Question

Please answer #s 7,11,12, and 13 thanks please really quick

Answer 1

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