All parallelogram are____ Rhombuses Squares Trapezoids Rectangles
2 answers:
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Answer:
Step-by-step explanation:
we know that
In a rectangle, the two diagonals are congruent and each diagonal bisects the other
so
----> equation A
-----> equation B
step 1
Find the value of x
solve equation A
substitute the given values
solve for x
step 2
Find the value of KE
substitute the value of x
Remember that
-----> by each diagonal bisects the other
therefore
This is an interesting question. I chose to tackle it using the Law of Cosines.
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355