Question

If sides of a right angled triangle conatining the right angle are 5x and 3x–1. If area of the triangle is 60 then what is the positive value of x?​

Answer 1

$\LARGE{ \underline{ \purple{ \sf{Required \: answer:}}}}$

The sides containing the right angled triangle are perpendicular and the base. And we known that:

• Area of the triangle = 1/2 × Base × Perpendicular

In the question, the measure of these two sides and the area of the triangle is given. From here, we can solve for x.

• Sides are 5x and 3x - 1
• Area of the triangle = 60

━━━━━━━━━━━━━━━━━━━━

Plugging the given values in the formula,

⇛ $\rm{ \dfrac{1}{2} \times 5x \times (3x - 1) = 60}$

Multiplying 2 on the RHS because the inverse of division is multiplication.

⇛ $\rm{5x(3x - 1) = 120}$

Expanding the parentheses,

⇛ $\rm{15 {x}^{2} - 5x = 120}$

⇛ $\rm{15 {x}^{2} - 5x - 120 = 0}$

Dividing the equation by 5,

⇛ $\rm{ 3{x}^{2} - x - 24 = 0}$

Now finding the zeroes by middle term factorisation,

⇛ $\rm{3 {x}^{2} - 9x + 8x - 24 = 0}$

⇛ $\rm{3x(x - 3) + 8(x - 3) = 0}$

Taking x - 3 common,

⇛ $\rm{(3x + 8)(x - 3) = 0}$

So, x = -8/3 or 3. The sides cannot be negative, thus the value for x is 3 (Answer)

Answer 2

Answer:

x=3

Step-by-step explanation:

let the ∆ABC right angled at B.

AB=5x, BC=3x-1

area of ∆ is 1/2×b×h.

1/2×(3x-1)×(5x)=60

15x²-5x=120

15x²-5x-120=0

5(3x²-x-24)=0

3x²-x-24=0

3x²-(9-8)x-24=0

3x²-9x+8x-24=0

3x(x-3)+8(x-3)=0

(3x+8)(x-3)=0

x=-8/3, x=3

Hence the positive value of x is 3

MARK AS BRAINLIST.