Sir Mike Ambrose is the author of the question.Calculating a.½((0, 6√(3)) (-3, 3√(3)) (a, (16a/3)) (0, 6√(3))) = ¼(9+20√(3))Therefore;a = x = ½√(3) unity = 16a/3 = (16/3)(√(3)/2)y = ⅓(8√(3)) unit.The...
Sir Mike Ambrose is the author of the question.Area green is;½*3*6sin120= ½(9√(3)) cm²Radius, r of the circle is;r = √(63)/√(3)r = √(21) cmTherefore;Area circle is;π(√(21))²= 21π cm²It implies;Area g...
Sir Mike Ambrose is the author of the question.Calculating Exactly, Golden Area ÷ Large Square Area.Let the side of the large square be 2 units.Therefore;Area large square is;2²= 4 square units.Area...
Calculating Area Green.Let theta be x.a = 90-39a = 51°x+x+27 = 512x = 51-272x = 24x = 12°b+x+39 = 180b = 180-39-xAnd x = 12°b = 180-39-12b = 129°Let c be the hypotenuse of the ascribed right-angled t...
Calculating The Red Inscribed Semi Circle Area.Let x be the area of the inscribed half circle.a² = 2x²a = √(2)x units.Calculating x.1² = x²+(√(2)x)²1 = x²+2x²1 = 3x²x = 1/√(3) units.Again, x is the r...
Total coloured area is;2(area triangle with height 6.69991708075 cm and base √(65)sin(atan(10)-atan(7/4)) cm= 2*½*6.69991708075*√(65)sin(atan(10)-atan(7/4))= 6.69991708075*3.28362272769= 22 cm²
Sir Mike Ambrose is the author of the question.Radius, r of the inscribed circle is;r = (3-√(3)) cmTherefore;Area purple exactly in square cm is;Area trapezium with parallel sides (3-√(3)) cm and 4 c...
Sir Mike Ambrose is the author of the question. P(2, (16√(5)/5))Gradient of the curve at the tangent is;= 24/5√(5)Therefore;Gradient of the curve at the normal is;= -5√(5)/24It implies;Coordinate Q i...
Calculating x, chord of the circle..a² = 1 square units.a = 1 unit.a is the side length of the 3 inscribed equal squares.b = 1+½(1)+yb = (1.5+y) units.c = 1+1c = 2 units.d² = 2²+0.5²d² = 4+0.25d = √(...
Sir Mike Ambrose is the author of the question.AB exactly is;= 16tan(3acos(8/9)) - 2√(17)= 2(8tan(3acos(8/9))-√(17)) cm.= 102.760478669 cm.
