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From the diagram above B\(^2\) = 10\(^2\) - 6\(^2\) = 64 B = \(\sqrt{64}\) = 8 cm( B represent half the second diagonal not given) So, the second diagonal not given = 8 + 8 = 16cm. But, Area of a rhombus = \(\frac{1}{2}\) x both diagonals that is, \(\frac{1}{2}\) x d\(_1\) x d\(_2\) Area of a rhombus = \(\frac{1}{2}\) x 12 x 16 = 6 x 16 = 96cm\(^2\)

No expanded working has been attached for this question yet. The verified answer is Option B.

In right angled \(\bigtriangleup\)QPR tan 30o = \(\frac{10}{x + 8}\) (x + 8) tan 30 = 10 x + 8 = \(\frac{10}{0.5773}\) x +8 = 17.3 x = 17.3 - 8 x = 9.32

xo = 86 x 2(angle at centre = 2 x angle at circumference) = 172o

If < SPR = 2m then < SQR = 2m but < AQR was n n = 2m m = \(\frac{n}{2}\)

m + y + x = 180o(sum of < s on straight line) y = n(vertically opp. angle) m + n + x = 180o

< NPQ = < PQB = 50º(alt. < s) < PQB = 50º < PQR = < PQR < PQB + < QBR = 72º < QBR = < PQR - < PQB = 72º - 50º = 22º < NQR = 180 - < QBR = 180º - 22 = 158º, the bearing of R from Q = 158º

No expanded working has been attached for this question yet. The verified answer is Option A.

in the diagram; x = 180º - ( 129 + 37)º = 14º ( sum of angles in a △) z = x = 14º ( alternate angles are equal ) y = 180º - ( 112 + z )º = 14º ( sum of angles in a △) y = 180º - ( 112 + 14)º = 54º

Reflex o(angles at a point) i.e. 2x + 8x = 360o 10x = 360o x = \(\frac{360}{10}\) = 36o o < PRS = 144

No expanded working has been attached for this question yet. The verified answer is Option C.

o (angle in the same segment) o(sum of interior angles of a \(\bigtriangleup\)) i.e. o + 48o = 180 o but o

Sum of interior angles = 3602 i.e 3x + 2x + 2x + x + 4x = 360o 12x = 360 x = \(\frac{360}{12}\) x = 30o x + c = 180 30o + c = 180 c = 180 - 30 c = 150o

(a) < UST = 100^\circ (exterior angle of a cyclic quad) < SUT = 36^\circ (angle in alternate segment) \(\therefore < UST + < SUT + < STU = 180^\circ\) (sum of angles in a triangle) \(\therefore 100^\circ + 36^\circ + < STU = 180^\circ\) \(< STU = 180^\circ - 136^\circ = 44^\circ\) (b) \(\frac{y}{6} = \frac{2}{5}\) \(\therefore 5y = 2(6) = 12\) \(\therefore y = \frac{12}{5} = 2.4 cm\) \(\therefore XQ + QZ = XZ\) \(QZ = XZ - XQ\) = \(6 cm - 2.4 cm = 3.6 cm\) \(\therefore QZ = 3.6 cm\).

(a) Let the angle subtended at the centre by XY = \(\theta\) Length of chord = \(2r \sin \frac{\theta}{2}\) \(\implies 6 cm = 2 \times 5 \times \frac{\theta}{2}\) \(\sin \frac{\theta}{2} = \frac{6}{10} = 0.6\) \(\frac{\theta}{2} = \sin^{-1} (0.6) = 36.87^\circ\) \(\theta = 2 \times 36.87^\circ = 73.74^\circ\) (b) Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{73.74}{360} \times \frac{22}{7} \times 5^{2}\) = \(16.094 cm^{2}\) Area of triangle = \(\sqrt{s(s - a)(s - b)(s - c)}\) where \(s = \frac{a + b + c}{2}\) \( a = 5 cm ; b = 5 cm ; c = 6 cm \therefore s = \frac{5 + 5 + 6}{2} = \frac{16}{2} = 8\) \(\therefore Area = \sqrt{8(8 - 5)(8 - 5)(8 - 6)}\) = \(\sqrt{8(3)(3)(2)}\) = \(\sqrt{144}\) = \(12 cm^{2}\) Area of shaded portion = \(16.094 cm^{2} - 12 cm^{2} = 4.094 cm^{2}\)