QRST is a rhombus
Let’s look at the diagonals
Length QS = sr(4^2 + 0) = 4
Length RT = sr(0 + 6^2) = 6
This means it cannot be a square or rectangle as they both have equal length diagonals.
QS is horizontal (y values same)
RT is horizontal (x values same)
This means it is a rhombus because diagonals are perpendicular and different lengths.
Squares have perpendicular diagonals of same length
Rectangles have diagonals of same length but not perpendicular
QN = 28
Solution:
Given MNPQ is a parallelogram.
QT = 4x + 6 and TN = 5x + 4
To find the length of QN:
Let us solve it using the property of parallelogram.
Property of Parallelogram:
Diagonals of the parallelogram bisect each other.
Therefore, QT = TN
⇒ 4x + 6 = 5x + 4
Arrange like terms together.
⇒ 6 – 4 = 5x – 4x
⇒ 2 = x
⇒ x = 2
Substitute x = 2 in QT and TN
QT = 4(2) + 6 = 14
TN = 5(2) + 4 = 14
QN = QT + TN
= 14 + 14
QN = 28
The length of QN is 28.
Answer:
D
Step-by-step explanation:
To be an identity we require both sides of the equation to equate
A
8x + 9 ≠ 8x - 3 ← not an identity
B
7x - 2 = 8x + 4 - x = 7x + 4 ≠ 7x - 2 ← not an identity
C
11 - (2x + 3) = 11 - 2x - 3 = - 2x + 8 ≠ - 2x - 8 ← not an identity
D
5x + 8 - x = 6x - 2(x - 4), that is
4x + 8 = 6x - 2x + 8 = 4x + 8 ← both sides equate thus an identity
Answer:
Step-by-step explanation:
If FG= 8
and GH=14
G is a midpoint
To find FH you would add 8 and 14.
