The problem asks us to find the area of a triangular plot given its side ratios and perimeter.
Let the sides of the triangle be $3x$, $4x$, and $5x$, where $x$ is a common multiplier.
The perimeter is the sum of these sides:
$ \text{Perimeter} = 3x + 4x + 5x $
$ \text{Perimeter} = 12x $
We are given that the perimeter is 60 m:
$ 12x = 60 \text{ m} $
Solve for $x$:
$ x = \frac{60}{12} $
$ x = 5 \text{ m} $
Now, calculate the actual lengths of the sides:
Since the sides are in the ratio 3:4:5, the triangle is a right-angled triangle. The sides 15 m and 20 m are the base and height (the two shorter sides perpendicular to each other).
The formula for the area of a right-angled triangle is:
$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $
Substitute the values of the base and height:
$ \text{Area} = \frac{1}{2} \times 15 \text{ m} \times 20 \text{ m} $
$ \text{Area} = \frac{1}{2} \times 300 \text{ m}^2 $
$ \text{Area} = 150 \text{ m}^2 $
The area of the triangular plot is 150 m².
The radius of the circumcircle of an equilateral triangle of √3 unit side, is:
If the ratio of the angles of a triangle is 3 : 5 : 7, find the value of the largest angle.
A. 36°
B. 60°
C. 84°
D. 15°
If ΔABC and Δ PQR are similar and \(\rm\frac{BC}{QR} = \frac{1}{3}\) , find \(\rm\frac{ar(\Delta PQR)} {ar(\Delta BCA)}\)
If the angles of a triangle are in the ratio of 2 : 5 : 8, then find the value of the smallest angle.
ABCD is a parallelogram. Side BC is produced to E such that BC = CE. Join AE which intersects side CD at P. The area of triangle ABE is: