Question

Can anyone solve this question ..................best answer will be marked as a brainlist.

An altitude of a triangle is 5/3 times the length of its corresponding base. If the altitude is increased by 4 cm and the base decreased by 2 cm, the area of the triangle remains the same. Find the altitude and the base of the triangle.

Answer 1

Here is your answer.
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Answer 2

let the length of base be x
and the length of altitude be y.
now according to question
y = (5/3)x = 5x/3 ____(1)
now if the altitude is increased by 4 cm then the new attitude will be = 5x/3 + 4
and the new base is increased by 2 centimetre = x - 2
initial area = (1/2)×b×h = 1/2 × X × (5x/3)
= 5x²/6 ____(2)
and final area = 1/2 × (x-2) × (5x/3 + 4)
= ((X-2)(5x + 12))/6
= (5x²+2x-24)/6 ____(3)
Now since the area is same.
so equation (2) and (3)are equal
→ 5x²/6 = (5x²+2x-24)/6
→ 0 = 2x-24
→ X = 24/2 = 12
so the length of base = 12 cm
and the length of altitude = (5(12))/3
= 20 cm


and now your second question
first let the side of first square be 'x'.
and the side of second square be 'y'
now going to question
x = 2y + 3 ___(1)
and your second equation is
x² - y² = 144 ___(2)
solve equation 1 and 2 we get:
(2y + 3)² - y² = 144
4y² + 9 + 12y - y² = 144
3y² + 12y -135 = 0
y² + 4y - 45 = 0
y² + 9y - 5y - 45 = 0
y(y+9) -5(y+9) = 0
(y-5) (y+9) = 0
y = 5 or 9
and from equation (1),we get x = 13 or 57