Respuesta :
special prpoerty of 45-45-90 triangle
if the legs are x then the hyptonuse is x√2
18=hyptonuse=x√2
18=x√2
divide both sides by √2
18/√2
make denomenator nice
times by √2/√2
(18√2)/(2)
9√2
one leg is 9√2 cm
if the legs are x then the hyptonuse is x√2
18=hyptonuse=x√2
18=x√2
divide both sides by √2
18/√2
make denomenator nice
times by √2/√2
(18√2)/(2)
9√2
one leg is 9√2 cm
The length of the one leg of the triangle is [tex]\boxed{\bf 9\sqrt{2}\text{\bf\ cm}}[/tex].
Further explanation:
Given:
The length of hypotenuse of a triangle is [tex]18\text{ cm}[/tex].
Calculation:
If one angle of the triangle is [tex]90^{\circ}[/tex] then the triangle is said to be a right angle triangle.
From attached Figure 1 it can be seen that [tex]AC[/tex] is the hypotenuse, [tex]BC[/tex] is the base side and [tex]AB[/tex] is perpendicular in a triangle [tex]\triangle\text{ ABC}[/tex].
In [tex]\triangle\text{ ABC}[/tex], the base length can be calculated as follows:
[tex]\text{cos}45^{\circ}=\dfrac{BC}{AC}[/tex] …… (1)
It is given that the value of [tex]AC[/tex] is [tex]18\text{ cm}[/tex]. The value of [tex]\text{cos}45^{\circ}[/tex] is [tex]\frac{1}{\sqrt{2}}[/tex].
Now, substitute [tex]AC=18[/tex] and [tex]\text{cos}45^{\circ}=\frac{1}{\sqrt{2}}[/tex] in equation (1) to obtain the value of length of [tex]BC[/tex].
[tex]\begin{aligned}\dfrac{1}{\sqrt{2}}&=\dfrac{BC}{18}\\BC&=\dfrac{18}{\sqrt{2}}\\&=\dfrac{18}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}\\&=\dfrac{18\sqrt{2}}{2}\\&=9\sqrt{2}\end{aligned}[/tex]
Therefore, the length of perpendicular side [tex]BC[/tex] is [tex]9\sqrt{2}\text{ cm}[/tex].
In [tex]\triangle\text{ ABC}[/tex], the perpendicular length can be calculated as,
[tex]\text{sin}45^{\circ}=\dfrac{AB}{AC}[/tex] …… (2)
Now, substitute [tex]AC=18[/tex] and [tex]\text{sin}45^{\circ}=\frac{1}{\sqrt{2}}[/tex] in equation (2) to obtain the length of side [tex]AB[/tex] as follows:
[tex]\begin{aligned}\dfrac{1}{\sqrt{2}}&=\dfrac{AB}{18}\\AB&=\dfrac{18}{\sqrt{2}}\\&=\dfrac{18}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}\\&=\dfrac{18\sqrt{2}}{2}\\&=9\sqrt{2}\end{aligned}[/tex]
Therefore, the length of perpendicular side [tex]AB[/tex] is [tex]\boxed{\bf 9\sqrt{2}\text{\bf cm}}[/tex].
Learn more:
1. Learn more about graph: https://brainly.com/question/2334270
2. Learn more about problem on triangle https://brainly.com/question/7437053
3. Learn more about problem on angle https://brainly.com/question/3717797
Answer details:
Grade: Middle school
Subject: Mathematics
Chapter: Triangles
Keywords: Triangle, perpendicular, base, hypotenuse, cos, sin, cos45, sin45, right angles triangle, figure, length, 45-45-90, 18cm, 9square root 2.
