Answer:
-271
Step-by-step explanation:
7) Area trapezoid = (B+b).H/2, but the Median is equal to (B+b)./ 2
84 =12 . H ==> H = 84/12 ==> H = 7
11) Area Equilateral triangle inscribe in a cercle with Radius R = 2√3
Area = (B.. Altitude) / 2. Calculate H, the altitude. The altitude in an equilateral triangle bisects the opposite side. Apply Pythagoras
(2√3)² = (√3)² + H² ==> 12 = 3 + H² ==> H² = 9 & H = 3
Hence Are = (2√3 x 3) /2 ==> 3√3 unit² or 5.2 unit²
12) Area of regular hexagone with perimeter =12
regular hexagone is formed with 6 equilateral triangles with
each side =12/6 = 2 units
Let's calculate the area of 1 equilateral triangle. Follow the same logic as in problem 11 & you will find that Altitude = √3, Area =(2.√3)/2 =√3 =1.73 Unit³
13) a) Circumference of a circle : 2πR==> 2π(30) =60π = 188.4
b) Area of a circle =πR³ ==> π(30)² = 900π = 2,826 unit²
The answer is C.31.2 HOPE IT HELPS
-3 is a negative number and 2 is a positive
The equation of a circle exists:
, where (h, k) be the center.
The center of the circle exists at (16, 30).
<h3>What is the equation of a circle?</h3>
Let, the equation of a circle exists:
, where (h, k) be the center.
We rewrite the equation and set them equal :
We solve for each coefficient meaning if the term on the LHS contains an x then its coefficient exists exactly as the one on the RHS containing the x or y.
-2hx = -32x
h = -32/-2
⇒ h = 16.
-2ky = -60y
k = -60/-2
⇒ k = 30.
The center of the circle exists at (16, 30).
To learn more about center of the circle refer to:
brainly.com/question/10633821
#SPJ4
