We are given the sides of a triangle: 44 cm, 33 cm, and 55 cm. To find the area efficiently, let's first determine if it's a special type of triangle, like a right-angled triangle. We can use the Pythagorean theorem ($a^2 + b^2 = c^2$) for this check.
The area of a right-angled triangle is given by the formula:
$Area = 1/2 * base * height$
In a right-angled triangle, the two shorter sides (legs) serve as the base and height.
Therefore, the area of the triangle is 726 cm$^2$.
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?