Formulas · Mathematics

Triangles

2 formulas, each with worked examples. Revision sheet · Practise these formulas

Foundation

The Baudhāyana-Pythagoras theorem

c = a^{2} + b^{2}

What each letter means
$c$	the hypotenuse: the longest side, opposite the right angle (length)
$a$	one of the two shorter sides (the legs) (length)
$b$	the other leg (length)

In a right-angled triangle, the square on the longest side equals the two squares on the other sides added together: a² + b² = c².

Why it works

Put four copies of the triangle inside a square of side a + b, with their hypotenuses making a tilted square of side c in the middle. The big square's area is (a + b)², and it is also the four triangles (4 × ab/2) plus c². So a² + 2ab + b² = 2ab + c², which leaves a² + b² = c². Baudhāyana's Śulbasūtra stated it for the diagonal of a rectangle around 800 BCE.

When to use it

Any right angle: the diagonal of a rectangle or a screen, a ladder against a wall, the straight-line distance after walking east then north, the distance between two points on a grid.

How to use it

Check there is a right angle, and find the side opposite it (the hypotenuse).
To find the hypotenuse: square the legs, add, take the square root.
To find a leg: square the hypotenuse, subtract the square of the other leg, take the square root.

To remember: Square, square, add, root for the longest side.

Other forms

Same formula, another formThe usual statement: a² + b² = c².
Rearranged $a = c^{2} - b^{2}$ A leg, from the hypotenuse and the other leg.
Special case $c = a 2$ When the two legs are equal (b = a): the diagonal of a square.

Worked examples

Story

Ravi walks 8 km east and then 6 km north. How far is he, in a straight line, from where he started?

East and north meet at a right angle.
$c = 8^{2} + 6^{2} = 64 + 36 = 100 = 10$
He is 10 km from the start.

Answer: 10

Picture

A rectangle is 5 cm by 12 cm. How long is its diagonal?

The diagonal and two sides make a right-angled triangle.
$c = 5^{2} + 1 2^{2} = 25 + 144 = 169 = 13$

Answer: 13

Direct

A right-angled triangle has legs of 3 cm and 4 cm. How long is its hypotenuse?

$c = 3^{2} + 4^{2}$
$= 9 + 16 = 25 = 5$

Answer: 5

Reverse

A 10 m ladder leans against a wall with its foot 6 m from the wall. How high up the wall does it reach?

Try it first, then show the working

The ladder is the hypotenuse; the wall and the ground are the legs.
$b = 1 0^{2} - 6^{2} = 100 - 36 = 64 = 8$
It reaches 8 m up the wall.

Answer: 8

Exam

Is a triangle with sides 7 cm, 24 cm and 25 cm right-angled? Give a reason.

Try it first, then show the working

If it is right-angled, the longest side, 25, is the hypotenuse.
$7^{2} + 2 4^{2} = 49 + 576 = 625$
$2 5^{2} = 625$
The two agree, so the triangle is right-angled (the converse of the theorem).

Answer: Yes, because 7² + 24² = 25²

Common mistake: Adding the squares when the unknown is a leg (then you must subtract), or using the theorem in a triangle with no right angle.

Class 10

Basic proportionality theorem: AD/DB = AE/EC

E C = \frac{A E \times D B}{A D}

What each letter means
$A D$	the part of side AB from A to D, where the parallel line crosses it (length)
$D B$	the rest of side AB, from D to B (length)
$A E$	the part of side AC from A to E, where the parallel line crosses it (length)
$E C$	the rest of side AC, from E to C (length)

A line drawn parallel to one side of a triangle cuts the other two sides in the same ratio: if DE is parallel to BC, then AD/DB = AE/EC. So three of the four parts give the fourth.

Why it works

Triangles ADE and BDE share the height from E, so their areas are in the ratio AD : DB. Triangles ADE and CED share the height from D, so their areas are in the ratio AE : EC. And triangles BDE and CED have equal areas, because they share the base DE and lie between the same parallels DE and BC. So the two ratios are equal.

When to use it

A triangle with a line parallel to one side (a shelf, a ladder rung, a crossbar): finding a missing length along a side; or, from equal ratios, showing that a line is parallel (the converse).

How to use it

Check that the line is parallel to the third side.
Write AD/DB = AE/EC with the lengths you know.
Cross multiply and solve for the missing length.

To remember: Parallel line, same split.

Other forms

Same formula, another formThe same result with whole sides: AD/AB = AE/AC, because AB = AD + DB and AC = AE + EC.
RearrangedThe converse: if AD/DB = AE/EC, then DE is parallel to BC.

Worked examples

Story

A triangular roof frame has a horizontal crossbar parallel to its base. Along one sloping edge the crossbar is 1.2 m from the top and 1.8 m from the bottom. Along the other edge it is 1.6 m from the top. How far is it from the bottom along that edge?

$E C = \frac{1.6 \times 1.8}{1.2} = \frac{2.88}{1.2} = 2.4$
It is 2.4 m from the bottom along the other edge.

Answer: 2.4

Picture

In triangle ABC, D is the midpoint of AB and DE is parallel to BC, with E on AC. Show that E is the midpoint of AC.

D is the midpoint, so AD = DB and AD/DB = 1.
By the theorem, AE/EC = AD/DB = 1, so AE = EC.
So E is the midpoint of AC.

Answer: AE = EC, so E is the midpoint

Direct

In triangle ABC, DE is parallel to BC with D on AB and E on AC. AD = 3 cm, DB = 6 cm and AE = 4 cm. Find EC.

$\frac{3}{6} = \frac{4}{E C}$
$E C = \frac{4 \times 6}{3} = 8$

Answer: 8

Reverse

DE is parallel to BC in triangle ABC. AE = 5 cm, EC = 10 cm and DB = 8 cm. Find AD.

Try it first, then show the working

$\frac{A D}{8} = \frac{5}{10}$
$A D = \frac{5 \times 8}{10} = 4$

Answer: 4

Exam

In triangle ABC, DE is parallel to BC, AD/DB = 3/4 and AC = 15 cm. Find AE.

Try it first, then show the working

AE/EC = 3/4, so AC is split into 3 + 4 = 7 equal parts.
$A E = \frac{3}{7} \times 15 = \frac{45}{7}$
AE is 45/7 cm, about 6.4 cm.

Answer: 45/7 cm

Common mistake: Mixing whole sides with parts: AD/DB = AE/EC compares parts of the sides; if you use a whole side (AB), the other side must be whole too: AD/AB = AE/AC.