an altitude of triangle is 5/3rd of the length of its corresponding base if their altitude be increased by 4cm and the base is decreased by 2cm, the area of the triangles remain the same. find the base and altitude of triangle ?????
math
let base=x
altitude=5x/3
area= (altitude*base)/2
area of original triangle= (5x^2/3)/2
according to the question
new base=x-2
new altitude=(5x 12)/3
now
area of original triangle=area of new triangle
5x^2/3=5x^2/3-10x/3 4x-8 (1/2 is cancelled from both sides)
now after solving we get
x=12 that is base
and altitude= 20
Let's assume the base of the triangle as "b" and the altitude as "h". We are given that the altitude of the triangle is 5/3 of the length of its corresponding base. Therefore, we can write:
h = (5/3)b
Now, we are also given that if the altitude of the triangle is increased by 4 cm (h + 4) and the base is decreased by 2 cm (b - 2), the area of the triangle remains the same.
The area of a triangle is given by the formula:
Area = (1/2) * base * altitude
Using this formula, we can write the equation for the area of the original triangle (A1) and the area of the modified triangle (A2):
A1 = (1/2) * b * h
A2 = (1/2) * (b - 2) * (h + 4)
Since the areas remain the same, we can set A1 = A2:
(1/2) * b * h = (1/2) * (b - 2) * (h + 4)
Now, substitute the value of h from the first equation into the second equation:
(1/2) * b * (5/3)b = (1/2) * (b - 2) * ((5/3)b + 4)
Simplify the equation:
(5/6)b^2 = (1/2) * (b - 2) * ((5/3)b + 4)
Multiply both sides of the equation by 6 to get rid of the fractions:
5b^2 = 3 * (b - 2) * ((5/3)b + 4)
Expand the equation further:
5b^2 = 3 * (5b^2 + 4b - 10b - 8)
5b^2 = 3 * (5b^2 - 6b - 8)
5b^2 = 15b^2 - 18b - 24
Combine like terms:
10b^2 +18b + 24 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula.
However, upon solving the equation, we find that it does not yield real solutions. Therefore, we can conclude that there is no base and altitude that satisfy the given conditions of the problem.
To find the base and altitude of the triangle, we can use the formula for the area of a triangle, which is given by:
Area = (1/2) * base * altitude
Let's denote the base of the triangle as "b" and the altitude as "h". According to the given information, we have two conditions:
Condition 1:
The altitude of the triangle is 5/3 times the length of its corresponding base. We can write this as:
h = (5/3) * b
Condition 2:
When the altitude is increased by 4 cm and the base is decreased by 2 cm, the area of the triangle remains the same. If we let the new altitude be "h + 4" and the new base be "b - 2", then the area remains constant. Mathematically, we can write this as:
(1/2) * (b - 2) * (h + 4) = (1/2) * b * h
Let's substitute the value of "h" from Condition 1 into Condition 2 and solve for "b":
(1/2) * (b - 2) * ((5/3) * b + 4) = (1/2) * b * ((5/3) * b)
To simplify, we can multiply through by 6 to get rid of the fractions:
(b - 2) * (5b + 12) = 5b^2
5b^2 + 12b - 10b - 24 = 5b^2
Simplifying further:
2b = 24
b = 12
Now, substitute the value of "b" back into Condition 1 to find the value of "h":
h = (5/3) * 12
h = 20
Therefore, the base of the triangle is 12 units and the altitude is 20 units.