Respuesta :
Area of a triangle=1/2bh
h=b-5
42=1/2*b*(b-5)
42=1/2*b^2-5b
84=b^2-5b
b^2-5b-84=0
b^2+7b-12b-84=0
b(b+7)-12(b+7)=0
b=12,b=-7
so width cant be negative so base =12 m
H=12-5=7 m
h=b-5
42=1/2*b*(b-5)
42=1/2*b^2-5b
84=b^2-5b
b^2-5b-84=0
b^2+7b-12b-84=0
b(b+7)-12(b+7)=0
b=12,b=-7
so width cant be negative so base =12 m
H=12-5=7 m
Answer:
The length of the base is 12 m
Step-by-step explanation:
Given : The height of a triangle is 5 m less than its base. The area of the triangle is 42 m².
We have to find the length of the base.
Given the height of triangle is 5 m less than base
Let base be x then height is ( x -5 ) m
Area of triangle is [tex]\frac{1}{2} \times base \times height[/tex]
also given area of triangle is 42 m²
substitute, we get,
[tex]42=\frac{1}{2} \times x \times (x-5)[/tex]
Solving , we get,
[tex]84=x(x-5)[/tex]
Simplify we get a quadratic equation as,
[tex]x^2-5x-84=0[/tex]
Using quadratic formula,
For the standard quadratic equation [tex]ax^2+bx+c=0[/tex] , the solution of roots is given by, [tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
For a = 1 , b= -5 , c = -84 , we get
[tex]x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\left(-84\right)}}{2\cdot \:1}[/tex]
Simplify, we get
[tex]=\frac{5\pm\sqrt{361}}{2\cdot \:1}[/tex]
We know [tex]\sqrt{361}=19[/tex] , we get
[tex]x_{1,2}=\frac{5\pm19}{2}[/tex]
[tex]x_1=\frac{5+19}{2}[/tex] and [tex]x_2=\frac{5-19}{2}[/tex]
[tex]x_1=12[/tex] and [tex]x_2=-7[/tex]
ignoring negative value as side cannot be negative.
thus, x = 12
base is 12 m then height is ( 12 -5 ) m = 4 m
Thus, the length of the base is 12 m.
