Every answer shows the matching of vertices, the criterion used, and the algebra. Check that your reasons name a property or theorem, not just a picture, and that proofs list the congruent parts in the same order as the triangle names.
Problem 1: Name corresponding parts
The letters match in order: A with D, B with E, C with F.
Sides: \(\overline{AB}\cong\overline{DE}\), \(\overline{BC}\cong\overline{EF}\), \(\overline{AC}\cong\overline{DF}\).
Angles: \(\angle A\cong\angle D\), \(\angle B\cong\angle E\), \(\angle C\cong\angle F\).
Problem 2: Read a congruence statement
P matches X, Q matches Y, and R matches Z.
\(\overline{QR}\) corresponds to \(\overline{YZ}\). \(\angle R\) corresponds to \(\angle Z\). \(\overline{PR}\) is congruent to \(\overline{XZ}\).
Problem 3: Find missing measures
Corresponding parts of congruent triangles are congruent. A matches D, B matches E, C matches F.
\(DE=AB=9\text{ in}\) and \(EF=BC=12\text{ in}\).
\(m\angle D=m\angle A=48^{\circ}\) and \(m\angle F=m\angle C=75^{\circ}\).
Answer: \(DE=9\text{ in}\), \(EF=12\text{ in}\), \(m\angle D=48^{\circ}\), \(m\angle F=75^{\circ}\).
Problem 4: Reflect a triangle
A reflection across the \(y\)-axis changes \((x,y)\) to \((-x,y)\): \(A^{\prime}(-1,2)\), \(B^{\prime}(-5,2)\), \(C^{\prime}(-1,5)\).
Original: \(AB=4\), \(AC=3\), and \(BC=\sqrt{4^{2}+3^{2}}=5\). Image: \(A^{\prime}B^{\prime}=4\), \(A^{\prime}C^{\prime}=3\), \(B^{\prime}C^{\prime}=5\).
All three pairs match, so the reflection produced a congruent triangle: \(\triangle ABC\cong\triangle A^{\prime}B^{\prime}C^{\prime}\).
Problem 5: Translate a triangle
Add 6 to each \(x\) and 2 to each \(y\): \(L^{\prime}(2,1)\), \(M^{\prime}(5,1)\), \(N^{\prime}(2,5)\).
Original: \(LM=3\), \(LN=4\), \(MN=\sqrt{3^{2}+4^{2}}=5\). Image: \(L^{\prime}M^{\prime}=3\), \(L^{\prime}N^{\prime}=4\), \(M^{\prime}N^{\prime}=5\).
A translation is a rigid motion and every side matches, so \(\triangle LMN\cong\triangle L^{\prime}M^{\prime}N^{\prime}\).
Problem 6: Name the criterion: pair 1
All three pairs of sides are marked congruent, so the criterion is SSS.
Problem 7: Name the criterion: pair 2
Sides \(\overline{AB}\cong\overline{DE}\) and \(\overline{AC}\cong\overline{DF}\) are marked, and the angle between them, \(\angle A\cong\angle D\), is marked. The criterion is SAS.
Problem 8: Name the criterion: pair 3
Angles \(\angle A\cong\angle D\) and \(\angle B\cong\angle E\) are marked, and the side between them, \(\overline{AB}\cong\overline{DE}\), is marked. The criterion is ASA.
Problem 9: Name the criterion: pair 4
Angles \(\angle A\cong\angle D\) and \(\angle B\cong\angle E\) are marked, but the marked side \(\overline{BC}\cong\overline{EF}\) is not between them. The criterion is AAS.
Problem 10: Name the criterion: pair 5
Each triangle has a right angle at C and F. The hypotenuses \(\overline{AB}\) and \(\overline{DE}\) are congruent, and the legs \(\overline{AC}\) and \(\overline{DF}\) are congruent. The criterion is HL.
Problem 11: Name the criterion: pair 6
\(\overline{AB}\cong\overline{DE}\), \(\overline{BC}\cong\overline{EF}\), and \(\angle A\cong\angle D\) are marked, but the angle is not between the two sides (SSA). The third sides clearly differ, so the answer is not enough information.
Problem 12: Solve for x with sides
Corresponding sides are equal: \(3x+2=5x-10\).
Subtract \(3x\): \(2=2x-10\). Add 10: \(12=2x\). Divide by 2: \(x=6\).
\(AB=3(6)+2=20\) and \(DE=5(6)-10=20\). Answer: \(x=6\), \(AB=20\text{ in}\).
Problem 13: Solve for x with angles
Corresponding angles are equal: \(4x+7=6x-13\).
Subtract \(4x\): \(7=2x-13\). Add 13: \(20=2x\). Divide: \(x=10\).
\(m\angle B=4(10)+7=47^{\circ}\) and \(m\angle E=6(10)-13=47^{\circ}\). Answer: \(x=10\), \(m\angle B=47^{\circ}\).
Problem 14: Triangles in a kite
- The shared side is \(\overline{AC}\); \(\overline{AC}\cong\overline{AC}\) by the reflexive property.
- \(\overline{AB}\cong\overline{AD}\), \(\overline{CB}\cong\overline{CD}\), and \(\overline{AC}\cong\overline{AC}\) give three pairs of congruent sides: SSS.
- B corresponds to D in \(\triangle ABC\cong\triangle ADC\), so \(\angle B\cong\angle D\) by CPCTC.
Problem 15: Proof with parallel segments
| Statements | Reasons |
|---|---|
| \(\overline{AC}\parallel\overline{BD}\) | Given |
| \(\angle CAM\cong\angle DBM\) | Alternate interior angles are congruent |
| \(M\) is the midpoint of \(\overline{AB}\) | Given |
| \(\overline{AM}\cong\overline{BM}\) | Definition of midpoint |
| \(\angle AMC\cong\angle BMD\) | Vertical angles are congruent |
| \(\triangle AMC\cong\triangle BMD\) | ASA (steps 2, 4, 5) |
Problem 16: Why SSA fails
Both triangles share angle A, the side \(\overline{AB}\), and a side of 9 inches opposite the angle. These are two sides and an angle that is not between them (SSA).
The third sides are 13.8 in and 4.6 in, which are different, so the triangles are not congruent.
SSA is not a congruence criterion: the given parts fit two different triangles.
Problem 17: Points on a perpendicular bisector
| Statements | Reasons |
|---|---|
| \(M\) is the midpoint of \(\overline{AB}\) | Given |
| \(\overline{AM}\cong\overline{BM}\) | Definition of midpoint |
| \(\ell\perp\overline{AB}\) at \(M\) | Given |
| \(\angle PMA\) and \(\angle PMB\) are right angles, so \(\angle PMA\cong\angle PMB\) | Perpendicular lines form right angles; right angles are congruent |
| \(\overline{PM}\cong\overline{PM}\) | Reflexive property |
| \(\triangle PMA\cong\triangle PMB\) | SAS (steps 2, 4, 5) |
| \(\overline{PA}\cong\overline{PB}\) | CPCTC |
Problem 18: Base angles of an isosceles triangle
| Statements | Reasons |
|---|---|
| \(\overline{AB}\cong\overline{AC}\) | Given |
| \(\overline{AD}\) bisects \(\angle BAC\) | Given |
| \(\angle BAD\cong\angle CAD\) | Definition of angle bisector |
| \(\overline{AD}\cong\overline{AD}\) | Reflexive property |
| \(\triangle ABD\cong\triangle ACD\) | SAS (steps 1, 3, 4) |
| \(\angle B\cong\angle C\) | CPCTC |
Problem 19: Flower beds
- All three pairs of sides match (12 with 12, 16 with 16, 20 with 20), so the beds are congruent by SSS.
- \(12^{2}+16^{2}=144+256=400=20^{2}\), so the converse of the Pythagorean theorem gives a right angle between the 12 ft and 16 ft sides. Area \(=\frac{1}{2}(12)(16)=96\) square feet.
- Both beds: \(2\times 96=192\) square feet. Cost: \(192\times 0.75=144\). The mulch costs $144.
Problem 20: Truss braces
- Both triangles are right triangles. The hypotenuses \(\overline{AB}\cong\overline{CB}\) (13 ft each) and the leg \(\overline{BD}\) is shared, so HL.
- By CPCTC, \(DC=AD=5\text{ ft}\).
- \(BD^{2}=13^{2}-5^{2}=169-25=144\), so \(BD=12\text{ ft}\). The base is \(AC=5+5=10\), so \(AC=10\text{ ft}\).
Problem 21: Distance across a pond
In \(\triangle ACB\) and \(\triangle DCE\): \(\overline{AC}\cong\overline{DC}\) (by construction), \(\angle ACB\cong\angle DCE\) (vertical angles), and \(\overline{BC}\cong\overline{EC}\) (by construction).
That is SAS, so \(\triangle ACB\cong\triangle DCE\). By CPCTC, \(\overline{AB}\cong\overline{DE}\).
Therefore \(AB=84\text{ yd}\).
Problem 22: Perimeter and angle from congruence
- \(2x+3=4x-9\) gives \(12=2x\), so \(x=6\). \(5y-4=3y+10\) gives \(2y=14\), so \(y=7\). \(3z+9=5z-21\) gives \(30=2z\), so \(z=15\). Answer: \(x=6\), \(y=7\), \(z=15\).
- \(DE=4(6)-9=15\), \(EF=3(7)+10=31\), and \(DF=AC=22\). Perimeter \(=15+31+22=68\): 68 in.
- \(m\angle A=3(15)+9=54^{\circ}\). Check: \(m\angle D=5(15)-21=54^{\circ}\). Answer: \(54^{\circ}\).
Problem 23: Congruent triangles on a grid
- Each point moves 2 units right and 1 unit up: \((0,0)\) moves to \((2,1)\), \((6,0)\) to \((8,1)\), and \((0,8)\) to \((2,9)\). A translation is a rigid motion, so \(\triangle ABC\cong\triangle DEF\).
- In \(\triangle ABC\): \(AB=6\) is horizontal and \(AC=8\) is vertical, so the angle at A is a right angle. In \(\triangle GHI\): \(GH=8-2=6\) is vertical and \(GI=5-(-3)=8\) is horizontal, so the angle at G is a right angle. Two sides and the included angle match (A with G): SAS.
- \(BC=\sqrt{6^{2}+8^{2}}=10\), \(EF=\sqrt{6^{2}+8^{2}}=10\), and \(HI=\sqrt{8^{2}+6^{2}}=10\). Each hypotenuse is 10 units.
Problem 24: Ethan's claim about equal angles
- Each triangle is equilateral: \(3\times 4=12\) ft and \(3\times 6=18\) ft.
- Equal angles (AAA) give triangles of the same shape but not necessarily the same size. The perimeters 12 ft and 18 ft differ, so the triangles are not congruent.
- If one pair of corresponding sides were congruent (for example both sides 4 ft), the triangles would be congruent by AAS.
See every chapter on the Geometry math page.