They diagonal through the interior of the cube is the longest. The hypotenuse is the longest side of a right triangle. The diagonal of the cube is the hypotenuse of a right angle with legs that are diagonal of a face and an edge. The edges are the shortest. They are legs of a right triangle with the diagonal of a face as the hypotenuse.
Answer:
<em>The perimeter of the rhombus is 20 cm</em>
Step-by-step explanation:
<u>Perimeter of a Rhombus</u>
Given the lengths of the two diagonals of a rhombus, let's call them a and b, the perimeter of the rhombus is given by:
![P=2\sqrt{a^2+b^2}](https://tex.z-dn.net/?f=P%3D2%5Csqrt%7Ba%5E2%2Bb%5E2%7D)
The values of the diagonals provided by the question are a=6 cm, b=8 cm, thus the perimeter is:
![P=2\sqrt{6^2+8^2}](https://tex.z-dn.net/?f=P%3D2%5Csqrt%7B6%5E2%2B8%5E2%7D)
![P=2\sqrt{36+64}](https://tex.z-dn.net/?f=P%3D2%5Csqrt%7B36%2B64%7D)
![P=2\sqrt{100}=2*10](https://tex.z-dn.net/?f=P%3D2%5Csqrt%7B100%7D%3D2%2A10)
![P=20\ cm](https://tex.z-dn.net/?f=P%3D20%5C%20cm)
The perimeter of the rhombus is 20 cm
Answer:
the secound triangle shows right angle
Step-by-step explanation:
But the question says it is 91 ⁰ so the question may be wrong
Answer:
![R=(\dfrac{b}{2},0)](https://tex.z-dn.net/?f=R%3D%28%5Cdfrac%7Bb%7D%7B2%7D%2C0%29)
![S=(\dfrac{b+c}{2},\dfrac{d}{2})](https://tex.z-dn.net/?f=S%3D%28%5Cdfrac%7Bb%2Bc%7D%7B2%7D%2C%5Cdfrac%7Bd%7D%7B2%7D%29)
![T=(\dfrac{c+e}{2},\dfrac{d+f}{2})](https://tex.z-dn.net/?f=T%3D%28%5Cdfrac%7Bc%2Be%7D%7B2%7D%2C%5Cdfrac%7Bd%2Bf%7D%7B2%7D%29)
![U=(\dfrac{e}{2},\dfrac{f}{2})](https://tex.z-dn.net/?f=U%3D%28%5Cdfrac%7Be%7D%7B2%7D%2C%5Cdfrac%7Bf%7D%7B2%7D%29)
![M=(\dfrac{b+c+e}{4},\dfrac{d+f}{4})](https://tex.z-dn.net/?f=M%3D%28%5Cdfrac%7Bb%2Bc%2Be%7D%7B4%7D%2C%5Cdfrac%7Bd%2Bf%7D%7B4%7D%29)
![M=(\dfrac{c+b+e}{4},\dfrac{d+f}{4})](https://tex.z-dn.net/?f=M%3D%28%5Cdfrac%7Bc%2Bb%2Be%7D%7B4%7D%2C%5Cdfrac%7Bd%2Bf%7D%7B4%7D%29)
Step-by-step explanation:
We are given coordinates as:
![A(0,0)\ ,\ B=(b,0)\ ,\ C=(c,d)\ ,\ D=(e,f)\\\\R=(\dfrac{b}{2},0)\ ,\ S=(\dfrac{b+c}{2},\dfrac{d}{2})\ ,\ U=(\dfrac{e}{2},\dfrac{f}{2})\ ,\ T=(\dfrac{c+e}{2},\dfrac{d+f}{2})](https://tex.z-dn.net/?f=A%280%2C0%29%5C%20%2C%5C%20B%3D%28b%2C0%29%5C%20%2C%5C%20C%3D%28c%2Cd%29%5C%20%2C%5C%20D%3D%28e%2Cf%29%5C%5C%5C%5CR%3D%28%5Cdfrac%7Bb%7D%7B2%7D%2C0%29%5C%20%2C%5C%20S%3D%28%5Cdfrac%7Bb%2Bc%7D%7B2%7D%2C%5Cdfrac%7Bd%7D%7B2%7D%29%5C%20%2C%5C%20U%3D%28%5Cdfrac%7Be%7D%7B2%7D%2C%5Cdfrac%7Bf%7D%7B2%7D%29%5C%20%2C%5C%20T%3D%28%5Cdfrac%7Bc%2Be%7D%7B2%7D%2C%5Cdfrac%7Bd%2Bf%7D%7B2%7D%29)
Now, it is given that M is the mid-point of the line segment RT and of US.
Hence, the coordinates of M is given as:
By taking the mid-point of side RT.
![M=(\dfrac{\dfrac{b}{2}+\dfrac{c+e}{2}}{2},\dfrac{0+\dfrac{d+f}{2}}{2}})\\\\\\i.e.\\\\\\M=(\dfrac{b+c+e}{4},\dfrac{d+f}{4})](https://tex.z-dn.net/?f=M%3D%28%5Cdfrac%7B%5Cdfrac%7Bb%7D%7B2%7D%2B%5Cdfrac%7Bc%2Be%7D%7B2%7D%7D%7B2%7D%2C%5Cdfrac%7B0%2B%5Cdfrac%7Bd%2Bf%7D%7B2%7D%7D%7B2%7D%7D%29%5C%5C%5C%5C%5C%5Ci.e.%5C%5C%5C%5C%5C%5CM%3D%28%5Cdfrac%7Bb%2Bc%2Be%7D%7B4%7D%2C%5Cdfrac%7Bd%2Bf%7D%7B4%7D%29)
By taking the mid-point of side US.
![M=(\dfrac{\dfrac{e}{2}+\dfrac{b+c}{2}}{2},\dfrac{\dfrac{f}{2}+\dfrac{d}{2}}{2})\\\\\\i.e.\\\\\\M=(\dfrac{b+c+e}{4},\dfrac{d+f}{4})](https://tex.z-dn.net/?f=M%3D%28%5Cdfrac%7B%5Cdfrac%7Be%7D%7B2%7D%2B%5Cdfrac%7Bb%2Bc%7D%7B2%7D%7D%7B2%7D%2C%5Cdfrac%7B%5Cdfrac%7Bf%7D%7B2%7D%2B%5Cdfrac%7Bd%7D%7B2%7D%7D%7B2%7D%29%5C%5C%5C%5C%5C%5Ci.e.%5C%5C%5C%5C%5C%5CM%3D%28%5Cdfrac%7Bb%2Bc%2Be%7D%7B4%7D%2C%5Cdfrac%7Bd%2Bf%7D%7B4%7D%29)
The completed equation that you can put into a graphing calculator or can graph is: y=3x-2