In Triangle XYZ, A is the midpoint of XY, B is the midpoint of YZ, and C is the midpoint of XZ.

Mathematics

1 answer:

Roman55 [17]3 years ago

5 0

The perimeter of triangle ABC is 24 units

Step-by-step explanation:

If a segment joining the mid points of two sides of a triangle, then

this segment is:

Parallel to the third side
Its length is half the length of the third side

In The triangle XYZ

∵ A is the mid point of XY

∵ B is the mid point of YZ

∴ AB = $\frac{1}{2}$ XZ

∵ XZ = 18 units

- Substitute the value of XZ in AB

∴ AB = $\frac{1}{2}$ × 18 = 9 units

∵ B is the mid point of YZ

∵ C is the mid point of XZ

∴ BC = $\frac{1}{2}$ XY

∵ AY = 7 units

∵ AY = $\frac{1}{2}$ XY

∴ XY = 2 × AY

∴ XY = 2 × 7

∴ XY = 14 units

∴ BC = $\frac{1}{2}$ × 14 = 7 units

∵ A is the mid point of XY

∵ C is the mid point of XZ

∴ AC = $\frac{1}{2}$ YZ

∵ BZ = 8 units

∵ BZ = $\frac{1}{2}$ YZ

∴ YZ = 2 × BZ

∴ YZ = 2 × 8

∴ YZ = 16 units

∴ AC = $\frac{1}{2}$ × 16 = 8 units

∵ The perimeter of a triangle = the sum of the lengths of its sides

∴ Perimeter Δ ABC = AB + BC + AC

∴ Perimeter Δ ABC = 9 + 7 + 8 = 24 units

The perimeter of triangle ABC is 24 units

Learn more:

You can learn more about triangles in brainly.com/question/5924921

#LearnwithBrainly

You might be interested in

State the vertical asymptote of the rational function. f(x) =((x-9)(x+7))/(x^2-4)

Lubov Fominskaja [6]

$D:x^2-4\not=0\\
D:x^2\not=4\\
D:x\not=-2 \wedge x\not =2\\\\
\displaystyle
\lim_{x\to-2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(-2-9)(-2+7)}{(-2^-)^2-4}=\dfrac{-11\cdot5}{4^+-4}=\dfrac{-55}{0^+}=-55\cdot\infty=-\infty\\
\dfrac{(-2-9)(-2+7)}{(-2^+)^2-4}=\dfrac{-11\cdot5}{4^--4}=\dfrac{-55}{0^-}=-55\cdot(-\infty)=\infty
$

$\displaystyle
\lim_{x\to2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(2-9)(2+7)}{(2^-)^2-4}=\dfrac{-7\cdot9}{4^--4}=\dfrac{-63}{0^-}=-63\cdot(-\infty)=\infty\\
\dfrac{(2-9)(2+7)}{(2^+)^2-4}=\dfrac{-7\cdot9}{4^+-4}=\dfrac{-63}{0^+}=-63\cdot\infty=-\infty\\$

So, the vertical asymptotes are $x=\pm 2$