A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm.
Answers
SOLUTION :
From the figure, In ∆ABC ,Let, AB = 5cm, BC = 12 cm and AC = 13 cm. Then, AC² = AB² + BC²
[By using Pythagoras theorem]
13² = 5² + 12²
169 = 25 + 144
169 = 169
This proves that ∆ABC is a right triangle, right angle at B. Let BD be the length of perpendicular from B on AC.
Now, Area ∆ABC = ½ × BC ×AB
[Area of ∆ = ½ × base × height]
= 1/2× (12 × 5)
= 6×5
Area of ∆ABC= 30 cm²
Also, Area of ∆ABC = 1/2× AC × BD
30 = ½ × 13 × BD
30 × 2 = 13 × BD
13 BD = 60
BD = 60/13 cm
BD = 4.6 cm
Hence, the length of the perpendicular from the opposite vertex to the side whose length is 13 cm is 4.6 cm (to 1 decimal place).
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By using Pythagoras theorem
13² = 5² + 12²
169 = 25 + 144
169 = 169
This proves that ∆ABC is a right triangle, right angle at B. Let BD be the length of perpendicular from B on AC.
Now, Area ∆ABC = ½ × BC ×AB
[Area of ∆ = ½ × base × height]
= 1/2× (12 × 5)
= 6×5
Area of ∆ABC= 30 cm²
Also, Area of ∆ABC = 1/2× AC × BD
30 = ½ × 13 × BD
30 × 2 = 13 × BD
13 BD = 60
BD = 60/13 cm
BD = 4.6 cm