⇒ s = a+b+c2=5+12+132 a + b + c 2 = 5 + 12 + 13 2 = 15 cm.
Area of the base = √s(s−a)(s−b)(s−c)
= √15×10×3×2 15 × 10 × 3 × 2 cm2 = 30 cm2
Given:
Line segment NY has endpoints N(-11, 5) and Y(3,-3).
To find:
The equation of the perpendicular bisector of NY.
Solution:
Midpoint point of NY is




Slope of lines NY is




Product of slopes of two perpendicular lines is -1. So,


The perpendicular bisector of NY passes through (-4,1) with slope
. So, the equation of perpendicular bisector of NY is




Add 1 on both sides.

Therefore, the equation of perpendicular bisector of NY is
.
Answer:
28 square centimeters
Step-by-step explanation:
The area of a trapezoid (formula):
(a + b) ÷ 2 x h
Where a & b are the bases, and h is height.
Use formula with given measurements:
(9 + 5) ÷ 2 x 4 = 28
Area is measured in square centimeters
(centimeters in this case)
Therefore the area if the trapezoid is 28 cm^2
I really hope this helps!