Area

Understand the concept of area as a 1x1 grid measurement, learn the formulas, and practice calculating shapes.

Area Study Guide

Understand the concept of area as a 1x1 grid measurement, learn the formulas, and practice calculating shapes.

Area Practice Topics

1. Pythagorean Theorem

Learn and practice the Pythagorean theorem: find hypotenuses, missing legs, and diagonals of squares and rectangles with step-by-step exercises.

2. Special Triangles

Learn about isosceles, equilateral, and 30-60-90 triangles. Practice finding their area, height, or sides using the HL / RHS congruence rule and Pythagorean theorem.

3. Inscribed Circle & Square

Learn about squares inscribed in circles and circles inscribed in squares. Understand why their area ratios are fixed and solve exercises finding area differences.

Learning Guide

1. Pythagorean Theorem

1. The Pythagorean Theorem and Square Areas

Consider a right-angled triangle with legs

a

(base) and

b

(height) and hypotenuse

c

.

Build a square on each side: the square on leg

a

has area

a^2

, the square on leg

b

has area

b^2

, and the square on the hypotenuse has area

c^2

.

The Pythagorean theorem states that the area of the hypotenuse square exactly equals the combined area of the two leg squares. This is the geometric reason why side lengths are related by squares — and since area is related to the side², we can derive the length of any side if we know the other two.

Each square's area equals the corresponding side squared:

a^2

(blue),

b^2

(red),

c^2

(purple).

2. The Formula

The Pythagorean theorem states:

a^2 + b^2 = c^2

where

a

and

b

are the two perpendicular legs and

c

is the hypotenuse (the side opposite the right angle — always the longest side).

We can rearrange it to solve for any side:
• Find the hypotenuse:

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

• Find a missing leg:

a = \sqrt{c^2 - b^2}

b = \sqrt{c^2 - a^2}

Pythagorean Theorem

a^2 + b^2 = c^2

For a right-angled triangle: the square of the hypotenuse (c) is equal to the sum of the squares of the legs (a and b).

3. Proof of the Theorem

Start with a right-angled triangle with legs

a

and

b

and hypotenuse

c

. Arrange 4 congruent copies of this triangle around a tilted inner square. The result is a large outer square with side

(a+b)

.

We can compute the total area in two ways:

• Directly:

A_{\text{big}} = (a + b)^2 = a^2 + 2ab + b^2

• By parts:

A_{\text{big}} = 4 \times \dfrac{ab}{2} + c^2 = 2ab + c^2

The big square has side

(a+b)

. The 4 green triangles are congruent to our original triangle. The purple inner square has side

c

Setting the two expressions equal:

a^2 + 2ab + b^2 = 2ab + c^2

Subtracting

2ab

from both sides:

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

This proves the Pythagorean theorem! ✓

4. Special Case: Isosceles Right Triangle ( $a = b$ )

Both legs equal: a = b

When both legs are equal (

a = b

), we substitute into the theorem:

c^2 = a^2 + a^2 = 2a^2

c = a\sqrt{2}

This means the hypotenuse of an isosceles right triangle is always

\sqrt{2}

times the leg length. For example:
• If

a = b = 3

, then

c = 3\sqrt{2} \approx 4.24

• If

a = b = 5

, then

c = 5\sqrt{2} \approx 7.07

This also gives us the diagonal of any square with side

s

: the diagonal

d = s\sqrt{2}

Mastering SealMath: Entering Square Roots

Many Pythagorean answers involve square roots like

\sqrt{50}

5\sqrt{2}

. To enter a square root in the math input:
• Keyboard shortcut: Type sqrt in the input box — MathLive creates

\sqrt{\square}

instantly. Then type your number inside.
• Virtual keyboard: Click the ⌨️ keyboard icon, go to the 123 tab, and press the √□ button.

For answers like

5\sqrt{2}

, type 5 then sqrt then 2 and close the root.

2. Special Triangles

1. Right-Angle Congruence Rule: HL / RHS

Two right-angled triangles are congruent (identical in size and shape) if they satisfy the HL or RHS rule:

• HL (Hypotenuse-Leg):
- H (Hypotenuse): The hypotenuses are equal in length.
- L (Leg): One of the other sides (legs) is equal in length.

• RHS (Right angle-Hypotenuse-Side):
- R (Right angle): Both triangles have a

90^\circ

angle.
- H (Hypotenuse): The hypotenuses are equal in length.
- S (Side): One of the other sides (legs) is equal in length.

This allows us to prove properties of other triangles by splitting them into two right-angled halves.

2. Isosceles Triangle (Altitude Bisection)

An isosceles triangle is a triangle with at least two equal legs of length

b

. In geometric diagrams, equal sides are often indicated by matching tick marks (small slash tags) drawn on them. For example, if you see a single slash mark on two sides, it indicates that their lengths are identical.

If we draw the altitude (height

h

) from the apex perpendicular to the base

a

, it splits the triangle into two right-angled triangles:
• Both halves share the altitude

h

as a common leg.
• Both halves have equal hypotenuses (the equal legs of length

b

).

By the HL / RHS rule, these two halves are congruent! This means the altitude splits the base

a

into two equal halves of length

\frac{a}{2}

Altitude of Isosceles Triangle Formula

Using the Pythagorean theorem on one half:

\left(\frac{a}{2}\right)^2 + h^2 = b^2

From this, we can derive the height

h

if we know base

a

and leg

b

, or find the base

a

a = 2\sqrt{b^2 - h^2}

Altitude of an Isosceles Triangle

h = \sqrt{b^2 - \left(\frac{a}{2}\right)^2}

3. Equilateral Triangle (Special Case of Isosceles)

An equilateral triangle is a special case of an isosceles triangle where all three sides are equal to length

a

.

Since it is isosceles, we can draw the height

h

from the apex, splitting the base into two equal halves of length

\frac{a}{2}

. The hypotenuse is the side length

a

Equilateral Triangle Law

1. If a triangle is equilateral, all its angles are

60^\circ

.
2. Conversely, if all angles of a triangle are

60^\circ

, the triangle is equilateral.

Height of Equilateral Triangle Formula

By applying the Pythagorean theorem to one half:

\left(\frac{a}{2}\right)^2 + h^2 = a^2 \implies \frac{a^2}{4} + h^2 = a^2

h^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}

Height of an Equilateral Triangle

h = a \frac{\sqrt{3}}{2}

Area of Equilateral Triangle Formula

We can now calculate the area of the equilateral triangle using the base

a

and derived height

h

\text{Area} = \frac{b \times h}{2} = \frac{a \times \left(a\dfrac{\sqrt{3}}{2}\right)}{2}

Area of an Equilateral Triangle

A = a^2 \frac{\sqrt{3}}{4}

4. The 30-60-90 Right Triangle

If we cut the equilateral triangle (with side

c

) in half using the altitude, we obtain a right-angled triangle with angles $30^\circ$ , $60^\circ$ and $90^\circ$ .

In this triangle, the hypotenuse is the original side length

c

, and the shortest leg

a

is exactly half of

c

(as established during our bisection derivation above).

30-60-90 Triangle Law

In any 30-60-90 triangle, the leg opposite the

30^\circ

angle (the shortest leg) is always half the length of the hypotenuse:

a = \frac{c}{2}

(or

c = 2a

30-60-90 Triangle Area Formula

By Pythagoras, the longer leg (opposite

60^\circ

) is

b = a\sqrt{3}

. The area is:

\text{Area} = \frac{a \times (a\sqrt{3})}{2}

Area of a 30-60-90 Triangle

A = a^2 \frac{\sqrt{3}}{2}

where a is the shortest leg.

Note on 4th Roots

When solving certain problems (such as finding the height of an isosceles triangle from its area and the leg-to-height ratio), you might encounter equations of the form $h^4 = x$ . To solve for $h$ , you must take the fourth root of both sides: $h = \sqrt[4]{x}$ . For example, if $h^4 = 625$ , then $h = \sqrt[4]{625} = 5$ .

Mastering SealMath: Entering Custom Roots

To enter a third root, fourth root, or any other nth root, you have several options:

Keyboard shortcut: Type root or nthroot in the input box. MathLive will instantly create the root symbol $\sqrt[\scriptstyle\square]{\square}$ with the cursor inside the index box — type the root index (e.g., 4), then press the right arrow key to move inside the root and type your number.
Virtual keyboard: Click the ⌨️ keyboard icon inside the input box to open the on-screen keyboard, then press the $\sqrt[\scriptstyle y]{x}$ button found under the math/symbols tab.
In the Scientific Calculator: Use the nth root function nrt(index, value). For example, to calculate the fourth root of 16, type nrt(4, 16). Alternatively, use fractional exponents: 16^(1/4). You can also copy and paste LaTeX like \sqrt[4]{16} directly into the calculator. There is also a dedicated button: press Shift, then find the third button from the right on the second row of the calculator.

3. Inscribed Circle & Square

1. Inscribed Shapes & Ratios

An inscribed shape is a geometric figure that is drawn inside another figure, so that their boundaries touch.
We will explore two fundamental inscribed setups:
• A square inscribed in a circle (its vertices lie on the circle).
• A circle inscribed in a square (it is tangent to all four sides of the square).

2. Square Inscribed in a Circle

Square inscribed in a circle: Diagonal

d = 2r

, side

a

. Area ratio is exactly 2/pi.

Let a circle have radius

r

. The square inscribed inside has vertices touching the circle boundary.

The diagonal of the square is equal to the diameter of the circle:

d = 2r

.
Using the Pythagorean theorem for side length

a

of the square:

a^2 + a^2 = d^2 \implies 2a^2 = (2r)^2 = 4r^2 \implies a^2 = 2r^2

• Area of the square:

A_{\text{square}} = a^2 = 2r^2

• Area of the circle:

A_{\text{circle}} = \pi r^2

The ratio of the area of the square to the circle is constant:

\frac{A_{\text{square}}}{A_{\text{circle}}} = \frac{2r^2}{\pi r^2} = \frac{2}{\pi} \approx 0.637

Notice that the radius

r^2

cancels out completely! The ratio is always exactly

\frac{2}{\pi}

regardless of the size.

Ratio of Inscribed Square to Circle

\frac{A_{\text{square}}}{A_{\text{circle}}} = \frac{2}{\pi}

3. Circle Inscribed in a Square

Circle inscribed in a square: Side length

a = 2r

. Area ratio is exactly pi/4.

Let a circle of radius

r

be inscribed in a square of side length

a

. The circle fits perfectly inside, tangent to all four sides.

The diameter of the circle is equal to the side length of the square:

d = a = 2r

.
• Area of the square:

A_{\text{square}} = a^2 = (2r)^2 = 4r^2

• Area of the circle:

A_{\text{circle}} = \pi r^2

The ratio of the area of the circle to the square is constant:

\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{\pi r^2}{4r^2} = \frac{\pi}{4} \approx 0.785

Again, the radius

r^2

cancels out! The ratio is always exactly

\frac{\pi}{4}

regardless of the size.

Ratio of Inscribed Circle to Square

\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{\pi}{4}

Learning Topics

Area

Area Study Guide

Area Practice Topics

1. Pythagorean Theorem

2. Special Triangles

3. Inscribed Circle & Square

Learning Guide

1. Pythagorean Theorem

1. The Pythagorean Theorem and Square Areas

2. The Formula

Pythagorean Theorem

3. Proof of the Theorem

4. Special Case: Isosceles Right Triangle (a=ba = ba=b)

Mastering SealMath: Entering Square Roots

2. Special Triangles

1. Right-Angle Congruence Rule: HL / RHS

2. Isosceles Triangle (Altitude Bisection)

Altitude of Isosceles Triangle Formula

Altitude of an Isosceles Triangle

3. Equilateral Triangle (Special Case of Isosceles)

Equilateral Triangle Law

Height of Equilateral Triangle Formula

Height of an Equilateral Triangle

Area of Equilateral Triangle Formula

Area of an Equilateral Triangle

4. The 30-60-90 Right Triangle

30-60-90 Triangle Law

30-60-90 Triangle Area Formula

Area of a 30-60-90 Triangle

Note on 4th Roots

Mastering SealMath: Entering Custom Roots

3. Inscribed Circle & Square

1. Inscribed Shapes & Ratios

2. Square Inscribed in a Circle

Ratio of Inscribed Square to Circle

3. Circle Inscribed in a Square

Ratio of Inscribed Circle to Square

4. Special Case: Isosceles Right Triangle ( $a = b$ )