Answer:1.17
Step-by-step explanation:
Answer:
Step-by-step explanation:
4). a). If the diagonals of a parallelogram are congruent, then it must be a RECTANGLE.
b). If the diagonals of a parallelogram are perpendicular, then it must be a SQUARE.
c). If the diagonals of a parallelogram bisect the angles of the parallelogram, then it must be a RHOMBUS.
d). If the diagonals of a parallelogram are perpendicular and congruent, then it must be a SQUARE.
e). If a parallelogram has four congruent sides, then it must be a SQUARE.
5). Given quadrilateral SELF is a rhombus.
a). All sides of a rhombus are equal,
Therefore, ES = EL = 25
b). Diagonals of a rhombus bisects the opposite angles,
Therefore, m∠ELS = m∠FLS
3x - 2 = 2x + 7
3x - 2x = 7 + 2
x = 9
c). Diagonals of the rhombus bisect the opposite angles, and adjacent angles are supplementary.
m∠ELF = 2(m∠ELS) = 2(2y - 9)
m∠LES = 2(m∠LEF) = 2(3y + 9)
And 2(2y - 9) + 2(3y + 9) = 180
(2y - 9) + (3y + 9) = 90
5y = 90
y = 18
check the picture below.
if ∡F = 90° and ∡D = 30°, then the ∡A = 60°, meaning the triangle is a 30-60-90 triangle and therefore we can use the 30-60-90 rule as you see in the picture.
![\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ h=7\sqrt{3}\\ b=7 \end{cases}\implies A=\cfrac{1}{2}(7)(7\sqrt{3}) \\\\\\ A=\cfrac{49\sqrt{3}}{2}\implies A\approx 42.43524478543749369142](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20triangle%7D%5C%5C%5C%5C%0AA%3D%5Ccfrac%7B1%7D%7B2%7Dbh~~%0A%5Cbegin%7Bcases%7D%0Ab%3Dbase%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Ah%3D7%5Csqrt%7B3%7D%5C%5C%0Ab%3D7%0A%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%287%29%287%5Csqrt%7B3%7D%29%0A%5C%5C%5C%5C%5C%5C%0AA%3D%5Ccfrac%7B49%5Csqrt%7B3%7D%7D%7B2%7D%5Cimplies%20A%5Capprox%2042.43524478543749369142)
