Triangles — Practice

Question 1

In △ABC and △PQR, AB = PQ, BC = QR, AC = PR. Which congruence rule applies?

Accepted Answer

Step 1 — Count the matched parts.
  Three pairs of sides are equal: AB-PQ, BC-QR, AC-PR.

Step 2 — Match to a criterion.
  Three sides matching → Side-Side-Side rule.

✦ Answer: SSS (Side-Side-Side).

Question 2

Two right triangles have hypotenuses 13 cm each and one leg 5 cm each. Which rule proves congruence?

Accepted Answer

Step 1 — Note the special data: right angle is implied (right triangles), hypotenuse + one other side equal.

Step 2 — Match to RHS.
  Right-Hypotenuse-Side criterion exactly fits.

✦ Answer: RHS (Right-Hypotenuse-Side).

RHS is special — it's the ONLY case where 'two sides + a non-included angle' gives congruence (because the angle is forced to be 90°).

Question 3

A triangle has all sides equal. Classify it by sides and by angles.

Accepted Answer

Step 1 — By sides: all equal → equilateral.

Step 2 — By angles: equal sides force equal opposite angles (Isosceles theorem applied to all three pairs). All three angles equal. Sum = 180° → each = 60°.
  60° < 90° → acute.

✦ Answer: Equilateral by sides; acute by angles.

Question 4

Can sides 4, 5, 10 form a triangle?

Accepted Answer

Step 1 — Apply Triangle Inequality: sum of any two sides must exceed the third.

Step 2 — Test the longest side (10) against the sum of the others.
  4 + 5 = 9. Is 9 > 10? NO.

Step 3 — Triangle inequality fails.

✦ Answer: NO, these sides cannot form a triangle.

Rule of thumb: only check the LONGEST side against the sum of the other two. If that fails, no triangle exists.

Question 5

In △ABC, ∠A = 90°, ∠B = 60°, ∠C = 30°. Order the sides from longest to shortest.

Accepted Answer

Step 1 — Larger angle ↔ longer opposite side.
  Largest ∠A = 90° → longest side BC (opposite A).
  Middle ∠B = 60° → middle side CA.
  Smallest ∠C = 30° → shortest side AB.

✦ Answer: BC > CA > AB.

Verification: 90° > 60° > 30° matches BC > CA > AB. ✓

Question 6

In △ABC and △PQR, AB = PQ, ∠B = ∠Q, BC = QR. Are the triangles congruent? State the rule.

Accepted Answer

Step 1 — Identify which parts are given.
  AB = PQ (one side).
  ∠B = ∠Q (the angle BETWEEN the two equal sides).
  BC = QR (the other side).

Step 2 — The angle ∠B is INCLUDED between sides AB and BC. Same for ∠Q between PQ and QR.

Step 3 — Two sides + included angle → SAS criterion.

✦ Answer: Yes, △ABC ≅ △PQR by SAS (Side-Angle-Side).

Key: angle is BETWEEN the two sides — that's what makes it SAS rather than SSA.

Question 7

In △ABC, AB = AC and ∠A = 50°. Find ∠B and ∠C.

Accepted Answer

Step 1 — Apply the Isosceles Theorem.
  AB = AC ⇒ ∠B = ∠C (angles opposite equal sides).

Step 2 — Use the angle sum.
  ∠A + ∠B + ∠C = 180°.
  50° + ∠B + ∠B = 180° (substituting ∠C = ∠B).
  2∠B = 130° → ∠B = 65°.

Step 3 — Conclude.
  ∠B = ∠C = 65°.

✦ Answer: ∠B = 65°, ∠C = 65°.

Question 8

In △ABC, ∠B = ∠C = 55°. Show that AB = AC.

Accepted Answer

Step 1 — Apply the Converse of the Isosceles Theorem.
  If two angles of a triangle are equal, the sides opposite them are equal.

Step 2 — Identify which sides are opposite ∠B and ∠C.
  Side opposite ∠B = AC.
  Side opposite ∠C = AB.

Step 3 — Conclude.
  ∠B = ∠C ⇒ AC = AB, i.e. AB = AC. ∎

✦ Answer: AB = AC, by the Converse of the Isosceles Theorem.

Question 9

Three sides of a triangle are 6 cm, 8 cm, x cm. Find the range of possible values of x.

Accepted Answer

Step 1 — Triangle inequality: each side must be less than the sum AND greater than the absolute difference of the other two.

Step 2 — Apply to x.
  |6 − 8| < x < 6 + 8
  2 < x < 14.

Step 3 — So x is strictly between 2 and 14 (excluding endpoints).

✦ Answer: 2 < x < 14 (in cm).

Note: x = 2 or x = 14 would give a DEGENERATE triangle (collinear points), not a real triangle.

Question 10

In △ABC, AD is the median to BC (so D is midpoint of BC). If AB = AC, prove △ABD ≅ △ACD and conclude ∠ABD = ∠ACD.

Accepted Answer

Step 1 — Set up the two triangles to compare.
  △ABD and △ACD share AD.

Step 2 — List the three pairs.
  AB = AC (given, isosceles).
  BD = DC (D is midpoint of BC).
  AD = AD (common side).

Step 3 — Apply SSS.
  Three pairs of sides equal ⇒ △ABD ≅ △ACD by SSS.

Step 4 — Apply CPCT.
  Corresponding parts of congruent triangles are equal.
  ∠ABD = ∠ACD (these are the angles at B and C, respectively).

✦ Answer: △ABD ≅ △ACD by SSS; ∠ABD = ∠ACD by CPCT.

Corollary: this re-proves the Isosceles theorem (∠B = ∠C from AB = AC) — useful when you have a median to work with.

Question 11

In two right triangles △ABC (right-angled at B) and △PQR (right-angled at Q), the hypotenuses AC and PR are equal, and one leg AB = PQ. Prove the triangles are congruent.

Accepted Answer

Step 1 — List what's known.
  ∠B = ∠Q = 90° (right angles).
  AC = PR (hypotenuses).
  AB = PQ (one leg).

Step 2 — Match to RHS criterion.
  Right angle ✓.
  Hypotenuse equal ✓.
  Side (one leg) equal ✓.

Step 3 — Apply RHS.
  △ABC ≅ △PQR by RHS.

✦ Answer: △ABC ≅ △PQR by RHS (Right-Hypotenuse-Side).

Why this works (but SSA doesn't in general): the right angle FORCES the third side via Pythagoras, removing the ambiguity.

Question 12

Prove: if a point P is equidistant from two points A and B, then P lies on the perpendicular bisector of AB.

Accepted Answer

Step 1 — Set up. Let M be the midpoint of AB. We want to show PM ⊥ AB.

Step 2 — Compare △PAM and △PBM.
  PA = PB (given P is equidistant from A, B).
  AM = BM (M is midpoint).
  PM = PM (common).

Step 3 — Apply SSS.
  △PAM ≅ △PBM by SSS.

Step 4 — Apply CPCT.
  ∠PMA = ∠PMB.
  Since ∠PMA + ∠PMB = 180° (they form a linear pair on line AB), each equals 90°.

Step 5 — Conclude.
  PM is perpendicular to AB and passes through the midpoint M ⇒ PM is the perpendicular bisector of AB. ∎

✦ Answer: P lies on the perpendicular bisector of AB.

This result is a CLASSIC; it's used in countless later proofs (circles, triangles, locus problems).

Question 13

In △ABC, AB = AC. The bisectors of ∠B and ∠C meet at point O. Prove that AO bisects ∠A.

Accepted Answer

Step 1 — From the Isosceles theorem (AB = AC ⇒ ∠B = ∠C).
  Their bisectors are equal: ∠OBC = (1/2)∠B = (1/2)∠C = ∠OCB.

Step 2 — In △OBC, two angles are equal (∠OBC = ∠OCB) ⇒ OB = OC (converse Isosceles).

Step 3 — Compare △AOB and △AOC.
  AB = AC (given).
  OB = OC (Step 2).
  AO = AO (common).

Step 4 — Apply SSS.
  △AOB ≅ △AOC.

Step 5 — Apply CPCT.
  ∠OAB = ∠OAC.
  AO therefore bisects ∠A. ∎

✦ Answer: AO bisects ∠A. The three angle bisectors of an isosceles triangle are concurrent at a point on the perpendicular bisector of the base.

Practice — Triangles

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