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First solve the length of side BC, CD, EF and FA
Since BC = CD = sqrt( 10^2 + 10^2)
BC = CD = 14.1421
FA = EF = sqrt(10^2 + 20^2)
= 23.3607
So the perimeter = 10 + 10 + 14.1421 + 14.1421 + 23.3607
= 93
The area is made up be triangle FAE, rectangle ABDE and
triangle BCD
A = 0.5(20)(20) + (10)(20) + 0.5(20)(10)
<span>A = 500 sq units</span>
Answer:
The length of AA' = √29 = 5.39
Step-by-step explanation:
* Lets revise how to find the length of a line joining between
any two points in the coordinates system
- If point A is (x1 , y1) and point B is (x2 , y2)
- The length of AB segment √[(x2 - x1)² + (y2 - y1)²]
* Lets use this rule to solve the problem
∵ Point A is (0 , 0)
∵ Point A' = (5 , 2)
∵ (x2 - x1)² = (5 - 0)² = 5² = 25
∵ (y2 - y1)² = (2 - 0)² = 2² = 4
∴ The length of AA' = √(25 + 4) = √29 = 5.39
7x - 5 = 0
7x = 5
x = 5/7
--------------------------
Answer:
b = 15.75
Step-by-step explanation:
Lets find the interception points of the curves
36 x² = 25
x² = 25/36 = 0.69444
|x| = √(25/36) = 5/6
thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).
The area of the bounded region is given by the integral

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = b^{1.5}/9
b = (62.5²)^{1/3} = 15.75