Question

The sides of a triangle are 9 cm ,40 cm and 41 cm . find its perimeter and its area . also find the length of the altitude drawn on the side length 41 cm.​

Answer 1

Given:

• The sides of a triangle are 9 cm, 40 cm and 41 cm.

To find:

• Perimeter of a triangle.
• Area of a triangle.
• Length of the altitude drawn on the side length 41 cm.

Solution:

• Perimeter of a triangle = sum of all the three sides of a triangle.
• Perimeter of a triangle = 9 + 40 + 41
• Perimeter of a triangle = 90

Let,

a = 9 cm

b = 40 cm

c = 41 cm

Semi-perimeter ( S )

• S = ( a + b + c )/2
• S = (9 + 40 + 41)/2
• S = 90/2
• S = 45

• Area of triangle = √s(s - a ) ( s - b ) ( s - c )
• Area of triangle = √45 ( 45 - 9 ) ( 45 - 40 ) ( 45 - 41 )
• Area of triangle = √45 × 36 × 5 × 4
• Area of triangle = √9 × 5 × 6 × 6 × 5 × 2 × 2
• Area of triangle = √3 × 3 × 5 × 2 × 3 × 2 × 3 × 5 × 2 × 2
• Area of triangle = √3 × 3 × 2 × 2 × 3 × 3 × 5 × 5 × 2 × 2
• Area of triangle = √3 × 2 × 3 × 5 × 2
• Area of triangle = √180
• Area of triangle = 13.42 cm²

Finally, find out altitude...

It is perpendicular to the side of length 41 cm.

• DC = 1/2 AC
• DC = 41/2
• DC = 20.5 cm

By using Pythagoras theorem,

• BC² = DC² + BD²
• 40² = 20.5² + BD²
• BD² = 1600 - 420.25
• BD² = 1179.75
• BD = √1179.75
• BD ≈ 34.35 cm

∴ The Length of the altitude drawn on the side length 41 cm = 34.35 cm

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