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Converse of the Pythagoras theorem · 5 marks

Here is a statement of the Baudhāyana–Pythagoras theorem.
Let a, b, c be the sidelengths of a triangle. If the triangle is right-angled, then a2 + b2 = c2.
What is its converse? Is this converse true?

  1. (a) What is its converse?
  2. (b) Let ∆ABC be the triangle with sidelengths a, b, c such that a2 + b2 = c2. Construct another right triangle ∆XYZ whose perpendicular sides YZ and XZ have sidelengths a and b. What can we say about the length of XY? (Hint: Use the Baudhāyana–Pythagoras Theorem.)
  3. (c) What can we say about ∆ABC and ∆XYZ? (Hint: Use the SSS criterian for congruence.)
  4. (d) Is this converse true?
BCAabcYZXab
Answer: (a) Converse: if a2 + b2 = c2, then the triangle is right-angled. (b) XY = c. (c) △ABC ≅ △XYZ (SSS), so ∠C = ∠Z = 90°. (d) Yes, the converse is true.

Step-by-step solution

Idea: Build a right triangle XYZ with the same two shorter sides a and b. The theorem itself gives its third side as c, so the two triangles have the same three sides and are congruent. The right angle of XYZ must then appear in ABC too.

(a) What is its converse?

  1. Keep the first sentence (it only names the sides). In the second sentence, swap the ‘if’ part and the ‘then’ part.½ mark
  2. Converse: Let a, b, c be the sidelengths of a triangle. If a2 + b2 = c2, then the triangle is right-angled.½ mark
Let a, b, c be the sidelengths of a triangle. If a2 + b2 = c2, then the triangle is right-angled.

(b) Let ∆ABC be the triangle with sidelengths a, b, c such that a2 + b2 = c2. Construct another right triangle ∆XYZ whose perpendicular sides YZ and XZ have sidelengths a and b. What can we say about the length of XY? (Hint: Use the Baudhāyana–Pythagoras Theorem.)

  1. △XYZ is right-angled at Z, with YZ = a and XZ = b. XY is the side opposite the right angle (the hypotenuse).½ mark
  2. By the Baudhāyana–Pythagoras theorem in △XYZ: XY2 = YZ2 + XZ2 = a2 + b2.½ mark
  3. But a2 + b2 = c2 (given for △ABC). So XY2 = c2, and since lengths are positive, XY = c.½ mark
XY = c, the same length as AB.

(c) What can we say about ∆ABC and ∆XYZ? (Hint: Use the SSS criterian for congruence.)

  1. Compare the sides: BC = a = YZ, CA = b = ZX, and AB = c = XY (from (b)).½ mark
  2. All three sides match, so △ABC ≅ △XYZ by the SSS rule, with B ↔ Y, C ↔ Z, A ↔ X.½ mark
  3. Matching angles of congruent triangles are equal, so ∠C = ∠Z = 90° (CPCT). (∠C is the angle between the sides a and b, just as ∠Z is.)½ mark
△ABC ≅ △XYZ (SSS), so ∠C = ∠Z = 90°.

(d) Is this converse true?

  1. From (c), ∠C = 90°, so △ABC is right-angled (at C, the angle opposite the side c).½ mark
  2. △ABC was any triangle with a2 + b2 = c2, so the converse is true for every such triangle.½ mark
Yes. △ABC had to be a right triangle after all, so the converse of the Baudhāyana–Pythagoras theorem is true.
The converse — ‘if a2 + b2 = c2, then the triangle is right-angled’ — is true: the right triangle XYZ with legs a, b has XY = c, so △ABC ≅ △XYZ (SSS) and ∠C = ∠Z = 90°.

Check: Sides 3, 4, 5: 9 + 16 = 25. A triangle drawn with sides 3 cm, 4 cm and 5 cm does have a right angle between the 3 cm and 4 cm sides.

Answer to write in the exam

(a)

Converse: Let a, b, c be the sidelengths of a triangle.

If a2 + b2 = c2, then the triangle is right-angled.

(b)

In △XYZ, ∠Z = 90°: XY2 = YZ2 + XZ2 (Baudhāyana–Pythagoras theorem)

XY2 = a2 + b2 = c2 (given)

∴ XY = c

(c)

BC = YZ = a, CA = ZX = b, AB = XY = c

∴ △ABC ≅ △XYZ (SSS)

∴ ∠C = ∠Z = 90° (CPCT)

(d)

∠C = 90°, so △ABC is right-angled.

∴ Yes, the converse is true: if a2 + b2 = c2, the triangle is right-angled (at the angle opposite c).

Common mistakes that cost marks

  • Writing the converse as ‘If the triangle is not right-angled, then a2 + b2 ≠ c2‘. That is a different statement, not the converse.
  • Using the Baudhāyana–Pythagoras theorem on △ABC itself. That assumes △ABC is right-angled, which is what we want to prove. It may only be used on △XYZ, which was built with a right angle.
  • Matching the wrong vertices: the right angle Z corresponds to C, the angle between the sides a and b.

How this can come in the exam

MCQ (1 mark)

Which set of sidelengths forms a right-angled triangle?

  1. 4, 5, 6
  2. 6, 8, 10
  3. 5, 6, 8
  4. 7, 8, 9
Show answer

(B) 6, 8, 10
62 + 82 = 36 + 64 = 100 = 102, so by the converse the triangle is right-angled. For the others the squares do not add up (16 + 25 ≠ 36, 25 + 36 ≠ 64, 49 + 64 ≠ 81).

Case-based (4 marks)

A carpenter wants to check that the corner of a wooden frame is a right angle. From the corner she marks 60 cm along one edge and 80 cm along the other, then measures the straight distance between the two marks.
(i) What distance should she get if the corner is a right angle?
(ii) If she does get that distance, which result lets her conclude the corner is a right angle?
(iii) On another frame she measures 98 cm. Is that corner a right angle? Give a reason.

Show answer(i) If the corner is 90°, the distance is √(602 + 802) = √(3600 + 6400) = √10000 = 100 cm (1 mark).
(ii) The converse of the Baudhāyana–Pythagoras theorem: 602 + 802 = 1002, so the triangle is right-angled at the corner (1 mark).
(iii) No (1 mark). If the corner were a right angle, the theorem would make the distance exactly 100 cm; 98 cm ≠ 100 cm (982 = 9604 ≠ 10000), so it is not a right angle (1 mark).

Try one yourself

Is a triangle with sides 9 cm, 12 cm and 15 cm right-angled? If so, which angle is the right angle?

Show answer

92 + 122 = 81 + 144 = 225 = 152, so yes. The right angle is opposite the 15 cm side, between the 9 cm and 12 cm sides.

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