Two rectangular picture frames are mathematically similar.
(a) The areas of the frames are 350 cm² and 1134 cm². The width of the smaller frame is 17.5 cm.
Calculate the width of the larger frame.
(b) A picture in the smaller frame has length 15 cm and width 10.5 cm, both correct to the nearest 5 mm.
Calculate the upper bound for the area of this picture.
(c) In a sale, the price of a large frame is reduced by 18%. Parthi pays $166.05 for 5 large frames in the sale.
Calculate the original price of one large frame.
(d) Parthi advertises a large frame for a price of $57 or 48.20 euros. The exchange rate is $1 = 0.88 euros.
Calculate the difference between these prices, in dollars and cents, correct to the nearest cent.
▶️ Answer/Explanation
(a) 31.5 cm
Area ratio = 1134/350 = 3.24. Linear ratio = √3.24 = 1.8. Larger width = 17.5 × 1.8 = 31.5 cm.
(b) 163.9375 cm²
Upper bounds: length = 15.25 cm, width = 10.75 cm. Max area = 15.25 × 10.75 = 163.9375 cm².
(c) $40.50
Sale price per frame = $166.05 ÷ 5 = $33.21. Original price = $33.21 ÷ 0.82 = $40.50.
(d) $2.23
Euro price in dollars = 48.20 ÷ 0.88 ≈ $54.77. Difference = $57 – $54.77 = $2.23.
(a) Find the size of an exterior angle of a regular polygon with 18 sides.
(b) 
In triangle ACD, B lies on AC and E lies on AD such that BE is parallel to CD.
AE = 5.2 cm and ED = 2.6 cm.
Calculate BE.
(c) Two solids are mathematically similar.
The smaller solid has height 2 cm and volume 32 cm³.
The larger solid has volume 780 cm³.
Calculate the height of the larger solid.
(d) 
PQ is parallel to RS, PNS is a straight line and N is the midpoint of RQ.
Explain, giving reasons, why triangle PQN is congruent to triangle SRN.
▶️ Answer/Explanation
(a) 20°
For any regular polygon, the exterior angle is 360° divided by number of sides. Here, 360° ÷ 18 = 20°.
(b) 4.5 cm
Since BE is parallel to CD, triangles ABE and ACD are similar. The ratio AE:AD is 5.2:(5.2+2.6) = 2:3. Therefore, BE:CD = 2:3, so BE = (2/3)×6.75 = 4.5 cm.
(c) 5.80 cm
For similar solids, the ratio of volumes is the cube of the ratio of heights. (780/32) = (h/2)³. Solving gives h = 2׳√(780/32) ≈ 5.80 cm.
(d)
1. QN = NR (given as N is midpoint)
2. Angle PQN = angle SRN (alternate angles as PQ∥RS)
3. Angle PNQ = angle SNR (vertically opposite angles)
Therefore, triangles are congruent by ASA (Angle-Side-Angle) condition.