Understanding the Right Triangle Problem
We are given a triangle named ABC, which has a right angle at vertex B. A point D is located on the hypotenuse AC such that the line segment BD is perpendicular to AC. This means BD is the altitude from the right angle to the hypotenuse. We are provided with the lengths of the two legs of the triangle:
- AB = \(p\)
- BC = \(\sqrt{3}p\)
Our goal is to find the length of the altitude BD.
Calculating the Hypotenuse AC
Since triangle ABC is a right-angled triangle at B, we can use the Pythagorean theorem to find the length of the hypotenuse AC. The theorem states that the square of the hypotenuse (AC) is equal to the sum of the squares of the other two sides (AB and BC).
Applying the theorem: \(AC^2 = AB^2 + BC^2\) Substitute the given values: \(AC^2 = (p)^2 + (\sqrt{3}p)^2\) \(AC^2 = p^2 + 3p^2\) \(AC^2 = 4p^2\) Now, take the square root of both sides to find AC: \(AC = \sqrt{4p^2}\) \(AC = 2p\)
Finding the Altitude BD using Triangle Area
There are two ways to calculate the area of triangle ABC:
- Using the two legs (AB and BC) as base and height.
- Using the hypotenuse (AC) as the base and the altitude (BD) as the height.
Let's calculate the area using the first method: Area \(= \frac{1}{2} \times \text{base} \times \text{height}\) Area \(= \frac{1}{2} \times AB \times BC\) Area \(= \frac{1}{2} \times p \times (\sqrt{3}p)\) Area \(= \frac{\sqrt{3}p^2}{2}\)
Now, let's use the second method, equating it to the area calculated above: Area \(= \frac{1}{2} \times AC \times BD\) \(\frac{\sqrt{3}p^2}{2} = \frac{1}{2} \times (2p) \times BD\)
To find BD, we can rearrange the equation: \(\frac{\sqrt{3}p^2}{2} = p \times BD\) \(BD = \frac{\sqrt{3}p^2}{2p}\) Simplify the expression by canceling out one \(p\): \(BD = \frac{\sqrt{3}p}{2}\)
Alternative Method: Similar Triangles
When an altitude is drawn from the right angle to the hypotenuse in a right-angled triangle, it divides the triangle into two smaller triangles that are similar to the original triangle and to each other. So, we have \(\triangle ABC \sim \triangle ADB \sim \triangle BDC\).
Let's use the similarity between \(\triangle ABC\) and \(\triangle ADB\): The ratio of corresponding sides are equal. \(\frac{AB}{AC} = \frac{BD}{BC}\) Substitute the known values: \(\frac{p}{2p} = \frac{BD}{\sqrt{3}p}\) \(\frac{1}{2} = \frac{BD}{\sqrt{3}p}\) Now, solve for BD: \(BD = \frac{1}{2} \times \sqrt{3}p\) \(BD = \frac{\sqrt{3}p}{2}\)
Both methods yield the same result for the length of the altitude BD.
Conclusion
The length of the altitude BD is \(\frac{\sqrt{3}p}{2}\).
Similar Questions
What is \(AB + BC\) equal to?
Let X, Y and Z be the midpoints of the sides BC, CA and AB of a triangle ABC respectively. Consider the following statements:
I. The quadrilateral AZXY is a parallelogram.
II. The area of the quadrilateral AZXY is half of the area of the triangle ABC.
Which of the statements given above is/are correct?- The sides of a triangle are 11 cm, 60 cm and 61 cm. What is the area of the triangle formed by joining the mid- points of the sides of the triangle?
- In a right-angled triangle ABC, AB = 15 cm, BC = 20 cm and AC = 25 cm. Further, BP is the perpendicular on AC. What is the difference in the area of triangles PAB and PCB?
- In a triangle ABC, \(\angle A = 30^\circ\), AB = 7 cm and AC = 12 cm. What is the area of the triangle ABC?
Important Questions from Triangles
Among the following options, which are NOT sides of a triangle?
In a Δ ABC, if ∠A = 120° and AB = AC, then the values of ∠B and ∠C are respectively:
In the equilateral Δ ABC, the base BC is trisected at D and E. The line through D, Parallel to AB, meets AC at F and the line through E parallel to AC meets AB at G. If EG and DF intersect at H, then what is the ratio of the sum of the area of parallelogram AGHF and the area of the Δ DHE to the area of the Δ ABC?
The product of the perimeter of a triangle, the radius of its in‐circle, and a number gives the area of the triangle. The number is
A man goes 24 m towards east and then 10 m towards north. How far is he away from his initial position?
