Congruent Triangles: practice solutions, Grade 10 – download the PDF

1 Third angle ★★★

The angles of a triangle add up to \(180^\circ\), so the third angle is \(180 - 52 - 71 = 57^\circ\).

2 An exterior angle ★★★

By the Exterior Angle Theorem, the exterior angle is \(38^\circ + 64^\circ = 102^\circ\).

Check: the adjacent interior angle is \(180 - 38 - 64 = 78^\circ\), and \(78 + 102 = 180\). It works.

3 True or false? ★★★

1. False. SSA is ambiguous: the same data can build two different triangles.
2. True. The third angles are equal by the angle sum theorem, so AAS becomes ASA.
3. False. Similar triangles of different sizes have equal angles but are not congruent.
4. True. Congruent triangles can be matched exactly, so they cover the same area.

4 Corresponding parts ★★★

Vertices match in order: \(P \leftrightarrow X\), \(Q \leftrightarrow Y\), \(R \leftrightarrow Z\).

So \(XY = PQ = 9\) cm, \(YZ = QR = 12\) cm and \(m\angle Z = m\angle R = 41^\circ\).

5 Name the criterion ★★★

1. SAS.
2. ASA.
3. AAS.
4. SSS.
5. None: SSA does not prove congruence.
6. HL.

6 Isosceles and equilateral ★★★

1. Base angles are equal: \((180 - 38) \div 2 = 71^\circ\) each.
2. All sides are equal: \(45 \div 3 = 15\) in, which is \(15 \times 2.54 = 38.1\) cm. Each angle measures \(60^\circ\).

7 Angles with algebra ★★★

The sum is \(6x = 180\), so \(x = 30\).

The angles are \(40^\circ\), \(40^\circ\) and \(100^\circ\). Check: \(40 + 40 + 100 = 180\).

Two angles are equal, so the triangle is isosceles (and obtuse), with the two sides opposite the \(40^\circ\) angles equal.

8 Solve for x ★★★

Corresponding parts of congruent triangles are equal.

1. \(3x + 2 = 5x - 10\) gives \(12 = 2x\), so \(x = 6\) and \(AB = 3 \times 6 + 2 = 20\) cm (check: \(5 \times 6 - 10 = 20\)).
2. \(2y + 9 = 3y - 11\) gives \(y = 20\), so \(m\angle B = 49^\circ\) (check: \(3 \times 20 - 11 = 49\)).

9 Exterior angle equation ★★★

Exterior angle = sum of remote interior angles: \(7x - 5 = 5x + 35\), so \(2x = 40\) and \(x = 20\).

The exterior angle is \(135^\circ\); the remote interior angles are \(60^\circ\) and \(75^\circ\) (sum \(135^\circ\)).

The adjacent interior angle is \(180 - 135 = 45^\circ\). Check: \(60 + 75 + 45 = 180\).

10 Ladders and HL ★★★

Both triangles are right triangles with \(AC = DF = 17\) ft and \(AB = DE = 8\) ft (hypotenuse and a leg), so \(\triangle ABC \cong \triangle DEF\) by HL.

By the Pythagorean theorem, \(BC = \sqrt{17^2 - 8^2} = \sqrt{225} = 15\) ft. By CPCTC, \(EF = 15\) ft as well. Each foot is 15 ft (about 4.57 m) from its wall.

11 Base angles with algebra ★★★

Base angles are equal: \(4x - 6 = 2x + 30\), so \(2x = 36\) and \(x = 18\).

Each base angle is \(4 \times 18 - 6 = 66^\circ\). The apex angle is \(180 - 66 - 66 = 48^\circ\).

12 Why SSA fails ★★★

The height from \(B\) is \(8 \sin 30^\circ = 4\). Since \(4 < 5 < 8\), the side \(BC = 5\) can reach line \(AC\) in two places.

The foot of the height is \(8 \cos 30^\circ \approx 6.93\) from \(A\), and \(\sqrt{5^2 - 4^2} = 3\). So \(AC \approx 6.93 + 3 = 9.93\) or \(AC \approx 6.93 - 3 = 3.93\).

Two different triangles share the same SSA data, so SSA cannot prove congruence.

13 The kite ★★★

Draw diagonal \(\overline{AC}\), shared by \(\triangle ABC\) and \(\triangle ADC\).

14 Congruent on a grid ★★★

\(AB = 4\) (same \(y\)), \(AC = 3\) (same \(x\)), \(BC = \sqrt{4^2 + 3^2} = 5\).

\(DE = 3\), \(DF = 4\), \(EF = \sqrt{3^2 + 4^2} = 5\).

The sides \(AB = 4\), \(BC = 5\), \(CA = 3\) match \(DF = 4\), \(FE = 5\), \(ED = 3\). By SSS, \(\triangle ABC \cong \triangle DFE\).

15 Points on an angle bisector ★★★

Compare \(\triangle PDB\) and \(\triangle PEB\).

16 The median of an isosceles triangle ★★★

\(AB = AC\) (given), \(BM = CM\) (midpoint), \(AM = AM\) (shared). So \(\triangle ABM \cong \triangle ACM\) by SSS.

By CPCTC, \(\angle B = \angle C\), and also \(\angle AMB = \angle AMC\).

These last two angles are adjacent and form a straight angle: \(\angle AMB + \angle AMC = 180^\circ\). Being equal, each is \(90^\circ\), so \(\overline{AM} \perp \overline{BC}\).

17 The converse ★★★

Compare \(\triangle ABD\) and \(\triangle ACD\).

18 Equal distances in an equilateral triangle ★★★

Compare \(\triangle DAC\) and \(\triangle EBA\).

\(DA = EB\) (given). \(AC = BA\) (sides of an equilateral triangle). \(\angle DAC = \angle EBA = 60^\circ\) (angles of an equilateral triangle), and these angles lie between the pairs of equal sides.

So \(\triangle DAC \cong \triangle EBA\) by SAS, with \(D \leftrightarrow E\), \(A \leftrightarrow B\), \(C \leftrightarrow A\). By CPCTC, \(DC = EA\).

19 A rotation on the grid ★★★

1. \(AB^2 = 6^2 + 2^2 = 40\), \(BC^2 = 4^2 + 3^2 = 25\), \(CA^2 = 2^2 + 5^2 = 29\).
   \(A'B'^2 = 2^2 + 6^2 = 40\), \(B'C'^2 = 3^2 + 4^2 = 25\), \(C'A'^2 = 5^2 + 2^2 = 29\).
   All three sides match, so \(\triangle ABC \cong \triangle A'B'C'\) by SSS.
2. Each image is \((x, y) \to (-y, x)\): \(A(1,1) \to (-1,1)\), \(B(7,3) \to (-3,7)\), \(C(3,6) \to (-6,3)\). It is a \(90^\circ\) counterclockwise rotation about the origin.

20 Across the pond ★★★

Compare \(\triangle ABC\) and \(\triangle DEC\): \(CA = CD\) and \(CB = CE\) (by construction), and \(\angle ACB = \angle DCE\) (vertical angles).

So \(\triangle ABC \cong \triangle DEC\) by SAS, and by CPCTC \(AB = DE = 46\) m.

In feet: \(46 \times 3.281 \approx 150.9\) ft.

21 Exterior angle and a ratio ★★★

The remote interior angles add up to the exterior angle: \(m\angle A + m\angle B = 130^\circ\). Split in the ratio \(3 : 2\) (five parts of \(26^\circ\)): \(m\angle A = 78^\circ\) and \(m\angle B = 52^\circ\).

The interior angle at \(C\) is \(180 - 130 = 50^\circ\). Check: \(78 + 52 + 50 = 180\).

22 Equal altitudes ★★★

Compare \(\triangle BEC\) and \(\triangle CFB\).