Ninth Grade Mathematics: Calculating Triangular & Quadrilateral Areas
While Heron's formula is derived for triangles, its practical utility extends across all of polygonal mensuration. By drawing non-overlapping diagonals, any irregular quadrilateral or multi-sided polygon can be partitioned into a collection of triangles, enabling you to calculate the area of complex plots of land using only linear boundary measurements.
The Triangulation Method for Quadrilaterals
An irregular quadrilateral ABCD has no simple area formula because its shape can flex without altering its four side lengths. However, if the length of a single diagonal (say diagonal AC) is measured alongside the four boundary sides, the quadrilateral is triangulated into two rigid triangles:
TRIANGULATION OF AN IRREGULAR QUADRILATERAL
B
*
/ | \
AB / | \ BC
/ | \
*-----+-------*
A Diagonal C
\ AC /
AD \ | / CD
\ | /
\| /
*
D
Triangulation Area Protocol:
1. Divide Quadrilateral ABCD into Triangle ABC and Triangle ADC using diagonal AC.
2. Calculate the area of Triangle ABC using Heron's formula with sides AB, BC, and AC.
3. Calculate the area of Triangle ADC using Heron's formula with sides AD, CD, and AC.
4. By the Area Addition Axiom, the total area of the quadrilateral is:
ar(ABCD) = ar(Triangle ABC) + ar(Triangle ADC).
This method is the standard protocol used by cadastral land surveyors worldwide to map irregular plots of land.
Triangles with Ratios and Perimeters
A common practical problem provides the ratio of a triangle's side lengths along with its total perimeter.
Problem Archetype:
A triangular field has sides in the ratio 3 : 5 : 7, and its perimeter is 300 meters.
Find its area.
Step 1: Set up the side representations:
Let the sides be 3*x, 5*x, and 7*x.
Step 2: Solve for x using the perimeter:
3*x + 5*x + 7*x = 300
15*x = 300 ===> x = 20 meters.
Step 3: Determine the actual side lengths:
a = 3 * 20 = 60 meters
b = 5 * 20 = 100 meters
c = 7 * 20 = 140 meters
Step 4: Compute semi-perimeter:
s = 300 / 2 = 150 meters.
Step 5: Compute differences:
s - a = 150 - 60 = 90
s - b = 150 - 100 = 50
s - c = 150 - 140 = 10
Step 6: Apply Heron's Formula:
Area = sqrt( 150 * 90 * 50 * 10 )
= sqrt( (15 * 10) * (9 * 10) * (5 * 10) * (1 * 10) )
= sqrt( 15 * 9 * 5 * 1 * 10^4 )
= sqrt( 675 * 10,000 ) = sqrt( 225 * 3 * 10,000 )
= 15 * 100 * sqrt(3) = 1,500 * sqrt(3) square meters.
Finding the Shortest and Longest Altitudes
Once the area of a triangle is determined via Heron's formula, the lengths of all three perpendicular altitudes can be found instantly using the relationship Area = (1/2) * Base * Altitude:
Altitude Formulas:
h_a = (2 * Area) / a (Altitude to side a)
h_b = (2 * Area) / b (Altitude to side b)
h_c = (2 * Area) / c (Altitude to side c)
Because the area is constant, the altitude is inversely proportional to the base: The longest altitude corresponds to the shortest side. The shortest altitude corresponds to the longest side.
Chapter Exercises
Exercise 1: Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and diagonal AC = 5 cm.
Exercise 2: The sides of a triangular park are 50 m, 65 m, and 65 m. Find the area of the park, and determine the length of the altitude drawn to the longest side.
Exercise 3: A field is in the shape of a trapezoid whose parallel sides are 25 meters and 10 meters, and non-parallel sides are 14 meters and 13 meters. Calculate the area of the field by partitioning it into a parallelogram and a triangle.
Solutions and Explanations
Solution 1: Divide quadrilateral ABCD into Triangle ABC and Triangle ADC using diagonal AC = 5 cm. For Triangle ABC: sides are 3 cm, 4 cm, and 5 cm. Since 3^2 + 4^2 = 9 + 16 = 25 = 5^2, triangle ABC is a right-angled triangle with right angle at B. Its area is (1/2) * 3 * 4 = 6 square cm. For Triangle ADC: sides are AD = 5 cm, CD = 4 cm, and AC = 5 cm. This is an isosceles triangle. Semi-perimeter s = (5 + 4 + 5) / 2 = 14 / 2 = 7 cm. Differences: s - a = 7 - 5 = 2; s - b = 7 - 4 = 3; s - c = 7 - 5 = 2. Apply Heron's formula: ar(Triangle ADC) = sqrt(7 * 2 * 3 * 2) = sqrt(84) = 2 * sqrt(21) approx 2 * 4.583 = 9.17 square cm. Total area of quadrilateral ABCD = ar(Triangle ABC) + ar(Triangle ADC) = 6 + 2 * sqrt(21) approx 15.17 square cm.
Solution 2: The triangular park has sides a = 50 m, b = 65 m, and c = 65 m. Semi-perimeter s = (50 + 65 + 65) / 2 = 180 / 2 = 90 meters. Differences: s - a = 90 - 50 = 40; s - b = 90 - 65 = 25; s - c = 90 - 65 = 25. Apply Heron's formula: Area = sqrt(90 * 40 * 25 * 25) = sqrt(3600 * 625) = 60 * 25 = 1,500 square meters. The longest sides are the two 65-meter sides. The altitude h to a 65-meter side is given by Area = (1/2) * base * h, so 1500 = (1/2) * 65 * h. Multiply by 2: 3000 = 65 * h, so h = 3000 / 65 = 600 / 13 approx 46.15 meters.
Solution 3: Let the trapezoid be ABCD with parallel bases AB = 25 m and CD = 10 m, and non-parallel sides AD = 13 m and BC = 14 m. Through vertex C, draw line CE parallel to side AD, meeting base AB at point E. This partitions the trapezoid into a parallelogram AECD and a triangle EBC. In parallelogram AECD: AE = CD = 10 m, and CE = AD = 13 m. The base of triangle EBC is EB = AB - AE = 25 - 10 = 15 m. The sides of triangle EBC are EB = 15 m, BC = 14 m, and CE = 13 m. Compute the area of triangle EBC using Heron's formula: semi-perimeter s = (15 + 14 + 13) / 2 = 42 / 2 = 21 m. Differences: 21 - 15 = 6; 21 - 14 = 7; 21 - 13 = 8. Area(EBC) = sqrt(21 * 6 * 7 * 8) = sqrt((37) * (23) * 7 * (2^3)) = sqrt(2^4 * 3^2 * 7^2) = 4 * 3 * 7 = 84 square meters. The area of triangle EBC is also (1/2) * base * height = (1/2) * 15 * h = 84, so h = (84 * 2) / 15 = 168 / 15 = 11.2 meters. The area of parallelogram AECD is Base * Height = AE * h = 10 * 11.2 = 112 square meters. The total area of the trapezoidal field is Area(AECD) + Area(EBC) = 112 + 84 = 196 square meters.