The question asks for the area of a triangle given its side lengths and inradius.
First, check if the triangle is a right-angled triangle using the Pythagorean theorem ($a^2 + b^2 = c^2$):
$5^2 + 12^2 = 25 + 144 = 169$ $13^2 = 169$Since $5^2 + 12^2 = 13^2$, the triangle is a right-angled triangle with legs $5\text{ cm}$ and $12\text{ cm}$.
The area of a right-angled triangle can be calculated as half the product of its legs:
$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $ $ \text{Area} = \frac{1}{2} \times 5\text{ cm} \times 12\text{ cm} = 30\text{ cm}^2 $Alternatively, the area of any triangle can be found using the formula $\text{Area} = r \times s$, where $r$ is the inradius and $s$ is the semi-perimeter.
Calculate the semi-perimeter ($s$):
$ s = \frac{a+b+c}{2} = \frac{5\text{ cm} + 12\text{ cm} + 13\text{ cm}}{2} = \frac{30\text{ cm}}{2} = 15\text{ cm} $Now, calculate the area:
$ \text{Area} = r \times s = 2\text{ cm} \times 15\text{ cm} = 30\text{ cm}^2 $Both methods confirm that the area of the triangle is $30\text{ cm}^2$. The given inradius value is consistent with the triangle's dimensions.
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.