Respuesta :
The base and height of the triangle whose area is 51 [tex]cm^2[/tex] and base of a triangle is twelve more than twice its height are 21.49 cm and 4.745 cm respectively.
Solution:
Given that
The base of a triangle is twelve more than twice its height
And area of triangle = 51 square centimeter
Let’s assume height of triangle = "x" cm
So base of triangle = 12 + (2 [tex]\times[/tex] height ) = 2x + 12
[tex]\text { Area of triangle }=\frac{1}{2} \times \text { Base } \times \text { height }[/tex]
On substituting the given value of area and assumed values of height and base in above formula we get
[tex]\begin{array}{l}{51=\frac{1}{2} \times(2 x+12) \times x=x^{2}+6 x} \\\\ {\Rightarrow x^{2}+6 x-51=0}\end{array}[/tex]
We can find solution of this equation using quadratic formula.
According to quadratic formula for general equation [tex]\mathrm{ax}^{2}+\mathrm{b} x+\mathrm{c}=0[/tex] solution of the equation is given by
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Our equation is [tex]x^{2}+6 x-51=0[/tex]
So in our case, a = 1 , b = 6 and c = -51
On applying quadratic formula we get
[tex]\begin{array}{l}{x=\frac{-6 \pm \sqrt{6^{2}-(4 \times 1 \times(-51))}}{2 \times 1}} \\\\ {x=\frac{-6 \pm \sqrt{36+204}}{2}} \\\\ {x=\frac{-6 \pm \sqrt{240}}{2}} \\\\ {x=\frac{-6 \pm \sqrt{240}}{2}=\frac{-6 \pm 15.49}{2}}\end{array}[/tex]
[tex]=>x=\frac{-6+15.49}{2} \text { or } x=\frac{-6-15.49}{2}[/tex]
As dimensions of triangle cannot be negative so neglect negative value
[tex]x=\frac{-6+15.49}{2}=\frac{9.49}{2}=4.745[/tex]
Height of triangle = x = 4.745 cm
Base of triangle = 12 + ( 2 x 4.745 ) = 12 + 9.49 = 21.49 cm
Hence base and height of the triangle whose area is 51 [tex]cm^2[/tex] and base of a triangle is twelve more than twice its height are 21.49 cm and 4.745 cm respectively.
