We have studied the properties of triangles and quadrilaterals throughout this course. The triangle is the building block for all polygons. In this section will explore the area formulas for these figures.
In the figures below, the right triangle is cut off from the right side of the rectangle and translated to the other side to form a parallelogram with the same base and height. This gives us the formula for the area of a parallelogram in the box below.
Example 1:
Find the perimeter and area of the parallelogram. Solution: Perimeter Since the opposite sides of a parallelogram are congruent, \(\overline{AB}\cong\overline{DC}\) and \(\overline{AD}\cong\overline{BC}\), then we know \(DC = 15\) cm and \(AD = 10\) cm. Perimeter of parallelogram \(ABCD\) = \(AB\) + \(BC\) + \(DC\) + \(AD\) = \(15 + 10 + 15 + 10\) = \(50\) cm Area We need to find the height. We can use the Pythagorean Theorem to find it because we have a right triangle with one leg missing. \(6^2 + h^2 = 10^2\) \(36+ h^2 = 100\) \(h^2 = 64\) \(h = 8\) cm \(A = b\cdot h = 15(8) = 120\) cm\(^2\) |
Area of Triangles
In the figures below, a parallelogram is cut in half along the diagonal which creates two congruent triangles by a \(180^{\circ}\) rotation with the center at the midpoint of the diagonal. The two congruent triangles have the same area. So, one triangle with base, \(b\), and height, \(h\) has half of the area of the parallelogram with base, \(b\), and height, \(h\).
In the figures below, a parallelogram is cut in half along the diagonal which creates two congruent triangles by a \(180^{\circ}\) rotation with the center at the midpoint of the diagonal. The two congruent triangles have the same area. So, one triangle with base, \(b\), and height, \(h\) has half of the area of the parallelogram with base, \(b\), and height, \(h\).
Equilateral Triangles
Equilateral triangles are special triangles because all of their sides and angles are congruent. They have a special formula to find their areas as well if you know its side length. This is because of when the altitude of the triangle is drawn (the height) it creates two congruent \(30-60-90\) triangles. So half of the base (opposite the \(30^{\circ}\) of the triangle is half the side length and the height (opposite the \(60^{\circ}\) is half the side length times the square root of three.
Equilateral triangles are special triangles because all of their sides and angles are congruent. They have a special formula to find their areas as well if you know its side length. This is because of when the altitude of the triangle is drawn (the height) it creates two congruent \(30-60-90\) triangles. So half of the base (opposite the \(30^{\circ}\) of the triangle is half the side length and the height (opposite the \(60^{\circ}\) is half the side length times the square root of three.
Example 5:
Find the of an equilateral triangle given the side length of \(7\) cm.
Solution:
Using the formula for an equilateral triangle \(A = \dfrac{ 7^2\sqrt{3}}{4} = \dfrac{49\sqrt 3}{4}\)
Find the of an equilateral triangle given the side length of \(7\) cm.
Solution:
Using the formula for an equilateral triangle \(A = \dfrac{ 7^2\sqrt{3}}{4} = \dfrac{49\sqrt 3}{4}\)
Area of Rhombuses
Remember that a rhombus is a parallelogram with all four sides congruent. Because of that, the formula of the area of a rhombus can be the same as the formula of the area of a parallelogram. However, there is another formula for the area of a rhombus as well. We will show both of those formulas here and look at the second formula when we explore kites.
Remember that a rhombus is a parallelogram with all four sides congruent. Because of that, the formula of the area of a rhombus can be the same as the formula of the area of a parallelogram. However, there is another formula for the area of a rhombus as well. We will show both of those formulas here and look at the second formula when we explore kites.
Area of Kites
A kite can be divided into two congruent triangles by a reflection in the end diagonal, \(d_1\). We use the diagonals, \(d_1\) and \(d_2\), to determine the area of the triangles. The base is \(d_1\) and the height is \(\dfrac{d_2}{2}\) By adding the areas of the two triangles together \(A=\dfrac{1}{2}(d_1)\left(\dfrac{d_2}{2}\right)+\dfrac{1}{2}(d_1)\left(\dfrac{d_2}{2}\right)\) and combining like terms we derive the formula below.
A kite can be divided into two congruent triangles by a reflection in the end diagonal, \(d_1\). We use the diagonals, \(d_1\) and \(d_2\), to determine the area of the triangles. The base is \(d_1\) and the height is \(\dfrac{d_2}{2}\) By adding the areas of the two triangles together \(A=\dfrac{1}{2}(d_1)\left(\dfrac{d_2}{2}\right)+\dfrac{1}{2}(d_1)\left(\dfrac{d_2}{2}\right)\) and combining like terms we derive the formula below.
Why does this formula also work for rhombuses? Remember from the Quadrilateral Tree that rhombuses are also kites and that the diagonals are perpendicular.
Area of Trapezoids
A trapezoid is a quadrilateral with one pair of parallel sides. These parallel sides are the bases. The height of the trapezoid is the perpendicular distance between the bases. By rotating the trapezoid \(180^{\circ}\) with the center at the midpoint of one of the legs, \(\overline{BD}\), we create a parallelogram. The area of the trapezoid is half of the area of the parallelogram \(AC'A'C\).
A trapezoid is a quadrilateral with one pair of parallel sides. These parallel sides are the bases. The height of the trapezoid is the perpendicular distance between the bases. By rotating the trapezoid \(180^{\circ}\) with the center at the midpoint of one of the legs, \(\overline{BD}\), we create a parallelogram. The area of the trapezoid is half of the area of the parallelogram \(AC'A'C\).
The midsegment or median of a trapezoid is one half of the sum of the lengths of the bases or the average of the lengths of the bases. The midsegment or median can be used in another formula for the area of a trapezoid.
Example 8:
Find the area of the isosceles trapezoid. Solution: We need to find the height. We can use the Pythagorean Theorem to find it because we have a right triangle with one leg missing. \(3^2 + h^2 = 5^2\) \(9+ h^2 = 25\) \(h^2 = 16\) \(h = 4\) cm Find length of base. We need to determine the length of the missing base. Since the figure is an isosceles trapezoid, The figure has line symmetry. Drawing in another height creates a rectangle and two congruent triangles. So the length of the base is \(3 + 8 + 3 = 14\) cm. Find the area \(A =\dfrac{(b_1+b_2)h}{2}=\dfrac{(14 + 8)(4)}{2}=\dfrac{88}{2}=44\) cm\(^2\) |