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Let x represent total vol. 2 : 3 = 2 + 3 = 5 \(\frac{3}{5}\)x = 351 x = \(\frac{351 \times 5}{3}\) = 585 Volume of smaller block = \(\frac{2}{5}\) x 585 = 234.00cm\(^3\)

The diagonal of a rhombus is a line segment that joins any two non-adjacent vertices. A rhombus has two diagonals that bisect each other at right angles. i.e this splits 6cm into 3cm each AND 8cm to 4cm Using Hyp\(^2\) = adj\(^2\) + opp\(^2\) Hyp\(^2\) = 3\(^2\) + 4\(^2\) Hyp\(^2\) = 25 Hyp = 5 ∴ Length (L) is 5cm because a rhombus is a quadrilateral with 4 equal lengths

(a) Area of △PQR = \(\frac{1}{2}\) * b * 12 = 20 b = \(\frac{2 x 20}{12}\) = 3.33cm |QR| = |UT| = 3.33cm IPQ| = \sqrt(12\(^2\) + 5\(^2\))\(^2\) = 13cm IRT| = |QU| = 12cm From △RTS sin 50^\circ= \(\frac{12}{|RS|}\) |RS| = \(\frac{12}{sin 50^\circ}\) = 15.6649 cm |TS| = \(\frac{12}{tan 50^\circ}\) = 10.069cm Perimeter = 13 + 3.33 + 15.6649 + 18.399 = 50.3939cm = 50cm (nearest whole number) (b)Area= \(\frac{1}{2}\) (3.33+ 18.4004)12 6 (21.7304) = 130.3824cm\(^2\) = 130 cm\(^2\) (nearest whole number)

Area of Trapezium = 1/2(sum of parallel sides) * h 91 = \(\frac{1}{2}\) (5 + 9)h cross multiply 91 = 7h h = \(\frac{91}{7}\) h = 13cm

Length of Arc AB = \(\frac{\theta}{360}\) 2\(\pi\)r = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\) = \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm Perimeter = 8.8 + 2r = 8.8 + 2(10.5) = 8.8 + 21 = 29.8cm

Perimeter of a Rhombus = 4L or 2\sqrt( d1\(^2\) + d2\(^2\)) P = 2\sqrt(12\(^2\) + 5\(^2\)) P = 2\sqrt(144 + 25) P = 2\sqrt169 = 2 * 13 P = 26cm

Where l\(^2\) = h\(^2\) + r\(^2\) l\(^2\) = 11\(^2\) + 8\(^2\) l = \sqrt185 l = 13.60cm The formula of CSA of Cone is \pirl \(\frac{22}{7}\) * 8 * 13.60 = 341.979 or 341.98 (2d.p)

Volume of a cylinder = \pir\(^2\)h First convert 3m to cm by multiplying by 100 Volume of External cylinder = \(\pi \times 13^2 \times 300\) Volume of Internal cylinder = \(\pi \times 10^2 \times 300\) Hence; Volume of External cylinder - Volume of Internal cylinder Total volume (v) = \(\pi (169 - 100) \times 300\) V = \(\pi \times 69 \times 300\) V = 20700\picm\(^3\)

Angle at the centre O is twice the angle at the circumference. i.e O = 72^\circ * 2 = 144^\circ ∠OPS = 180 - 144 = 36^\circ

the form y=mx+c where m is the gradient and c is the y-intercept. the equation to gives 5y=3x - 7. divide both side by 5 the coefficient of y → \(\frac{3}{5}\)x - \(\frac{7}{5}\) comparing this with the general equation y=mx+c, you can see that m= the gradient= \(\frac{3}{5}\).

Each interior angle = \(\frac{(n - 2)180}{n}\) 140 = \(\frac{(n - 2)180}{n}\) Cross multiply: 140n = 180n - 360 40n = 360 n = 9 sides ( A Nonagon)

Sum of all 4 angles of a quadrilateral = 360^\circ (5x-30) + (3x + 61) + (60-x) + (4x+ 60) = 360^\circ 11x + 151 = 360^\circ 11x = 360 - 151 = 209 x = 209/11 = 19^\circ Each angles is : 5x - 30 = 65^\circ 4x+ 60 = 136^\circ 60 - x =41^\circ 3x + 61 = 118^\circ Smallest of these angles is 41^\circ

In a polygon with n sides, the sum of the angles = (n - 2)180, where n = the number of sides. In our problem, n = k. So, we have: (3k - 10)90 = (k - 2)180 270k - 900 = 180k - 360 Simplifying: 90k = 540 k = 6. So, we have a regular hexagon. Now, each exterior angle = 360/n We have: 360/6 = 60 Conclusion: Each exterior angle is 60 degrees.

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

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