The longest side of the obtuse triangle is 7 cm and the other two sides of the triangle are 4 cm and 5 cm. Find the area of the triangle.
Calculating the Area of an Obtuse Triangle
The question asks us to find the area of an obtuse triangle with side lengths of 4 cm, 5 cm, and 7 cm. We are given that the longest side is 7 cm.
Identifying an Obtuse Triangle
Before calculating the area, let's verify if this triangle is indeed obtuse. We can use the side lengths and the property relating the square of the longest side to the sum of the squares of the other two sides.
- Let the sides be \(a=4\) cm, \(b=5\) cm, and \(c=7\) cm. Side \(c\) is the longest side.
- Calculate \(a^2 + b^2\): \(4^2 + 5^2 = 16 + 25 = 41\).
- Calculate \(c^2\): \(7^2 = 49\).
- Compare: Since \(c^2 > a^2 + b^2\) (\(49 > 41\)), the angle opposite the longest side (7 cm) is obtuse. Thus, it is an obtuse triangle.
Using Heron's Formula for Triangle Area
When we know the lengths of all three sides of a triangle, we can use Heron's formula to find its area. Heron's formula is particularly useful when the height is not given.
Heron's formula states that the area \(A\) of a triangle with sides \(a, b, c\) is:
\(A = \sqrt{s(s-a)(s-b)(s-c)}\)
where \(s\) is the semi-perimeter of the triangle, calculated as:
\(s = \frac{a+b+c}{2}\)
Step-by-Step Calculation of Obtuse Triangle Area
Let's apply Heron's formula to find the area of the obtuse triangle with sides 4 cm, 5 cm, and 7 cm.
- Calculate the semi-perimeter (s):
\(s = \frac{4 + 5 + 7}{2} = \frac{16}{2} = 8 \text{ cm}\)
- Calculate the differences \(s-a\), \(s-b\), and \(s-c\):
- \(s-a = 8 - 4 = 4 \text{ cm}\)
- \(s-b = 8 - 5 = 3 \text{ cm}\)
- \(s-c = 8 - 7 = 1 \text{ cm}\)
- Substitute these values into Heron's formula:
\(A = \sqrt{s(s-a)(s-b)(s-c)}\)
\(A = \sqrt{8 \times 4 \times 3 \times 1}\)
\(A = \sqrt{96}\)
- Simplify the square root:
To simplify \(\sqrt{96}\), we look for perfect square factors of 96.
\(96 = 16 \times 6\)
\(A = \sqrt{16 \times 6}\)
\(A = \sqrt{16} \times \sqrt{6}\)
\(A = 4\sqrt{6} \text{ cm}^2\)
The area of the obtuse triangle is \(4\sqrt{6}\) cm2.
Comparing with Options
The calculated area \(4\sqrt{6}\) cm2 matches one of the given options.
| Calculated Area | Option 1 | Option 2 | Option 3 | Option 4 |
|---|---|---|---|---|
| \(4\sqrt{6}\) cm2 | \(1\sqrt{3}\) cm2 | \(6\sqrt{3}\) cm2 | \(3\sqrt{2}\) cm2 | \(4\sqrt{6}\) cm2 |
The calculated area matches Option 4.
Revision Table: Key Concepts for Triangle Area
| Concept | Description | Formula | When to Use |
|---|---|---|---|
| Area (Base and Height) | Half the product of the base and its corresponding height. | \(A = \frac{1}{2} \times \text{base} \times \text{height}\) | When base and height are known. |
| Area (Two Sides and Included Angle) | Half the product of two sides and the sine of the included angle. | \(A = \frac{1}{2}ab\sin{C}\) | When two sides and the angle between them are known. |
| Heron's Formula | Area calculated using only the lengths of the three sides. | \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac{a+b+c}{2}\) | When all three side lengths are known. |
| Right Triangle Area | Half the product of the two perpendicular sides (legs). | \(A = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2\) | Special case of base and height formula for right triangles. |
Additional Information: Properties of Obtuse Triangles
An obtuse triangle is a triangle in which one of the angles is greater than 90 degrees. Here are some key properties:
- Only one angle can be obtuse. The other two angles must be acute (less than 90 degrees).
- The side opposite the obtuse angle is the longest side of the triangle. This is a direct consequence of the Law of Sines and the relationship between angle size and opposite side length.
- In an obtuse triangle with sides \(a, b\), and longest side \(c\), the relationship \(c^2 > a^2 + b^2\) holds true. This is derived from the Law of Cosines (\(c^2 = a^2 + b^2 - 2ab\cos{C}\)). If \(C\) is obtuse, \(\cos{C}\) is negative, making \(-2ab\cos{C}\) positive, so \(c^2\) is greater than \(a^2 + b^2\).
- The orthocenter (intersection of altitudes) of an obtuse triangle lies outside the triangle.
- The circumcenter (center of the circumscribed circle) of an obtuse triangle lies outside the triangle.
Knowing these properties helps in understanding the geometry of obtuse triangles.
Important Questions from Plane Figures
If length of a rectangle is increased to its three times and breadth is decreased to its half, then the ratio of the area of given rectangle to the area of new rectangle is:
The width of the path around a square field is 4.5 m and its area is 105.75 m 2. Find the cost of fencing the field at the rate of Rs. 100 per meter.
What is the area of the square (in cm 2) whose vertices lie on a circle of radius 5 cm?
The circumcentre of an equilateral triangle is at a distance of 3.2 cm from the base of the triangle. What is the length (in cm) of each of its altitudes?
The perimeter of a circular lawn is 1232 m. There is 7 m wide path around the lawn. The area (in m 2) of the path is:
Take \(\left(\pi=\frac{22}{7}\right)\)
