+129845 Here's the second one
MIR is an inscribed angle
So....since it subtends minor arc MR, the angular measure of this arc is twice 18° = 36°
So......the length of MR is given by :
2 pi * radius * (36 / 360) =
2pi * 80 * (1 /10) =
(160 / 10) pi =
16 pi cm
+129845 Here's the first one....there might be an easier way to do it, however !!!
The area of ABC is (1/2) (16)^2 (√3/2) = 64√3 cm
So...the area of equilateral triangle PQR is (1/4) of this = 16√3 cm^2
And we can find the length of PR by using the area formula for PQR
(1/2) (PR)^2 (√3 /2 ) = 16√3
(1/2) PR^2 (1/2) = 16
PR^2 / 4 = 16
PR^2 = 64
PR = 8 cm
And note that we have trapezoid BPRC which has the same area as the small equilateral triangle
So area of trapezoid BPRC = area of triangle PQR
also
(1/2) (height) ( sum of the bases) = 16√3
height ( 16 + 8) = 32√3
height (24) = 32√3
height = 32√3 / 24 = (4/3)√3 cm
So...note that if we draw a perpendicular from P to BC this will be the height of the trapezoid and will also form one leg of a right triangle with BP as the hypotenuse
And the other leg length will be ( BC - PR) / 2 = (16 - 8) /2 = 8/2 = 4
So....using the Pythagorean Theorem...we can find BP as
BP = √ [ 4^2 + [ 4/3 * √3 ]^2 ] = √ [ 16 + 16/3 ] = 4 √ [ 1 + 1/3] =
4 √ [ 4/3 ] = 8√ [ 1/3 ] = 8 / √3 cm = 8√3 / 3 cm
+129845 Here's the third one
Orient the paper so that 22 is the width and 39 is the height
Note that the number of retangles of (2 * 3) that can be cut from this sheet is
(39/3) * ( 22/2) =
13 * 11 =
143
But each of the rectangles can be divided into two triangles with bases of 2 and heights of 3
So....the total number of triangles is 143 * 2 = 286
