What is the area of quadrilateral ABCD?

Question

Prepp Team · Accepted Answer

Calculating the Area of Quadrilateral ABCD

The problem asks us to find the area of a specific quadrilateral ABCD. We are given the lengths of its sides: AB = 9 cm, BC = 40 cm, CD = 28 cm, and DA = 15 cm. We are also told that angle ABC is a right angle, which means it measures 90 degrees.

To find the area of the quadrilateral, we can divide it into two triangles by drawing a diagonal. Since angle ABC is a right angle, drawing the diagonal AC seems logical because it creates a right-angled triangle ABC.

So, the area of quadrilateral ABCD will be the sum of the areas of triangle ABC and triangle ADC.

Step 1: Calculate the Area of Triangle ABC

Triangle ABC is a right-angled triangle with the right angle at B. The sides AB and BC are the base and height of this triangle.

The area of a right-angled triangle is given by the formula:

Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)

Using AB and BC as base and height:

Area of \(\triangle\) ABC = \(\frac{1}{2} \times \text{AB} \times \text{BC}\)

Area of \(\triangle\) ABC = \(\frac{1}{2} \times 9 \text{ cm} \times 40 \text{ cm}\)

Area of \(\triangle\) ABC = \(\frac{1}{2} \times 360 \text{ cm}^2\)

Area of \(\triangle\) ABC = \(180 \text{ cm}^2\)

Step 2: Calculate the Length of Diagonal AC

We need the length of the diagonal AC to find the area of triangle ADC. Since triangle ABC is a right-angled triangle, we can use the Pythagorean theorem to find the length of the hypotenuse AC.

According to the Pythagorean theorem:

AC\(^2\) = AB\(^2\) + BC\(^2\)

AC\(^2\) = \((9 \text{ cm})^2\) + \((40 \text{ cm})^2\)

AC\(^2\) = \(81 \text{ cm}^2\) + \(1600 \text{ cm}^2\)

AC\(^2\) = \(1681 \text{ cm}^2\)

AC = \(\sqrt{1681} \text{ cm}\)

To find the square root of 1681, we can test numbers. \(40^2 = 1600\). Let's try \(41^2\). \(41 \times 41 = 1681\).

So, AC = \(41 \text{ cm}\).

Step 3: Calculate the Area of Triangle ADC

Now we have triangle ADC with side lengths AD = 15 cm, CD = 28 cm, and AC = 41 cm. This is a general triangle, and we can calculate its area using Heron's formula.

Heron's formula for the area of a triangle with sides a, b, and c is \(\sqrt{s(s-a)(s-b)(s-c)}\), where s is the semi-perimeter of the triangle, calculated as \(s = \frac{a+b+c}{2}\).

For triangle ADC, let a = AD = 15 cm, b = CD = 28 cm, and c = AC = 41 cm.

First, calculate the semi-perimeter s:

s = \(\frac{15 + 28 + 41}{2} \text{ cm}\)

s = \(\frac{84}{2} \text{ cm}\)

s = \(42 \text{ cm}\)

Now, apply Heron's formula:

Area of \(\triangle\) ADC = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Area of \(\triangle\) ADC = \(\sqrt{42(42-15)(42-28)(42-41)} \text{ cm}^2\)

Area of \(\triangle\) ADC = \(\sqrt{42(27)(14)(1)} \text{ cm}^2\)

Let's simplify the terms inside the square root:

\(42 = 2 \times 3 \times 7\)
\(27 = 3 \times 3 \times 3 = 3^3\)
\(14 = 2 \times 7\)
\(1 = 1\)

Area of \(\triangle\) ADC = \(\sqrt{(2 \times 3 \times 7) \times (3^3) \times (2 \times 7) \times 1} \text{ cm}^2\)

Group the prime factors:

Area of \(\triangle\) ADC = \(\sqrt{2^2 \times 3^{(1+3)} \times 7^2} \text{ cm}^2\)

Area of \(\triangle\) ADC = \(\sqrt{2^2 \times 3^4 \times 7^2} \text{ cm}^2\)

Take the square root by halving the exponents:

Area of \(\triangle\) ADC = \(2^{\frac{2}{2}} \times 3^{\frac{4}{2}} \times 7^{\frac{2}{2}} \text{ cm}^2\)

Area of \(\triangle\) ADC = \(2^1 \times 3^2 \times 7^1 \text{ cm}^2\)

Area of \(\triangle\) ADC = \(2 \times 9 \times 7 \text{ cm}^2\)

Area of \(\triangle\) ADC = \(18 \times 7 \text{ cm}^2\)

Area of \(\triangle\) ADC = \(126 \text{ cm}^2\)

Step 4: Calculate the Total Area of Quadrilateral ABCD

The area of quadrilateral ABCD is the sum of the areas of the two triangles it was divided into:

Area of Quadrilateral ABCD = Area of \(\triangle\) ABC + Area of \(\triangle\) ADC

Area of Quadrilateral ABCD = \(180 \text{ cm}^2 + 126 \text{ cm}^2\)

Area of Quadrilateral ABCD = \(306 \text{ cm}^2\)

Thus, the area of quadrilateral ABCD is 306 cm\(^2\).

Revision Table: Key Steps for Quadrilateral Area

Step	Description	Formula/Method Used	Result
1	Divide quadrilateral into triangles using diagonal AC.	Conceptual division	\(\triangle\) ABC, \(\triangle\) ADC
2	Calculate Area of \(\triangle\) ABC (Right-angled).	Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)	\(180 \text{ cm}^2\)
3	Calculate length of diagonal AC.	Pythagorean Theorem (\(a^2+b^2=c^2\))	\(41 \text{ cm}\)
4	Calculate Area of \(\triangle\) ADC (General Triangle).	Heron's Formula (\(\sqrt{s(s-a)(s-b)(s-c)}\))	\(126 \text{ cm}^2\)
5	Sum areas of \(\triangle\) ABC and \(\triangle\) ADC.	Total Area = Area(\(\triangle\) ABC) + Area(\(\triangle\) ADC)	\(306 \text{ cm}^2\)

Additional Information on Quadrilateral Area Calculation

A quadrilateral is a polygon with four sides and four vertices. There are many types of quadrilaterals, such as squares, rectangles, parallelograms, trapezoids, rhombuses, and kites. The method for calculating the area of a quadrilateral depends on its type and the information available.

Simple Quadrilaterals: For basic shapes like squares or rectangles, specific formulas using side lengths are used (e.g., side\(^2\) for a square, length \(\times\) width for a rectangle).
Parallelograms: Area = base \(\times\) height.
Trapezoids: Area = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\).
General Quadrilaterals: If a quadrilateral doesn't fit a specific type, its area can often be found by dividing it into two triangles using a diagonal. The area of each triangle is then calculated (using base and height if possible, or Heron's formula if side lengths are known), and the areas are added together.
Using Diagonals: If the lengths of the two diagonals (p and q) and the angle (\(\theta\)) between them are known, the area of a general convex quadrilateral can be found using the formula: Area = \(\frac{1}{2} pq \sin(\theta)\).

In this problem, dividing the quadrilateral into two triangles was necessary because it's not a standard type like a rectangle or parallelogram where simple formulas apply directly to the given sides and angle.