If the altitude of an equilateral triangle is √6 cm. Then its area is:
2√3 cm2
Finding the Area of an Equilateral Triangle Given Altitude
Let's find the area of an equilateral triangle when its altitude is given as $\sqrt{6}$ cm. We need to use the relationship between the altitude, side length, and area of an equilateral triangle.
Key Formulas for Equilateral Triangles
- The altitude (h) of an equilateral triangle with side length 'a' is given by the formula: $h = \frac{\sqrt{3}}{2} a$
- The area (A) of an equilateral triangle with side length 'a' is given by the formula: $A = \frac{\sqrt{3}}{4} a^2$
Step-by-Step Calculation
Step 1: Find the side length 'a' using the given altitude
We are given the altitude $h = \sqrt{6}$ cm. Using the altitude formula:
$\sqrt{6} = \frac{\sqrt{3}}{2} a$
To find 'a', we can rearrange the formula:
$a = \frac{2 \times \sqrt{6}}{\sqrt{3}}$
We can simplify $\sqrt{6}$ as $\sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$.
$a = \frac{2 \times \sqrt{2} \times \sqrt{3}}{\sqrt{3}}$
Cancel out $\sqrt{3}$ from the numerator and denominator:
$a = 2 \sqrt{2}$ cm
So, the side length of the equilateral triangle is $2 \sqrt{2}$ cm.
Step 2: Calculate the area using the side length
Now that we have the side length $a = 2 \sqrt{2}$ cm, we can use the area formula for an equilateral triangle:
$A = \frac{\sqrt{3}}{4} a^2$
Substitute the value of 'a':
$A = \frac{\sqrt{3}}{4} (2 \sqrt{2})^2$
Calculate $(2 \sqrt{2})^2$:
$(2 \sqrt{2})^2 = (2)^2 \times (\sqrt{2})^2 = 4 \times 2 = 8$
Substitute this back into the area formula:
$A = \frac{\sqrt{3}}{4} \times 8$
Simplify the expression:
$A = \sqrt{3} \times \frac{8}{4}$
$A = \sqrt{3} \times 2$
$A = 2 \sqrt{3}$ cm$^2$
Thus, the area of the equilateral triangle with an altitude of $\sqrt{6}$ cm is $2 \sqrt{3}$ cm$^2$. This matches one of the given options.
Comparing with Options
Let's look at the given options:
- $2\sqrt{3}$ cm2
- $2\sqrt{2}$ cm2
- $3\sqrt{3}$ cm2
- $3\sqrt{2}$ cm2
Our calculated area is $2 \sqrt{3}$ cm$^2$, which corresponds to the first option.
| Given | Formula Used | Calculation | Result |
|---|---|---|---|
| Altitude $h = \sqrt{6}$ cm | $h = \frac{\sqrt{3}}{2} a$ | $a = \frac{2h}{\sqrt{3}} = \frac{2\sqrt{6}}{\sqrt{3}} = 2\sqrt{2}$ cm | Side $a = 2\sqrt{2}$ cm |
| Side $a = 2\sqrt{2}$ cm | $A = \frac{\sqrt{3}}{4} a^2$ | $A = \frac{\sqrt{3}}{4} (2\sqrt{2})^2 = \frac{\sqrt{3}}{4} \times 8 = 2\sqrt{3}$ cm$^2$ | Area $A = 2\sqrt{3}$ cm$^2$ |
Revision Table: Equilateral Triangle Properties
| Property | Formula (side 'a') |
|---|---|
| Perimeter | $3a$ |
| Altitude (h) | $\frac{\sqrt{3}}{2}a$ |
| Area (A) | $\frac{\sqrt{3}}{4}a^2$ |
| Relationship between Area and Altitude | $A = \frac{h^2}{\sqrt{3}}$ (derived from $h = \frac{\sqrt{3}}{2}a \Rightarrow a = \frac{2h}{\sqrt{3}}$, so $A = \frac{\sqrt{3}}{4} (\frac{2h}{\sqrt{3}})^2 = \frac{\sqrt{3}}{4} \frac{4h^2}{3} = \frac{\sqrt{3}h^2}{3} = \frac{h^2}{\sqrt{3}}$) |
Additional Information on Triangle Calculations
An equilateral triangle is a special type of triangle where all three sides are equal in length, and all three interior angles are equal, each measuring 60 degrees. Because of its symmetry, the altitude from any vertex bisects the opposite side and also bisects the angle at that vertex. The altitude, median, and angle bisector from a vertex are all the same line segment in an equilateral triangle.
Understanding the formulas for altitude and area is crucial for solving problems involving equilateral triangles. The Pythagorean theorem is implicitly used in deriving the altitude formula. If you drop an altitude, it forms two right-angled triangles with hypotenuse 'a' and one leg 'a/2'. The altitude 'h' is the other leg: $h^2 + (\frac{a}{2})^2 = a^2$, which simplifies to $h = \frac{\sqrt{3}}{2} a$.
Being able to quickly derive or recall these formulas helps in solving problems efficiently in exams.
Important Questions from Triangles
What is the circumcenter of the triangle ABC?
What is the centroid of the triangle ABC?
What is the foot of the altitude from the vertex A of the triangle ABC?
In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?
In ΔABC, ∠A = 66° and ∠B = 50 °. If the bisectors of ∠B and ∠C meet at P, then ∠BPC – ∠PCA = ?
