Consider the following for the next three (03) items :
What is the area of quadrilateral ABCD?
306 cm 2
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.
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 = \(180 \text{ cm}^2\)
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 = \(41 \text{ cm}\).
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:
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 = \(126 \text{ cm}^2\)
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\(^2\).
| 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) | \(306 \text{ cm}^2\) |
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.
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.
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