Answer:
Length of the shorter diagonal is 8.68 cm.
Step-by-step explanation:
Length of the side AB = 11 cm
Length of side BC = 5 cm
Angle between these sides, m∠ABC = 50°
By cosine rule in ΔABC, 
AC² = AB² + BC² - 2AB.BC.cos B
AC² = (11)² + 5² - 2(11)(5)cos50°
AC² = 121 + 25 - 70.71
AC = √(75.29)
AC = 8.68 cm
By the property of parallelogram,
m∠B + m∠C = 180° [Interior consecutive angles]
50° + m∠C = 180°
m∠C = 130°
Similarly, In ΔBCD,
BD² = BC² + CD² - 2BC.CD.cos130°
BD² = (5)² + (11)² - 2(5)(11)cos130°
BD² = 25 + 121 + 70.71
BD² = 216.71
BD = 14.72 cm
Therefore, length of the shorter diagonal will be 14.72 cm.
 
        
             
        
        
        
<span>Vector Equation
(Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5) 
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t
y = -2 + 5t
;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t 
y = -3 + 8t
;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations: 
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t 
(-8 - 5)/4 = t
-2 = t
For y sub in -8 
-8 = -2 + 3t
(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations: 
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t 
(-1 - 5)/4 = t
-1 = t
For y sub in -7 
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points: 
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int) 
Sub t = 2/3 into x = 5 + 4t 
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3) 
x = 23/3
The x-int is (23/3, 0)</span>What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?
        
             
        
        
        
610-187 is 423 but when you estimate it would be 400 this is your answer
        
                    
             
        
        
        
Well actually that is in the value place of the tenths not tens.