Triangle A″B″C″ is formed by a reflection over y = −3 and dilation by a scale factor of 2 from the origin. Which equation shows
the correct relationship between ΔABC and ΔA″B″C″
2 answers:
You might be interested in
Answer:
Height = 11 inches
Base = 33 inches
Step-by-step explanation:
In this question, we are asked to find the height and base of a parallelogram given the area of the parallelogram.
From the question, we are told that height is 1/3 of base. This means that base is 3 times the height
So if we represent the height with variable h, then the base is 3h
Mathematically the area of a parallelogram can be calculated by the formula
A = bh
Thus ,
363 = h * 3h
3h^2 = 363
divide through by 3
h^2 = 121
h = √121
h = 11 inches
since b = 3h , then b = 3 * 11 = 33 inches
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