Application of Heron’s Formula: Area of Quadrilaterals

Finding the area of a quadrilateral can be simplified by dividing it into two triangles. Since the area of a triangle is relatively easy to calculate (using the formula involving base and height or Heron’s formula), we can leverage this to find the area of more complex shapes like quadrilaterals. The process involves drawing a diagonal across the quadrilateral, which effectively splits it into two triangles. Then, calculate the area of each triangle separately and add them to obtain the total area of the quadrilateral.

Formulae

The primary formulae used here are:

• Area of a triangle (base and height): $Area = \frac{1}{2} \times base \times height$
• Area of a triangle (using Heron’s formula – when all three sides are known): $Area = \sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semi-perimeter ($s = \frac{a+b+c}{2}$), and $a$, $b$, and $c$ are the side lengths.
• Area of a quadrilateral: $Area_{quadrilateral} = Area_{triangle1} + Area_{triangle2}$

Examples

Example-1: Find the area of a quadrilateral ABCD where AC = 10 cm, the perpendicular distance from B to AC is 4 cm, and the perpendicular distance from D to AC is 6 cm.

Solution:

1. Divide the quadrilateral into two triangles: $\triangle ABC$ and $\triangle ADC$.
2. Area of $\triangle ABC = \frac{1}{2} \times AC \times (perpendicular \space from \space B \space to \space AC) = \frac{1}{2} \times 10 \times 4 = 20 \space cm^2$.
3. Area of $\triangle ADC = \frac{1}{2} \times AC \times (perpendicular \space from \space D \space to \space AC) = \frac{1}{2} \times 10 \times 6 = 30 \space cm^2$.
4. Area of quadrilateral ABCD = Area of $\triangle ABC$ + Area of $\triangle ADC$ = $20 + 30 = 50 \space cm^2$.

Example-2: A quadrilateral has vertices A(1,1), B(4,5), C(7,1) and D(4,-3). Find its area.

Solution:

1. Draw a diagonal, say AC. The coordinates of A and C are (1,1) and (7,1), respectively. The length AC = 7-1 = 6 units.
2. The perpendicular distance from B to AC is the y-coordinate difference, 5-1=4. Therefore, the area of $\triangle ABC = (1/2)*6*4 = 12$ square units.
3. The perpendicular distance from D to AC is the y-coordinate difference, 1-(-3)=4. Therefore, the area of $\triangle ADC = (1/2)*6*(-3-1) = 12$ square units.
4. The area of quadrilateral ABCD = Area of $\triangle ABC$ + Area of $\triangle ADC$ = $12 + 12 = 24 \space units^2$.

Theorem with Proof

Theorem: The area of a quadrilateral can be found by dividing it into two triangles and summing their areas.

Proof:

Let’s consider a quadrilateral ABCD. Draw a diagonal AC, dividing the quadrilateral into two triangles, $\triangle ABC$ and $\triangle ADC$.

1. The area of $\triangle ABC$ is given by $Area_{\triangle ABC} = \frac{1}{2} \times base_{AC} \times height_{B}$. (Let $h_B$ be the height from B to AC)
2. The area of $\triangle ADC$ is given by $Area_{\triangle ADC} = \frac{1}{2} \times base_{AC} \times height_{D}$. (Let $h_D$ be the height from D to AC)
3. The area of the quadrilateral ABCD is the sum of the areas of these two triangles: $Area_{ABCD} = Area_{\triangle ABC} + Area_{\triangle ADC}$ $Area_{ABCD} = (\frac{1}{2} \times base_{AC} \times h_B) + (\frac{1}{2} \times base_{AC} \times h_D)$ $Area_{ABCD} = \frac{1}{2} \times base_{AC} \times (h_B + h_D)$
4. This proves that dividing a quadrilateral into two triangles allows us to calculate its area by summing the areas of the individual triangles. Therefore, area of quadrilateral is simply the addition of areas of triangles.

Common mistakes by students

Common mistakes include:

• Incorrectly identifying the base and height of the triangles.
• Forgetting to divide by 2 when calculating the area of a triangle.
• Confusing the lengths of the sides of the triangles with the height.
• Not correctly understanding the co-ordinate system to calculate height of the triangles.

Real Life Application

This concept is used in various real-life situations, such as:

• Land Surveying: Surveyors often divide irregular land plots (which can be considered as quadrilaterals) into triangles to calculate their areas and determine property boundaries.
• Architecture and Construction: Architects and builders use this method to calculate the area of floors, walls, and roofs of buildings. They might divide complex shapes into simpler ones, like quadrilaterals, and then into triangles.
• Computer Graphics: In computer graphics, complex shapes are often rendered using triangles (triangulation), and the area calculation is crucial for rendering and other calculations.
• Navigation: Calculating the areas of regions on maps to estimate the area of territories or plan routes.

Fun Fact

Did you know that any polygon can be broken down into triangles? This is a fundamental concept in geometry, and it’s why triangles are so important in various fields like computer graphics and engineering. You can always decompose a polygon into triangles, making area calculations and other geometric analysis much easier.

Recommended YouTube Videos for Deeper Understanding

Q.1 A quadrilateral $ABCD$ has vertices $A(1,1)$, $B(4,2)$, $C(5,5)$, and $D(2,4)$. What is the area of quadrilateral $ABCD$?

Check Solution

Ans: A

Divide the quadrilateral into two triangles, $\triangle ABC$ and $\triangle ADC$. Area of $\triangle ABC = \frac{1}{2} |(1(2-5) + 4(5-1) + 5(1-2))| = \frac{1}{2}|-3+16-5| = \frac{1}{2}|8| = 4$. Area of $\triangle ADC = \frac{1}{2} |(1(5-4) + 5(4-1) + 2(1-5))| = \frac{1}{2}|1+15-8| = \frac{1}{2}|8| = 4$. Area of quadrilateral $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ADC$ = $4 + 4 = 8$.

Q.2 A quadrilateral has vertices at $(0,0)$, $(4,0)$, $(5,3)$, and $(1,3)$. Find the area of the quadrilateral.

Check Solution

Ans: A

Divide the quadrilateral into two triangles, $\triangle ABC$ and $\triangle ADC$. Let the points be A(0,0), B(4,0), C(5,3) and D(1,3). Area of $\triangle ABC = \frac{1}{2}|0(0-3) + 4(3-0) + 5(0-0)| = \frac{1}{2}|12| = 6$. Area of $\triangle ADC = \frac{1}{2}|0(3-3) + 5(3-0) + 1(0-3)| = \frac{1}{2}|15-3| = \frac{1}{2}|12| = 6$. Area of quadrilateral = $6+6 = 12$.

Q.3 The lengths of the diagonals of a rhombus are $10$ cm and $24$ cm. What is the area of the rhombus?

Check Solution

Ans: A

The area of a rhombus can be found by $\frac{1}{2} d_1 d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals. Area = $\frac{1}{2} \times 10 \times 24 = 120$ square cm.

Q.4 A quadrilateral $PQRS$ has diagonals $PR$ and $QS$. The length of $PR$ is $14$ cm, and the perpendicular distance from $Q$ to $PR$ is $6$ cm, and the perpendicular distance from $S$ to $PR$ is $8$ cm. What is the area of quadrilateral $PQRS$?

Check Solution

Ans: A

The area of quadrilateral $PQRS$ is the sum of the areas of the two triangles, $\triangle PQR$ and $\triangle PSR$. Area of $\triangle PQR = \frac{1}{2} \times PR \times h_1 = \frac{1}{2} \times 14 \times 6 = 42$. Area of $\triangle PSR = \frac{1}{2} \times PR \times h_2 = \frac{1}{2} \times 14 \times 8 = 56$. Area of quadrilateral $PQRS$ = $42 + 56 = 98$.

Q.5 If the sides of a quadrilateral are tangent to a circle, and two opposite sides are of length 10 cm and 14 cm, what is the perimeter of the quadrilateral?

Check Solution

Ans: A

A quadrilateral circumscribing a circle has the sum of opposite sides equal. Let the lengths of the sides be $a, b, c, d$. We are given that $a + c = b + d$. Let the sides be $a = 10$ and $c = 14$. Then $b + d = 10+14=24$. Therefore the perimeter is $a+b+c+d = (a+c) + (b+d) = 10+14+24= 2(10+14) = 48$.

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