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Given that, a * b = a\(^2\)b and a ^ b = 2a + b (-4 * 2) = (-4)\(^2\) x 2 = 16 x 2 = 32 (7 * -1) = 7\(^2\) x (-1) = 49 x (-1) = -49 ∴ (-4 * 2) ^ (7 * -1) = 2(32) + (-49) = 64 - 49 = 15

A binary operation '*' defined on a set A is said to be commutative only if a*b=b*a, ∀a, b∈A. If (a*b)*c=a*(b*c), then the operation is said to associative ∀ a, b∈ A. If (b ο c)*a=(b*a) ο (c*a), then the operation is said to be distributive ∀ a, b, c ∈ A.

Let the length of the longer side = \(x\) cm ∴ The length of the shorter side = (\(x\) - 6) cm If we increase each side's length by 2 cm, it becomes (\(x\) + 2) cm and (\(x\) - 4) cm respectively Area of a rectangle = L x B \(A_1 = x(x - 6) = x^2 - 6x\) \(A_2 = (x + 2)(x - 4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8\) \(A_1 + 68 = A_2\) (Given) ⇒ \(x^2 - 6x + 68 = x^2 - 2x - 8\) ⇒ \(x^2 - x^2 - 6x + 2x\) = -8 - 68 ⇒ -4\(x\) = -76 ⇒ \(x\) = \(\frac{-76}{-4}\) = 19cm ∴ The length of the shorter side = \(x\) - 6 = 19 - 6 = 13 cm

Area of the composite figure = Area of semi circle + Area of rectangle + Area of triangle Area of semi circle = \(\frac{1}{2}\pi r^2 = \frac{1}{2}\times\pi\times\frac{d^2}{4} = \frac{1}{2}\times\frac{22}{7}\times\frac{42^2}{4} = 693 m^2\) Area of rectangle = l x b = 42 x 60 =2520 m\(^2\) Area of triangle = \(\frac{1}{2}\times b \times h = \frac{1}{2}\times 36 \times 42 = 756 m^2\) ∴ Area of the composite figure = 693 + 2520 + 756 = 3969 m\(^2\)

Area of trapezium = \(\frac{1}{2}(a + b) h\) ⇒ \(\frac{1}{2} (a + b)\times 16 = 200\) ⇒ 8(a + b) = 200 ⇒ a + b = \(\frac{200}{8}\) = 25 -----(i) ⇒ a : b = 2 : 3 ⇒ \(\frac{a}{b} = \frac{2}{3}\) ⇒ 3a = 2b ⇒ a = \(\frac{2b}{3}\) -------(ii) Substitute \(\frac{2b}{3}\) for a in equation (i) ⇒ \(\frac{2b}{3}\) + b = 25 \(\frac{5b}{3}\) = 25 ⇒ b = 25 \div \(\frac{5}{3} = 25\times\frac{3}{5} = 15cm\) From equation (ii) ⇒ a = \(\frac{2 \times 15}{3} = 2\times5 = 10cm\) ∴ Lengths of each parallel sides are 10cm and 15cm

Perimeter of a triangle = sum of all sides ⇒ \(P = y + x + x = 2\) ⇒ \(y + 2x = 2\) ⇒ \(y= 2 - 2x\)-----(i) Using Pythagoras theorem \(y^2 = x^2 + x^2\) ⇒ \(y^2 = 2x^2\) ⇒ \(y = \sqrt2x^2\) ⇒ \(y = x\sqrt2\)-----(ii) Equate \(y\) ⇒ \(2 - 2x = x\sqrt2\) Square both sides ⇒ \((2 -2x) ^2 = (x\sqrt2)^2\) ⇒ \(4 - 8x + 4x^2 = 2x^2\) ⇒ \(4 - 8x + 4x^2 - 2x^2 = 0\) ⇒ \(2x^2 - 8x + 4 = 0\) ⇒ \(x = \frac{-(-8)\pm\sqrt(-8)^2 - 4\times2\times4}{2\times2}\) ⇒ \(x = \frac{8\pm\sqrt32}{4}\) ⇒ \(x = \frac{8\pm4\sqrt2}{4}\) ⇒ \(x = 2\pm\sqrt2\) ⇒ \(x = 2 + \sqrt2\) or \(2 - \sqrt2\) ∴ \(x = 2 - \sqrt2\) (for \(x\) has to be less than its perimeter) ∴ \(y = 2 - 2x = 2 - 2(2 - \sqrt2) = -2 + 2 \sqrt2\) ∴ The length of the longer side = -2 + 2\(\sqrt2\)m

\(\theta = 115^\circ , radius = 3.4cm^2\) Area of a sector = \(\frac{\theta}{360} \times \pi r^2\) = \(\frac{115}{360} \times \frac{22}{7} \times 3.4\times 3.4\) = \(\frac{29246.4}{2520}\) = \(11.6cm^2\)

using Pythagoras theorem H = \(\sqrt{ 2^2 - 1^2}\) = \(\sqrt{ 4 - 1}\) = \(\sqrt3\)cm

Let each side be l, then area = \(l^2\) Using Pythagoras theorem \(l^2 + l^2 = 12^2\) \(2l^2 = 144\) divide both sides by 2 \(l^2 = 72\) Therefore, the Area of the square is \(72 cm^2\)

In a rhombus, the diagonals are perpendicular bisectors of each other, and they bisect the angles of the rhombus. This means that a rhombus is essentially made up of four congruent right-angled triangles. We can use the Pythagorean theorem to find the length of one side of the rhombus (s) \(s^2 = 8^2 + 6^2\) \(s^2 = 64 + 36\) \(s^2 = 100\) s = \(\sqrt{100}\) s =10 cm So, the length of each side of the rhombus is 10 cm.

∆PQR is an equilateral triangle and M is the midpoint of Line QR. Hence, PM is the radius of the circle likewise Line PA and Line PB Using Pythagoras theorem to find Line PM : \( 18^2 = PM^2 + 9^2\) \( PM^2 = 324 - 81 = 243\) PM = \(\sqrt{243} = 9\sqrt{3}\) ∠QPR = \(\frac{180^0}{3} = 60^0\) ( each ∠ of an equilateral △ are equal) Area of △,PQR = \(\frac{1}{2}absin\theta\), but \theta = 60^\circ = \(\frac{1}{2}\times18\times18sin60 = 81\sqrt{3} = 140.30 cm^2\) Area of PAB = \(\frac{\theta}{360} \times\pi r^2\) = \(\frac{60}{360}\times \frac{22}{7} \times(9\sqrt{3})^2\) = \(\frac{1}{6} \times \frac{22}{7} \times243\) = \(127.29cm^2\) therefore, the Area of the shaded portion = area of △ PQR - area of sector PAB = 140. 30 - 127.29 = \(13.01cm^2\) (to 2 d.p)

a. Let the number of sides of polygon Q be x therefore, No. of sides of polygon P = 2x then, \(\frac{360}{x} - \frac{360}{2x}\) = 45, \(\frac{360}{x} - \frac{180}{x}\) = 45 L. C. M = x 360 - 180 = 45x x = \(\frac{180}{4}\) = 4 therefore, the number of P polygon = 2x = 2 x 4 = 8 b. Area of semi-circle = \(\frac{\pi r^2}{2}\) but Area of the semi-circle = 32\pi \(\frac{\pi r^2}{2}\) = 32\pi \(r^2 = 32\times2 = 64\) \(r^2 = \sqrt{64}\) = 8cm. Circumference of semi-circle = \pir + 2r = \pi(8) + 2(8) = 8\pi + 16 = 8(\pi + 2 ) cm

a. Area of sector MON = \(\frac{MON}{360} \times\pi r^2\) = \(\frac{60}{360}\times\frac{22}{7}\times7^2\) = \(\frac{77}{3}cm^2\) \(\frac{NT}{7}\) = sin60^\circ NT = 7sin60^\circ = \(7\times\frac{\sqrt{3}}{2}\) = \(7\times\frac{\sqrt{3}}{2}\)cm. \(\frac{OT}{9} = cos60^0\) OT = 7cos60^\circ = \(7\frac{1}{2}\)cm Area of △ONT = \(\frac{1}{2}\times OT\times NT\) = \(\frac{1}{2}\times\frac{7}{2}\times\frac{7\sqrt{3}}{2} = 49\sqrt{3}{8}cm^2\) Area of shaded portion = Area of sector MON - Area of △ONT = \(\frac{77}{3} - \frac{49\sqrt{3}}{8}\) = 25.67 - 10.61 = \(15.06cm^2\) = 15.1\(cm^2\) b. Given : y = mx + c; hence ; x = \(\frac{ y - c }{ m } = \frac{ y }{ m } - \frac{ c }{ m }\) x-intercept = \(\frac{ - c }{ m } = \frac{ - 3 }{ 4 }\) Therefore, m = \(\frac{4}{3}\)c ....................... (1) But y-intercept, c = \(\frac{2}{7}\) Therefore, m = \(\frac{4}{3}\times\frac{2}{7} = \frac{8}{21}\) Equation of straight: y = mx + c y = \(\frac{8}{21}x + \frac{2}{7}\) OR 21y = 8x + 6 = 8x - 21y = - 6

Using Pythagoras theorem ⇒ \((20\sqrt2)2 = x^2 + x^2\) ⇒ 800 = \(2x^2\) ⇒ 400 = \(x^2\) ⇒ x = \(\sqrt400\) = 20 cm ∴ Area of a square = \(x^2 = 20^2 = 400 cm^2\) ∴ Perimeter of a square = 4x = 4 x 20 = 80 cm

The sides have been given to the nearest meter, so 51.5 m \le length < 52.5 37.5 m \le width < 38.5 Minimum area = 37.5 x 51.5 = 1931.25 m\(^2\) Maximum area = 38.5 x 52.5 = 2021.25 m\(^2\) ∴ The range of the area = 1931.25 m\(^2\) \le A < 2021.25 m\(^2\)