MATH SOLVE

4 months ago

Q:
# In isosceles triangle the length of the base is twice shorter than the length of a leg. The perimeter of this triangle is 50cm. Find the lengths of all sides.

Accepted Solution

A:

x - length of a leg

1/2*xΒ - length of the base

so:

[tex]P=2\cdot x+\dfrac{1}{2}x\\\\\\50=2x+\dfrac{1}{2}x\\\\\\50=2\dfrac{1}{2}x\\\\\\50=\dfrac{5}{2}x\quad|\cdot\dfrac{2}{5}\\\\\\x=50\cdot\dfrac{2}{5}\\\\\\x=\dfrac{100}{5}\\\\\\\boxed{x=20\text{ cm}}\qquad\qquad\text{(leg)}[/tex]

and

[tex]\dfrac{1}{2}\cdot x=\dfrac{1}{2}\cdot20=\boxed{10\text{ cm}}\qquad\qquad\text{(base)}[/tex]

1/2*xΒ - length of the base

so:

[tex]P=2\cdot x+\dfrac{1}{2}x\\\\\\50=2x+\dfrac{1}{2}x\\\\\\50=2\dfrac{1}{2}x\\\\\\50=\dfrac{5}{2}x\quad|\cdot\dfrac{2}{5}\\\\\\x=50\cdot\dfrac{2}{5}\\\\\\x=\dfrac{100}{5}\\\\\\\boxed{x=20\text{ cm}}\qquad\qquad\text{(leg)}[/tex]

and

[tex]\dfrac{1}{2}\cdot x=\dfrac{1}{2}\cdot20=\boxed{10\text{ cm}}\qquad\qquad\text{(base)}[/tex]