Congruent Triangles: math test solutions, Grade 10 – download the PDF

1 Angles of a triangle / 3 pts

1. \(6x + 30 = 180\) gives \(x = 25\) (1 pt). The angles are \(55^\circ\), \(70^\circ\) and \(55^\circ\) (1 pt).
2. The exterior angle at \(C\) is \(180 - 55 = 125^\circ\), which equals \(m\angle A + m\angle B = 55 + 70\) (1 pt).

2 Which criterion? / 3 pts

1. SAS: the angle at \(B\) is between sides \(AB\) and \(BC\) (1 pt).
2. AAS: two angles and a side that is not between them (1 pt).
3. Not enough: the angle at \(A\) is not between the sides \(AB\) and \(BC\), so this is SSA (1 pt).

3 Hypotenuse-Leg / 3 pts

Both are right triangles with equal hypotenuses and one pair of equal legs: \(\triangle JKL \cong \triangle MNP\) by HL (2 pts).

\(LK = \sqrt{25^2 - 7^2} = \sqrt{576} = 24\) in, and \(PN = 24\) in by CPCTC; that is \(24 \times 2.54 = 60.96\) cm (1 pt).

4 Isosceles triangle / 4 pts

1. Base angles are equal: \(5y + 1 = 3y + 17\) (1 pt), so \(y = 8\) and each base angle is \(41^\circ\) (1 pt).
2. \(m\angle A = 180 - 41 - 41 = 98^\circ\) (2 pts).

5 Proof in a parallelogram / 4 pts

Compare \(\triangle ABC\) and \(\triangle CDA\).

\(AB = CD\) (given) (1 pt). \(\angle BAC = \angle DCA\), alternate interior angles for parallel lines \(AB\) and \(DC\) cut by \(\overline{AC}\) (1 pt). \(AC = CA\), shared side (1 pt).

So \(\triangle ABC \cong \triangle CDA\) by SAS, and \(BC = DA\) by CPCTC (1 pt).

6 Congruence by reflection / 3 pts

\(RS^2 = 36 + 4 = 40\), \(ST^2 = 16 + 9 = 25\), \(RT^2 = 4 + 25 = 29\) (1 pt).

\(R'S'^2 = 4 + 36 = 40\), \(S'T'^2 = 9 + 16 = 25\), \(R'T'^2 = 25 + 4 = 29\) (1 pt).

All three pairs of sides are equal, so \(\triangle RST \cong \triangle R'S'T'\) by SSS (1 pt).