5.3b Congruence Worksheets with geometric diagrams and questions on triangle congruence and coordinates.

Let’s solve each problem step by step.

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5.2 Practice K – Review: Fill in the missing blanks

1. The circumcenter is the point of concurrency of the perpendicular bisectors of a triangle.
→ *Reasoning:* The circumcenter is where the perpendicular bisectors of the sides meet. It’s also the center of the circle that passes through all three vertices (circumcircle).

2. Where is the circumcenter for an acute triangle? Inside the triangle
→ *Reasoning:* In an acute triangle, all angles are less than 90°, so the perpendicular bisectors intersect inside the triangle.

3. Where is the circumcenter for an obtuse triangle? Outside the triangle
→ *Reasoning:* In an obtuse triangle, one angle is greater than 90°, so the perpendicular bisectors intersect outside the triangle.

4. Find the coordinates of the circumcenter of each triangle.

Problem 8: Triangle with points at (0,0), (6,0), and (0,4) — right triangle at origin.

→ For a right triangle, the circumcenter is at the midpoint of the hypotenuse.

Hypotenuse is from (6,0) to (0,4). Midpoint = ((6+0)/2, (0+4)/2) = (3, 2)

✔ Answer: (3, 2)

Problem 9: Triangle with points at (-2, -2), (2, -2), and (0, 2) — isosceles triangle.

Let’s find perpendicular bisector of base from (-2,-2) to (2,-2): horizontal line y = -2 → midpoint is (0, -2). Perpendicular bisector is vertical line x = 0.

Now take side from (2,-2) to (0,2). Midpoint = ((2+0)/2, (-2+2)/2) = (1, 0)

Slope of this side: (2 - (-2)) / (0 - 2) = 4 / (-2) = -2 → perpendicular slope = 1/2

Equation of perpendicular bisector: y - 0 = (1/2)(x - 1) → y = (1/2)x - 1/2

Intersect with x = 0: y = (1/2)(0) - 1/2 = -1/2

So circumcenter is at (0, -0.5) or (0, -1/2)

✔ Answer: (0, -1/2)

Problem 10: Given O is circumcenter of △ABC, AO = 7, find CO.

→ Since O is circumcenter, it is equidistant from all vertices: AO = BO = CO

So if AO = 7, then CO = 7

✔ Answer: 7

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5.1 Extra H – Measurements of Triangles

5. Write an equation to show that the length of the midsegment is half the length of a parallel segment. Then find the value of x.

Diagram shows midsegment = 3x + 1, parallel side = 8x - 14

Midsegment theorem: midsegment = ½ × parallel side

So:
3x + 1 = ½ (8x - 14)

Multiply both sides by 2:
2(3x + 1) = 8x - 14
6x + 2 = 8x - 14

Subtract 6x:
2 = 2x - 14

Add 14:
16 = 2x → x = 8

✔ Answer: x = 8

6. The midpoints of the sides of △ABC are P, Q, and R. The perimeter of △ABC is 14. What is the perimeter of △PQR?

→ Triangle formed by connecting midpoints is called the medial triangle. Its sides are each half the length of the corresponding side of the original triangle.

So perimeter of △PQR = ½ × perimeter of △ABC = ½ × 14 = 7

✔ Answer: 7

7. X is the midpoint of AC̅. Y is the endpoint of BC̅ such that XY is a midsegment.

Given:
AB = 7x + 1
XY = 4x - 2
BC = 10x - 9
AC = 6x + 3

Since XY is a midsegment parallel to AB, then XY = ½ AB

So:
4x - 2 = ½ (7x + 1)

Multiply both sides by 2:
8x - 4 = 7x + 1

Subtract 7x:
x - 4 = 1 → x = 5

Now check if this makes sense with other info? Not needed since question only asks for x based on midsegment relation.

✔ Answer: x = 5

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Use the diagram at the right for Exercises 26 and 27.

Diagram: Right triangle ABC, right angle at C. D is foot of altitude from C to AB. CD = 6, AD = 8, DB = ? , AB = ?, AC = ?, BC = ?

Actually, looking at labels:

- Point A top, B bottom left, C bottom right → right angle at C.
- Altitude from C to AB meets at D.
- CD = 6
- AD = 8
- DB = ? → labeled as “?” but we can compute.

Wait — actually, in the diagram, it says:

CD = 6
AD = 8
DB = ?
AB = ?
AC = ?
BC = ?

But also, there's a label “h” near CD, and “a”, “b” near legs.

Actually, standard geometric mean relationships in right triangles with altitude to hypotenuse:

In right triangle ABC, right-angled at C, with altitude CD to hypotenuse AB:

Then:
- CD² = AD × DB
- AC² = AD × AB
- BC² = BD × AB

Given: CD = 6, AD = 8

So:
CD² = AD × DB
→ 6² = 8 × DB
→ 36 = 8 × DB
→ DB = 36 ÷ 8 = 4.5

✔ So DB = 4.5

Now AB = AD + DB = 8 + 4.5 = 12.5

Now AC² = AD × AB = 8 × 12.5 = 100 → AC = √100 = 10

BC² = BD × AB = 4.5 × 12.5 = let’s compute:
4.5 × 12.5 = (9/2) × (25/2) = 225/4 = 56.25 → BC = √56.25 = 7.5

Check: Does 10² + 7.5² = 12.5²?
100 + 56.25 = 156.25
12.5² = 156.25 → yes!

So:

26. What is the shortest distance from the river?
→ Assuming “river” is along AB, and point C is the location, then shortest distance is the perpendicular distance → which is CD = 6

✔ Answer: 6

27. What is the shortest distance from the school at point B to Euclid Ave?
→ Assuming Euclid Ave is along AC? Wait — need to interpret.

Looking back: “What is the shortest distance from the school at point B to Euclid Ave?”

If Euclid Ave is the line containing AC, then shortest distance from B to line AC is the perpendicular distance.

But in triangle ABC, right-angled at C, so AC and BC are legs.

Distance from B to line AC: since angle C is 90°, BC is perpendicular to AC → so distance is just length of BC = 7.5

Alternatively, if Euclid Ave is AB, then distance from B to AB is zero — doesn’t make sense.

More likely: Euclid Ave is the line containing AC, and school is at B → shortest distance is perpendicular from B to AC → which is BC, since angle C is 90°.

So answer is BC = 7.5

But let’s confirm with area method to be sure.

Area of triangle ABC = ½ × AC × BC = ½ × 10 × 7.5 = 37.5

Also, area = ½ × AB × CD = ½ × 12.5 × 6 = 37.5 → matches.

Now, distance from B to line AC: since AC is one leg, and BC is perpendicular to it, yes — distance is BC = 7.5

✔ Answer: 7.5

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Final Answers Summary:

5.2 Practice K:
1. perpendicular bisectors
2. Inside the triangle
3. Outside the triangle
8. (3, 2)
9. (0, -1/2)
10. 7

5.1 Extra H:
5. x = 8
6. 7
7. x = 5

Exercises 26 & 27:
26. 6
27. 7.5

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Final Answer:
5.2 Practice K:
1. perpendicular bisectors
2. Inside the triangle
3. Outside the triangle
8. (3, 2)
9. (0, -1/2)
10. 7

5.1 Extra H:
5. x = 8
6. 7
7. x = 5

26. 6
27. 7.5