The problem asks for the area of an equilateral triangle given its perimeter.
$A = \frac{\sqrt{3}}{4} (7.4 \text{ cm})^2$
$A = \frac{\sqrt{3}}{4} \times 54.76 \text{ cm}^2$
$A = \frac{54.76}{4} \sqrt{3} \text{ cm}^2$
$A = 13.69\sqrt{3} \text{ cm}^2$
The calculated area of the equilateral triangle is $13.69\sqrt{3}$ $cm^2$, which corresponds to Option A.
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 = ?
In a triangle ABC, points P and Q are on AB and AC, respectively, such that AP = 4 cm, PB = 6 cm, AQ = 5 cm and QC = 7.5 cm. If PQ = 6 cm, then find BC (in cm).
The perimeters of two similar ΔABC and Δ PQR are 48.4 cm and 12.1 cm, respectively. What is the ratio of the areas of Δ ABC and Δ PQR?