Geometric Twins · Lesson 7 of 10
Measuring Two Sides in a Right Triangle
“Identify the hypotenuse and use the right angle, hypotenuse, and one side to establish congruence.”
• Identify the hypotenuse of a right triangle. • Construct a right triangle from its hypotenuse and one other side. • Apply the RHS condition with the right vertices matched. • Distinguish RHS from generic SSA and from other sufficient conditions.
The right angle supplies special information
A right triangle has one angle equal to 90°. The side opposite that angle is called its hypotenuse. The other two sides meet at the right angle. The hypotenuse is the longest side, and it is not one of the arms of the right angle. Identifying it correctly is essential before using the right-triangle congruence condition.
The side opposite the 90° angle in a right triangle.
For triangle ABC with angle B = 90°, AC is the hypotenuse. If another right triangle XYZ has angle Y = 90°, its hypotenuse is XZ. Suppose AC = XZ = 5 cm and BC = YZ = 4 cm. These matching lengths, together with the right angles, force the triangles to be congruent.
Two right triangles are congruent if their hypotenuses and one pair of corresponding sides other than the hypotenuse are equal. RHS stands for Right angle–Hypotenuse–Side.
Problem
In triangle LMN, angle M = 90°. Which side must match another triangle’s hypotenuse?
- 1.Locate the right angle at M.
- 2.The side that does not touch M is LN; it lies opposite the right angle.
- 3.LN is the hypotenuse. LM and MN are the two other sides. A hypotenuse must be paired with a hypotenuse in RHS.
A perpendicular and a circle
Construct a right triangle PQR with angle Q = 90°, QR = 4 cm, and PR = 5 cm. First draw QR and draw a line perpendicular to it at Q. The third vertex P must lie on that perpendicular. It must also lie on the circle centred at R with radius 5 cm, because its distance from R is the hypotenuse length.
The circle meets the perpendicular at two points, one on each side of QR. The resulting triangles are reflections across QR, so they have the same shape and size. Unlike the generic SSA example, the two intersections do not give different distances from Q. The perpendicular at the right angle makes the two locations symmetric.
Problem
Angle B = angle Y = 90°, AC = XZ = 5 cm, and BC = YZ = 4 cm. What can we prove?
- 1.AC and XZ are opposite the right angles, so they are the hypotenuses.
- 2.BC and YZ are matching non-hypotenuse sides. Both triangles are right triangles.
- 3.The match is B ↔ Y, C ↔ Z, and A ↔ X. Therefore ABC ≅ XYZ by RHS.
- 4.It follows that AB = XY and the remaining corresponding angles are equal. No calculation of the missing side is needed for this proof.
Choose the condition supported by the facts
The right angle is non-included between the hypotenuse and the given leg, so the arrangement resembles SSA. RHS is a special sufficient case with extra structure: both triangles are right triangles and the measured opposite side is their hypotenuse. The earlier warning remains correct for generic SSA.
Two matching legs in right triangles also establish congruence, but the direct reason is SAS: the two legs include the equal 90° angles. RHS is used when the equal lengths are a hypotenuse and one leg. Giving just a right angle and a hypotenuse is insufficient because the right triangle can have different proportions.
Problem
Triangle ABC has angle B = 90°; triangle FDE has angle D = 90°. AB = FD and AC = FE. Prove congruence.
- 1.The right vertices match B ↔ D. Their hypotenuses are AC and FE, which are equal.
- 2.The given legs AB and FD match, placing A ↔ F. The remaining vertices match C ↔ E.
- 3.Therefore ABC ≅ FDE by RHS. The equality BC = DE follows as a corresponding side equality.
An equal hypotenuse and one equal leg establish RHS only when both triangles are known to be right triangles. A square-looking corner in an unmarked sketch is not enough.
Quiz
If angle C = 90° in triangle ABC, which is the hypotenuse?
What must be known for RHS?
Two right triangles have matching legs. Which condition uses those legs and the right angles directly?
Why are the two construction triangles congruent?
Angle B = D = 90°, AC = FE, AB = FD. What is the correct statement?
Practice Problems
- Draw three differently oriented right triangles. Mark each right angle and identify its opposite hypotenuse.
- Construct angle Q = 90°, QR = 4 cm, and PR = 5 cm using a perpendicular and a circle.
- Construct the reflected triangle on the other side of QR and test congruence with tracing paper.
- Angle M = angle B = 90°, LN = AC, and MN = BC. Prove LMN ≅ ABC and identify one further side equality.
- Explain why equal right angles and equal hypotenuses alone do not supply RHS.
- If AB = XY and BC = YZ with angle B = angle Y = 90°, use SAS and explain why no hypotenuse measurement was needed.
Key Takeaways
• The hypotenuse is opposite the right angle. • RHS requires right triangles, equal hypotenuses, and one equal leg pair. • The perpendicular-and-circle construction gives congruent reflected triangles. • RHS is a special right-triangle condition; generic SSA remains insufficient. • Two matching legs and the right angle instead give SAS.