How does using the converse differ from using Pythagoras' theorem itself?

Pythagoras' theorem finds a missing side , when you already know the triangle is right-angled. The converse works the other way, starting from all three sides and deciding whether the right angle is there at all.

Define the space diagonal of a cuboid.

The space diagonal is the line joining two opposite corners of a cuboid, passing through the inside of it. It is sometimes called an interior or a body diagonal.

Why is it worth redrawing a triangle from a 3D diagram flat on the page?

A triangle drawn inside a 3D picture is foreshortened, so the right angle is hard to see and the sides look the wrong lengths. Drawn flat it becomes an ordinary right-angled triangle with three labelled sides, and Pythagoras then applies straightforwardly.

National 5MathsSQAFlashcardsGeometric SkillsPythagoras' Theorem

Pythagoras' Theorem (SQA National 5 Maths): Flashcards

Exam code: X847 75

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FrontConverse of Pythagoras' Theorem
Complete Pythagoras' theorem for a right-angled triangle.
$a^{2} +______=______$

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Cards in this collection (14)

Complete Pythagoras' theorem for a right-angled triangle.
$a^{2} +______=______$
The completed theorem is:
$a^{2} + b^{2} = c^{2}$
The letters $a$ and $b$ are the two shorter sides, in either order, and $c$ is the longest.
Define the hypotenuse of a right-angled triangle.
The hypotenuse is the longest side of a right-angled triangle, and it always lies opposite the right angle.
It is the side labelled $c$ in $a^{2} + b^{2} = c^{2}$ .
How does using the converse differ from using Pythagoras' theorem itself?
Pythagoras' theorem finds a missing side, when you already know the triangle is right-angled.
The converse works the other way, starting from all three sides and deciding whether the right angle is there at all.
True or False?
The converse of Pythagoras' theorem can show that a triangle is not right-angled, as well as that it is.
True.
Working out $a^{2} + b^{2}$ and $c^{2}$ separately settles it either way, since the two results either match or they do not.
For sides $5$ , $12$ and $14$ they come to $169$ and $196$ , so that triangle is not right-angled.
Which of the three sides must be used as $c$ in the check?
Always use the longest side as $c$ , since that is the only one that could be the hypotenuse.
Which of the other two is $a$ and which is $b$ makes no difference to the result.
Why must the two sides of the check be worked out separately rather than written as one equation?
Writing $a^{2} + b^{2} = c^{2}$ at the start assumes the very thing you are trying to find out.
Working out $a^{2} + b^{2}$ and $c^{2}$ as two separate values leaves the question genuinely open until you compare them.
A question asks whether a wall is perpendicular to the ground, so what must the answer say?
The conclusion has to be about the wall and the ground, not merely about the triangle.
Finding that $45 \neq 49$ shows that angle ACB is not a right angle, so the wall is not perpendicular to the ground.
Complete the 3D version of Pythagoras' theorem.
$d^{2} = x^{2} +______+______$
The completed formula is:
$d^{2} = x^{2} + y^{2} + z^{2}$
Here $d$ is the distance between the two points you are joining.
Define the space diagonal of a cuboid.
The space diagonal is the line joining two opposite corners of a cuboid, passing through the inside of it.
It is sometimes called an interior or a body diagonal.
What do $x$ , $y$ and $z$ measure in the 3D Pythagoras formula?
The three letters are the distances between the two points in directions that are all perpendicular to each other.
For a cuboid they are simply its length, its width and its height.
True or False?
A space diagonal can only be found by using the 3D formula $d^{2} = x^{2} + y^{2} + z^{2}$ .
False.
Every 3D problem can be broken into two 2D right-angled triangles instead, which needs only ordinary Pythagoras.
The 3D formula is quicker where it applies, but it is not given in the exam, so the two-triangle route is always worth having.
How do you find a space diagonal without the 3D formula?
Use Pythagoras on one face to find a face diagonal, then use it again together with the remaining edge.
For a $3$ by $4$ by $6$ cuboid the face gives $4^{2} + 6^{2} = 52$ , and then $3^{2} + 52 = 61$ , so the diagonal is $\sqrt{61}$ .
Why is it worth redrawing a triangle from a 3D diagram flat on the page?
A triangle drawn inside a 3D picture is foreshortened, so the right angle is hard to see and the sides look the wrong lengths.
Drawn flat it becomes an ordinary right-angled triangle with three labelled sides, and Pythagoras then applies straightforwardly.
How do you find the slant height of a cone from its radius and perpendicular height?
The radius, the perpendicular height and the slant height form a right-angled triangle, with the slant height as the hypotenuse.
A cone of radius $5$ cm and perpendicular height $12$ cm has slant height $\sqrt{5^{2} + 12^{2}} = 13$ cm.

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