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.
Angle between the internal bisectors of two angles ∠B and ∠C of a ΔABC is 132°, then the value of ∠A is
In ΔPQR, PQ = PR and S is a point on QR such that ∠PSQ = 96° + ∠QPS and ∠QPR = 132°. What is the measure of ∠PSR?
In Δ ABC, ∠A = 50°. If the bisectors of the angle B and angle C, meet at a point O, then ∠BOC is equal to:
Triangle ABC is right angled at B. BD is an altitude intersecting AC at D. If AC = 9 cm and CD = 3 cm. then find the measure of AB (in cm).
The base and altitude of an isosceles triangle are 10 cm and 12 cm respectively. Then the length of each equal side is: