the area of rhobus is 540m² and length of one diagonal is 36cm then find other diagonal
Answers
Answer:
1. The area of the rhombus can be found by the formula
A=\dfrac{d_1\cdot d_2}{2},A=
2
d
1
⋅d
2
, where d_1,\ d_2d
1
, d
2
are rhombus's diagonals.
Note that d_1=4.5\ dm=45\ cm,d
1
=4.5 dm=45 cm, then
\begin{gathered}540=\dfrac{45\cdot d_2}{2},\\ \\540\cdot 2=45d_2,\\ \\d_2=24\ cm.\end{gathered}
540=
2
45⋅d
2
,
540⋅2=45d
2
,
d
2
=24 cm.
2. The diagonals of rhombus are perpendicular and are bisectors of each other. Then the triangle formed with halfs of diagonals is right triangles with legs
\dfrac{d_1}{2}=22.5\ cm,\ \dfrac{d_2}{2}=12\ cm.
2
d
1
=22.5 cm,
2
d
2
=12 cm.
The hypotenuse of this triangle is the rhombus's side. By the Pythagorean theorem
\begin{gathered}\text{rhombus's side}^2=(22.5)^2+12^2=506.25+144=650.25,\\ \\\text{rhombus's side}=25.5\ cm.\end{gathered}
rhombus’s side
2
=(22.5)
2
+12
2
=506.25+144=650.25,
rhombus’s side=25.5 cm.
3. The distance between the point of intersection of the diagonals and the side of the rhombus is the height of right triangle considered above.
Use twice the Pythagorean theorem to find this height:
\begin{gathered}\left\{\begin{array}{l}x^2+h^2=12^2\\(25.5-x)^2+h^2=22.5^2,\end{array}\right.\end{gathered}
{
x
2
+h
2
=12
2
(25.5−x)
2
+h
2
=22.5
2
,
where x is projection of leg 12 cm and h is height.
Subtract the first equation from the second:
\begin{gathered}(25.5-x)^2+h^2-x^2-h^2=22.5^2-12^2,\\ \\650.25-51x=506.25-144,\\ \\51x=650.25-362.25=288,\\ \\x=\dfrac{96}{17}\ cm.\end{gathered}
(25.5−x)
2
+h
2
−x
2
−h
2
=22.5
2
−12
2
,
650.25−51x=506.25−144,
51x=650.25−362.25=288,
x=
17
96
cm.
Then
\begin{gathered}h^2=144-\left(\dfrac{96}{17}\right)^2=144-\dfrac{9216}{289}=\dfrac{32400}{289},\\ \\h=\dfrac{180}{17}\ cm.\end{gathered}
h
2
=144−(
17
96
)
2
=144−
289
9216
=
289
32400
,
h= 1718 cm.
Answer: h=\dfrac{180}{17}\ cm.h=
17
180
cm.