Given that JKLM is a rhombus and the length of diagonal KM=10 na d JL=24, the perimeter will be found as follows;
the length of one side of the rhombus will be given by Pythagorean theorem, the reason being at the point the diagonals intersect, they form a perpendicular angles;
thus
c^2=a^2+b^2
hence;
c^2=5^2+12^2
c^2=144+25
c^2=169
thus;
c=sqrt169
c=13 units;
thus the perimeter of the rhombus will be:
P=L+L+L+L
P=13+13+13+13
P=52 units
Answer:
Equation
is in the standard dorm.
Step-by-step explanation:
Here, the x - intercept = 4 , so ( 4,0) is a point on the line.
and the y - intercept = 5, so ( 0,5) is a point on the line.
So, the slope of the equation is given as = 
Now, the SLOPE INTERCEPT FORM of an equation is :
y = mx + b: here m = slope and b = y- intercept
or, 
Now, standard form is Ax +By = C
So, the standard form is
Hence, the above equation
is in the standard dorm.
Answer:
7 units by 8 units
Step-by-step explanation:
ALL POSSIBLE PERIMETERS:
1x14
2x13
3x12
4x11
5x10
6x9
7x8
Answer:
<A = 35 degrees
Step-by-step explanation:
<A = x
<B = 2x
<C = 2x + 5
X + 2x + 2x + 5 = 180 degrees
5x + 5 = 180
5x = 180 -5
x = 175/5
x = 35
