Solving for Area Difference in Right-Angled Triangle ABC
We are given a right-angled triangle ABC with side lengths AB = 15 cm, BC = 20 cm, and AC = 25 cm. We need to find the difference between the areas of triangles PAB and PCB, where BP is the perpendicular drawn from vertex B to the hypotenuse AC.
Step 1: Verifying the Right Angle
First, let's confirm that the triangle is right-angled using the Pythagorean theorem: \(a^2 + b^2 = c^2\). Here, the sides are 15 cm, 20 cm, and 25 cm. Let's check if the sum of the squares of the two shorter sides equals the square of the longest side:
- \( AB^2 = 15^2 = 225 \)
- \( BC^2 = 20^2 = 400 \)
- \( AC^2 = 25^2 = 625 \)
Since \( AB^2 + BC^2 = 225 + 400 = 625 \), which is equal to \( AC^2 \), the triangle ABC is indeed a right-angled triangle with the right angle at vertex B.
Step 2: Calculating the Area of Triangle ABC
The area of a right-angled triangle is given by \( \frac{1}{2} \times \text{base} \times \text{height} \). Using the sides adjacent to the right angle (AB and BC):
Area(ABC) = \( \frac{1}{2} \times AB \times BC \)
Area(ABC) = \( \frac{1}{2} \times 15 \text{ cm} \times 20 \text{ cm} = 150 \text{ square cm} \)
Step 3: Calculating the Length of Perpendicular BP
The area of triangle ABC can also be calculated using the hypotenuse (AC) as the base and BP as the height (since BP is perpendicular to AC):
Area(ABC) = \( \frac{1}{2} \times AC \times BP \)
We know the area is 150 sq cm and AC is 25 cm. So,
\( 150 = \frac{1}{2} \times 25 \times BP \)
\( 300 = 25 \times BP \)
BP = \( \frac{300}{25} = 12 \) cm
Step 4: Calculating Lengths AP and PC
Now we have two smaller right-angled triangles, PAB and PCB, formed by the altitude BP.
Using Triangle PAB:
Triangle PAB is right-angled at P. We know AB = 15 cm and BP = 12 cm.
Using the Pythagorean theorem in triangle PAB:
\( AP^2 + BP^2 = AB^2 \)
\( AP^2 + 12^2 = 15^2 \)
\( AP^2 + 144 = 225 \)
\( AP^2 = 225 - 144 = 81 \)
AP = \( \sqrt{81} = 9 \) cm
Calculating PC:
Since P lies on AC, we have AC = AP + PC.
\( 25 = 9 + PC \)
PC = \( 25 - 9 = 16 \) cm
Step 5: Calculating the Areas of Triangles PAB and PCB
Area of Triangle PAB:
Base = AP = 9 cm, Height = BP = 12 cm
Area(PAB) = \( \frac{1}{2} \times AP \times BP \)
Area(PAB) = \( \frac{1}{2} \times 9 \text{ cm} \times 12 \text{ cm} = 54 \text{ square cm} \)
Area of Triangle PCB:
Base = PC = 16 cm, Height = BP = 12 cm
Area(PCB) = \( \frac{1}{2} \times PC \times BP \)
Area(PCB) = \( \frac{1}{2} \times 16 \text{ cm} \times 12 \text{ cm} = 96 \text{ square cm} \)
Step 6: Finding the Difference in Areas
The question asks for the difference in the areas of triangles PAB and PCB.
Difference = | Area(PAB) - Area(PCB) |
Difference = | 54 sq cm - 96 sq cm |
Difference = | -42 sq cm |
Difference = 42 square cm
Therefore, the difference in the area of triangles PAB and PCB is 42 square cm.
Similar Questions
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I. The quadrilateral AZXY is a parallelogram.
II. The area of the quadrilateral AZXY is half of the area of the triangle ABC.
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