Calculating Equilateral Triangle Area from Perimeter
The problem asks for the area of an equilateral triangle given its perimeter.
Finding the Side Length
- An equilateral triangle has three equal sides.
- Let the side length be denoted by '$s$'.
- The perimeter '$P$' is given as 22.2 cm.
- The formula for the perimeter is $P = 3s$.
- Substituting the given perimeter: $22.2 \text{ cm} = 3s$.
- Solving for '$s$': $s = \frac{22.2 \text{ cm}}{3} = 7.4 \text{ cm}$.
Calculating the Area
- The formula for the area '$A$' of an equilateral triangle is $A = \frac{\sqrt{3}}{4} s^2$.
- Substitute the calculated side length '$s = 7.4$ cm' into the formula:
$A = \frac{\sqrt{3}}{4} (7.4 \text{ cm})^2$
- Calculate the square of the side length: $(7.4)^2 = 54.76$.
- Now calculate the area:
$A = \frac{\sqrt{3}}{4} \times 54.76 \text{ cm}^2$
$A = \frac{54.76}{4} \sqrt{3} \text{ cm}^2$
$A = 13.69\sqrt{3} \text{ cm}^2$
Conclusion
The calculated area of the equilateral triangle is $13.69\sqrt{3}$ $cm^2$, which corresponds to Option A.
Similar Questions
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- In $\Delta ABC$, $BD \perp AC$ at D and $\angle DBC = 71^\circ$. E is a point on BC such that $\angle CAE = 17^\circ$. What is the measure of $\angle AEB$?
- In triangle ABC, bisector of $\angle\text{ABC}$ and $\angle\text{ACB}$ meet at O. If $\angle\text{BAC} = 60^\circ$, then find the measure of $\angle\text{BOC}$.
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- In $\Delta\text{PQR}$, $\text{QR}$ is extended up to $\text{S}$ so that $\text{RS} = \text{RP}$. If $\angle\text{PRQ} = 70^\circ$ and $\angle\text{QPS} = 110^\circ$ then find the measure of $\angle\text{PQS}$.
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- In a triangle, if angle $A = 30^\circ$ and angle $B = 45^\circ$, then what is the angle of C?
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 = ?
