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Greeley [361]
3 years ago
6

In the figure, BD¯¯¯¯¯ bisects ∠ABC .

Mathematics
1 answer:
pashok25 [27]3 years ago
5 0
The bisector BD is perpendicular to the line segment AC then the angle formed in between is 90°. We can solve for DC using SOH CAH TOA theorem or Pythagorean theorem.
Since ∠BDA is 90°, we can examine that AD is equal to 2units. Proving this, we have BD^2=BD^2+AD^2.
3^2=DB^2+2^2
DB=5units
Therefore, DC=AC-AD and the answer is 2 units.

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1. The area of the rhombus can be found by the formula

A=\dfrac{d_1\cdot d_2}{2}, where d_1,\ d_2 are rhombus's diagonals.

Note that d_1=4.5\ dm=45\ cm, then

540=\dfrac{45\cdot d_2}{2},\\ \\540\cdot 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.

The hypotenuse of this triangle is the rhombus's side. By the Pythagorean theorem

\text{rhombus's side}^2=(22.5)^2+12^2=506.25+144=650.25,\\ \\\text{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:

\left\{\begin{array}{l}x^2+h^2=12^2\\(25.5-x)^2+h^2=22.5^2,\end{array}\right.

where x is projection of leg 12 cm and h is height.

Subtract the first equation from the second:

(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.

Then

h^2=144-\left(\dfrac{96}{17}\right)^2=144-\dfrac{9216}{289}=\dfrac{32400}{289},\\ \\h=\dfrac{180}{17}\ cm.

Answer: h=\dfrac{180}{17}\ cm.

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