Geometry path · G08

Before this lesson: Similar triangles, Equations and quadratic tools

Your goal: Choose a length identity from a verified geometric configuration.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Names you may know: Pythagoras’ theorem; Pythagorean theorem; Apollonius’ theorem; Stewart’s theorem.

Pythagorean, Apollonius and Stewart’s Theorems: the key idea

In a right triangle with legs a,b and hypotenuse c, a²+b²=c². Conversely, if the largest side c satisfies this equation, the triangle is right. Comparing c² with a²+b² classifies the opposite angle as acute or obtuse through the cosine rule. For a median to side a in a triangle with other sides b,c, Apollonius gives mₐ²=(2b²+2c²−a²)/4. More generally, if D lies on BC with BD=m,DC=n, AD=d, AB=c,AC=b and a=m+n, Stewart’s theorem is b²m+c²n=a(d²+mn). Keep the segment-side pairing explicit.

Right triangle with legs 3 and 4 and hypotenuse 5.ABC345
A 3–4–5 right triangle. The right-angle marker identifies the two legs.

A worked example

A right triangle has legs 6 and 8. Find its hypotenuse.

c²=36+64=100. A length is positive, so c=10.

Your turn: change one thing

Find the median to the side of length 6 in a triangle whose other sides are both 5.

Try this on paper before opening the explanation.

Compare your reasoning

Apollonius gives m²=(2·25+2·25−36)/4=16, so m=4. In this isosceles triangle the median is also the altitude.

Pause and check

A trap to avoid: Applying Pythagoras when the triangle is not right-angled.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove Apollonius’s formula using coordinates.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Place the midpoint M at (0,0), B=(−a/2,0), C=(a/2,0), and A=(x,y). Then mₐ²=x²+y². The other squared sides sum to (x+a/2)²+y²+(x−a/2)²+y²=2mₐ²+a²/2. Rearranging gives 4mₐ²=2b²+2c²−a².

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. In Pythagoras, c must be which side?

Foundation

  1. A median
  2. The hypotenuse
  3. Any side
  4. Always the shortest side
Hint

It is opposite the right angle.

Answer and reasoning

The hypotenuse. Only the hypotenuse squared equals the sum of the leg squares.

2. Can the negative square root represent a side length?

Foundation

  1. Only in an obtuse triangle
  2. Only with coordinates
  3. No
  4. Yes always
Hint

Lengths are positive.

Answer and reasoning

No. Coordinates may be negative, but segment lengths are nonnegative.

3. Right triangle legs 6 and 8: hypotenuse?

Core

  1. 14
  2. 7
  3. 100
  4. 10
Hint

Use 36+64\sqrt{36+64}.

Answer and reasoning

10. The positive square root of 100 is 10.

4. Sides 5,5,6: median to side 6?

Core

  1. 3
  2. 5
  3. 8
  4. 4
Hint

Use Apollonius or split the isosceles triangle.

Answer and reasoning

4. m²=25−3²=16, so m=4.

5. For largest side c, c²>a²+b² implies which opposite angle?

Stretch

  1. Right
  2. Acute
  3. Zero
  4. Obtuse
Hint

Use the cosine rule.

Answer and reasoning

Obtuse. cos C=(a²+b²−c²)/(2ab)<0, so C>90°.

6. In Stewart’s formula with BD=m,DC=n, AB=c,AC=b, which left side is correct?

Stretch

  1. b²+c²
  2. b²m+c²n
  3. b²n+c²m
  4. a²m+d²n
Hint

The adjacent base segment pairs with the far side square.

Answer and reasoning

b²m+c²n. The standard identity is b²m+c²n=(m+n)(d²+mn).

7. The hypotenuse lies opposite what angle?

Foundation

  1. Any chosen angle
  2. 90°
  3. 60°
  4. The smallest angle
Hint

It is the side opposite the right angle.

Answer and reasoning

90°. That side is also the longest side of the right triangle.

8. A right triangle has hypotenuse 13 and one leg 5. Find the other leg.

Core

  1. 194\sqrt{194}
  2. 12
  3. 8
  4. 18
Hint

Subtract squared lengths.

Answer and reasoning

12. The other leg squared is 169−25=144, so its length is 12.

9. Can Pythagoras be applied to a triangle merely because it looks right-angled?

Stretch

  1. No, the right angle must be given or proved
  2. Yes
  3. Only if the sides are integers
  4. Only if the drawing is large
Hint

Check the theorem’s geometric hypothesis.

Answer and reasoning

No, the right angle must be given or proved. A schematic appearance is not a valid right-angle condition.

Choose your next step

Continue to Quadrilaterals and their diagonals. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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