Respuesta :
Answer:
Base of triangle is 9 inches and altitude of triangle is 14 inches.
Step-by-step explanation:
Given:
Area of Triangle = 63 sq. in.
Let base of the triangle be 'b'.
Let altitude of triangle be 'a'.
Now according to question;
altitude is 5 inches longer than the base.
hence equation can be framed as;
[tex]a=b+5[/tex]
Now we know that Area of triangle is half times base and altitude.
Hence we get;
[tex]\frac{1}{2} \times b \times a =\textrm{Area of Triangle}[/tex]
Substituting the values we get;
[tex]\frac{1}{2} \times b \times (b+5) =63\\\\b(b+5)=63\times2\\\\b^2+5b=126\\\\b^2+5b-126=0[/tex]
Now finding the roots for given equation we get;
[tex]b^2+14b-9b-126=0\\\\b(b+14)-9(b+14)=0\\\\(b+14)(b-9)=0[/tex]
Hence there are 2 values of b[tex]b-9 = 0\\b=9\\\\b+14=0\\b=-14[/tex]
Since base of triangle cannot be negative hence we can say [tex]b=9\ inches[/tex]
So Base of triangle = 9 inches.
Altitude = 5 + base = 5 + 9 = 14 in.
Hence Base of triangle is 9 inches and altitude of triangle is 14 inches.
