as long as you remember to show sPR @ sPR, so we have what we need.

Proof:

 

1. sPQ @ sRS
2. sPR @ sPR
3. Ð1 > Ð2
4. PS > QR

1. Given
2. Refl.
3. Ext. Ð always > any remote int. Ð
4. Hinge Thm.

9.

Since x + 100 is the sum of measures of ÐA and ÐC, solve to get x = 20. Find all angles and use notion of bigger side opp. bigger Ð to get sBC largest, then sAC, then sAB smallest.

10.

60 < x < 150 (explained in class)

11.

(a) sAB (since opp. the obtuse Ð, which must be the largest Ð)
(b) 12 < perim. < 6 + 6Ö2 (explanation: perim. is smallest if ÐC approaches 180°, largest if ÐC approaches 90°)

13.

(a) PQ = Ö34; QR = Ö13; PR = 7
(b) obtuse
(c) ÐP, ÐR, ÐQ

14.

We did this in class using forward chaining. From the given AB < AD, you should immediately say to yourself that the big angle at B is larger than the big angle at D. This remains true even after bisection, so that in DBED, the longer side (ED) is opp. the larger Ð (namely, ÐEBD).

Proof:

 

1. AB < AD
2. ÐABD > ÐADB
3. rBE bis. ÐABC, rDE bis. ÐADC
4. ÐEBC > ÐEDC
5. ED > EB

1. Given
2. Larger Ð opp. longer side (DABD)
3. Given
4. Div. prop. (technically, div. prop. ¹)
5. Longer side opp. larger Ð (DBED)

16.

Use backward chaining, recognizing that Converse Hinge Thm. is needed.

Proof: 

 

1. sPQ @ sSR
2. sQR @ sQR
3. PR < QS
4. ÐPQR < ÐSRQ

1. Given
2. Refl.
3. Given
4. Converse Hinge Thm. (OK to say HT)

17.

Again, use backward chaining, recognizing that Hinge Thm. is needed. This one is interesting because you have to generate "pre-goals" more than once. Showing ÐABE < ÐCBE (a pre-goal) requires using ÐBEC as an ext. Ð, which in turn requires showing a relationship between ÐBEC and ÐCBE. Luckily, this last (first?) step is no problem, because we know that DBEC is isosceles. Whew!

Proof: 

 

1. sBC @ sEC
2. ÐBEC @ ÐCBE
3. ÐABE < ÐBEC
4. ÐABE < ÐCBE
5. ÐA @ ÐC
6. sBA @ sBC
7. sBE @ sBE
8. AE < EC

1. Given
2. ITT (aka BAILC)
3. Rem. int. Ð always < ext. Ð (DABE)
4. Subst.
5. Given
6. ITT (aka BAILC)
7. Refl. (don’t forget this step)
8. Hinge Thm. (steps 6, 7, 4)

 

pp.683-684

1.

{points on ^ bisector of sAB} \ {midpoint of sAB}
In other words, the locus is the ^ bisector of sAB, excluding the midpoint of sAB. The problem should have specified locus of points in a plane.

2.

^ bisector of the segment joing the two points
Again, the problem should have specified locus of points in a plane.

4.

open-ended circular cylinder

5.

(Assume locus of points in a plane, which should have been specified.) Answer: the center (1 point) in union with a concentric circle of radius 4 in.

6.

a larger similar equilateral D

8.

 (a) "ABIC": Construct any two Ð bisectors (or use three to cross-check yourself). Intersection is the incenter, which incidentally is the point equidistant from all three sides. The incenter (obviously) is always inside the D.
(b) "PBCC": Construct any two ^ bisectors (or use three to cross-check yourself). Intersection is the circumcenter, which incidentally is the point equidistant from all three vertices. The circumcenter, as you discovered in your Sketchpad lab, may lie outside the D (if D is obtuse) or on the hypotenuse (in the case of a right D).
(c) "MCCG": Construct any two midpoints and use straightedge to construct the two medians. (You can do all three if you wish to cross-check yourself.) Centroid is at intersection of medians. Incidentally, the centroid is always inside the D.
(d) "ALTO": Construct ^ segments from any two vertices to their opposite sides, where the opposite sides may be extended if necessary. (You can do all three if you wish to cross-check yourself.) Orthocenter is at intersection. Incidentally, the orthocenter is at the right angle's vertex in a right triangle. In an obtuse triangle, the orthocenter lies outside the triangle.

9.

The locus is a "racetrack" or "pill" 5 cm long and 2 cm across. Again, the problem should have specified locus of points in a plane, since otherwise you would have a 3-dimensional capsule 5 cm long and 2 cm across.

12.

Problem should specify in a plane. Locus #1 is a circle. Locus #2 is a line (namely, the ^ bisector of the segment joining the two points). These loci can intersect in 0, 1, or 2 points.

14.

The locus is the line passing through the center but ^ to the plane of the circle. Note: If the problem had said "in a plane" instead of "in space," the locus would simply be the center.

15.

{points on ^ bisector of sPQ} \ {midpoint of sPQ}
In other words, the locus is the ^ bisector of sPQ, excluding the midpoint of sPQ. The problem should have specified locus of points in a plane.

16.

The locus is a cicle with center P and radius 8 cm.

17.

The locus is a circle internally tangent to the given circle at the given point, excluding the point itself. The radius of the small circle is half that of the original circle.

24.

The problem should really clarify that A, B, and C are distinct points and that the loci are in a plane. With those assumptions, the intersection of loci will be either Æ (if A, B, and C are collinear) or otherwise a single point (where the ^ bisectors of sAB and sBC intersect).