Question

The height of the triangle is 7 cm longer than its base. the area of the triangle is 60 cm squared. what is the base of the triangle?

Answer 1

Answer:

8 cm

Step-by-step explanation:

Let x cm be the length of the base, then the length of the height is x+7 cm. Use formula for the area of the triangle:

$A_{triangle}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}$

In your case,

$60=\dfrac{1}{2}\cdot x\cdot (x+7),\\ \\120=x(x+7),\\ \\x^2+7x-120=0,\\ \\D=7^2 -4\cdot 1\cdot (-120)=49+480=529,\\ \\x_{1,2}=\dfrac{-7\pm\sqrt{529}}{2\cdot 1}=\dfrac{-7\pm 23}{2}=-15,\ 8$

Since the base cannot have negative length, then x=8 cm.

Answer 2

Answer:

Base = 8 cm

Step-by-step explanation:

We are given that the height of a triangle is 7 cm longer than its base and its area is 60 cm squared.

We are to find the base of the triangle.

Area of the triangle = $\frac{1}{2} \times base \times height$

Assuming the base to be $x$, so height will be $x+7$

Substituting the values in the above formula to get:

$60=\frac{1}{2} \times x \times (x+7)$

$120= x(x+7)$

$x^2+7x-120=0$

Factorizing the quadratic equation to get:

$x^2+15x-8x-120=0$

$x(x+15)-8(x+15)=0$

$(x-8)(x+15)=0$

$x = 8$ and $x = -15$

Since base cannot be negative so it will be 8 cm.