This page shows how to construct a right triangle that has both the given leg lengths.

The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.

It is possible to draw more than one triangle has the side lengths as given. You can use the triangle to the left or right of the initial perpendicular, and also draw them below the initial line. All four are correct in that they satisfy the requirements, and are congruent to each other.

This construction works by effectively building two congruent triangles. The image below is the final drawing above with the blue lines PQ and QA added

Argument | Reason | |
---|---|---|

We first prove that ∆BCA is a right triangle | ||

1 | CP is congruent to CA | They were both drawn with the same compass width |

2 | PQ is congruent to AQ | They were both drawn with the same compass width |

3 | CQ is common to both triangles ∆PQC and ∆AQC | Common side |

4 | Triangles ∆PQC and ∆AQC are congruent | Three sides congruent (SSS). |

5 | ∠QCP, ∠QCA are congruent | CPCTC. Corresponding parts of congruent triangles are congruent |

6 | m∠QCA = 90° | ∠QCA and ∠QCP are a linear pair and (so add to 180°) and congruent so each must be 90° |

7 | ∆BCA is a right triangle | ∠BCA = 90°. |

We now prove the triangle is the right size | ||

8 | CA is congruent to the given leg L1 | CA copied from L1. See Copying a segment. |

9 | BC is congruent to the given leg L2 | Drawn with same compass width |

10 | ∆BCA is a right triangle with the desired side lengths | (7), (8), (9) |

- Q.E.D

- Introduction to constructions
- Copy a line segment
- Sum of n line segments
- Difference of two line segments
- Perpendicular bisector of a line segment
- Perpendicular from a line at a point
- Perpendicular from a line through a point
- Perpendicular from endpoint of a ray
- Divide a segment into n equal parts
- Parallel line through a point (angle copy)
- Parallel line through a point (rhombus)
- Parallel line through a point (translation)

- Bisecting an angle
- Copy an angle
- Construct a 30° angle
- Construct a 45° angle
- Construct a 60° angle
- Construct a 90° angle (right angle)
- Sum of n angles
- Difference of two angles
- Supplementary angle
- Complementary angle
- Constructing 75° 105° 120° 135° 150° angles and more

- Copy a triangle
- Isosceles triangle, given base and side
- Isosceles triangle, given base and altitude
- Isosceles triangle, given leg and apex angle
- Equilateral triangle
- 30-60-90 triangle, given the hypotenuse
- Triangle, given 3 sides (sss)
- Triangle, given one side and adjacent angles (asa)
- Triangle, given two angles and non-included side (aas)
- Triangle, given two sides and included angle (sas)
- Triangle medians
- Triangle midsegment
- Triangle altitude
- Triangle altitude (outside case)

- Right Triangle, given one leg and hypotenuse (HL)
- Right Triangle, given both legs (LL)
- Right Triangle, given hypotenuse and one angle (HA)
- Right Triangle, given one leg and one angle (LA)

- Finding the center of a circle
- Circle given 3 points
- Tangent at a point on the circle
- Tangents through an external point
- Tangents to two circles (external)
- Tangents to two circles (internal)
- Incircle of a triangle
- Focus points of a given ellipse
- Circumcircle of a triangle

- Square given one side
- Square inscribed in a circle
- Hexagon given one side
- Hexagon inscribed in a given circle
- Pentagon inscribed in a given circle

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