Rectangle
A rectangle is a quadrilateral with opposite sides equal and all angles equal to \(90\) degrees.Area of rectangle
Area = Base × Height
To find the area of a rectangle, we use the formula
\(\square= bh\)
This means we multiply the length of the rectangle by its breadth (width).
This is a rectangle, it has
- opposite sides equal
- each angle equal to \(90^0\)
- diagonals are equal
- diagonals bisect each other
Rectangle Area Quiz
Drag points A or C to change the rectangle. Calculate the area and enter your answer!
Parallelogram
A parallelogram is a quadrilateral in which opposite sides are equal and parallel.Area of parallelogram
To find its area, we use the formula
\(\square= bh\)
Here, the base is any one of its sides, and the height is the perpendicular distance from the base to the opposite side (not the slanted side).
\(\square= bh\)
works because a parallelogram can be rearranged into a rectangle without changing its area. NOTE: Drag the point C
Triangle
A triangle is a closed plane figure having three sides and three angles.Area of Triangle
To find its area, we use the formula
\(\triangle= \frac{1}{2}bh\)
Here, the base is any one of its sides, and the height is the perpendicular distance from the base to the opposite vertex.
\(\triangle=\frac{1}{2} bh\)
works because a triangle is always half of parallelogram with same base and same height. NOTE: Drag the point C
Area of Triangle: Pick's Formula
\(\triangle= I + \frac{B}{2}-1\)
यहाँ \(I\) भनेको त्रिभुज भित्र पर्ने बिन्दुहरूको संख्या हो, र \(B\) भनेको त्रिभुजको किनारमा पर्ने बिन्दुहरूको संख्या हो।
Right Angled Triangle
A right-angled triangle has one angle equal to 90°. The area of right angled triangle is
\(\triangle = \frac{1}{2} \times b \times h\)
Equilateral Triangle
In equilateral triangle, all sides equal, and all angles 60°. The Area of an equilateral triangle is
\(\triangle = \frac{\sqrt{3}}{4} \times a^2\)
\( h = \sqrt{a^2 - \left( \frac{a}{2} \right)^2} \)
or\( h = \sqrt{a^2 - \frac{a^2}{4}} = \sqrt{\frac{4a^2 - a^2}{4}} = \sqrt{\frac{3a^2}{4}} = \frac{\sqrt{3}}{2} a \)
Now, in the triangle \( \triangle ABC \)
base of the triangle \( a \)
height of the triangle \( h = \frac{\sqrt{3}}{2} a \)
Therefore
Area of the triangle \( \triangle = \frac{1}{2} bh = \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{4} a^2 \)
Isosceles Triangle
In an isosceles triangle, any two sides are equal. The Area of isosceles triangle is
\(\triangle =\frac{a}{4} \sqrt{4b^2-a^2} \)
In the right angled triangle \( \triangle ADC \), using Pythagoras theorem, we have
\( h = \sqrt{b^2 - \left( \frac{a}{2} \right)^2} \)
or\( h = \sqrt{b^2 - \frac{a^2}{4}} = \sqrt{\frac{4b^2 - a^2}{4}} = \frac{1}{2} \sqrt{4b^2 - a^2} \)
Now, in the triangle \( \triangle ABC \)
base of the triangle \( a \)
height of the triangle \( h = \frac{1}{2} \sqrt{4b^2 - a^2} \)
Therefore
Area of the triangle \( \triangle = \frac{1}{2} bh = \frac{1}{2} \times a \times \frac{1}{2} \sqrt{4b^2 - a^2} = \frac{a}{4} \sqrt{4b^2 - a^2} \)
Heron's Formula
For triangle with sides \(a, b, c\) and semi-perimeter \(s = \frac{(a+b+c)}{2}\), the area of triangle is
\(\triangle =\sqrt{s(s−a)(s−b)(s−c)}\)
Proof
Consider a triangle \( \triangle ABC \) have sides \( a, b, c \) opposite to vertices \( A, B, C \) respectively, where \( s \) is the semi-perimeter
\( s = \dfrac{a+b+c}{2} \)
In right angled triangle \( \triangle ADC \), we write
\( b^2 = h^2 + d^2 \)
(1)
In right angled triangle \( \triangle BDC \), we write
\( a^2 = h^2 + (c - d)^2 \)
(2)
Subtracting (1) from (2), we get
\( a^2 - b^2 = c^2 - 2cd \)
(3)
This equation (3) simplifies \( d \) in terms of the sides of the triangle
\( d = \dfrac{b^2 + c^2 - a^2}{2c} \)
Now, the height of the triangle is
\( h^2 = b^2 - \left( \dfrac{b^2 + c^2 - a^2}{2c} \right)^2 \)
or\( h^2 = \dfrac{(2bc - a^2 + b^2 + c^2)(2bc + a^2 - b^2 - c^2)}{4c^2} \)
or\( h^2 = \dfrac{((b+c)^2 - a^2)(a^2 - (b - c)^2)}{4c^2} \)
or\( h^2 = \dfrac{(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^2} \)
or\( h^2 = \dfrac{2(s-a) \cdot 2s \cdot 2(s-c) \cdot 2(s-b)}{4c^2} = \dfrac{4s(s-a)(s-b)(s-c)}{c^2} \)
Now we apply formula for the area of triangle, we get
\( A = \dfrac{1}{2} ch = \dfrac{1}{2} c \sqrt{ \dfrac{4s(s-a)(s-b)(s-c)}{c^2} } = \sqrt{s(s-a)(s-b)(s-c)} \)
Area of Triangle: Summary
| Properties | Area Formula | Remarks |
|---|---|---|
| Base and Height | \(\triangle = \frac{1}{2} \times b \times h\) | b = base, h = height |
| Three sides | \(\triangle =\sqrt{s(s−a)(s−b)(s−c)}\) | \(s = \frac{(a+b+c)}{2}\) |
| Two sides and included angle | \(\triangle = \frac{1}{2} bc \sin A\) | A between sides b and c |
| Equilateral Triangle | \(\triangle = \frac{\sqrt{3}}{4} \times a^2\) | a = side length |
| Isosceles Triangle | \(\triangle =\frac{a}{4} \sqrt{4b^2-a^2} \) | a = one side b=two sides |
| Three Vertices (x₁,y₁), etc. | \(\triangle = \frac{1}{2} |det|\) | Determinant form |
| Two co-initial vectors | \(\triangle = \frac{1}{2} |a \times b|\) | Cross product magnitude |
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