Two rectangular picture frames are mathematically similar.
(a) The areas of the frames are \(350 \text{ cm}^{2}\) and \(1134 \text{ cm}^{2}\). 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) Ans: 31.5 cm
Since the frames are similar, the ratio of their areas is the square of the ratio of their corresponding sides. Let the width of the larger frame be \( w \). Then:
\[ \left(\frac{w}{17.5}\right)^2 = \frac{1134}{350} \implies \frac{w}{17.5} = \sqrt{\frac{1134}{350}} = 1.8 \implies w = 17.5 \times 1.8 = 31.5 \text{ cm}. \]
(b) Ans: 163.9375 cm²
The upper bounds for the length and width are 15.25 cm and 10.75 cm, respectively. The upper bound for the area is:
\[ 15.25 \times 10.75 = 163.9375 \text{ cm}^2. \]
(c) Ans: \$40.50
Let the original price of one large frame be \( P \). After an 18% discount, the sale price is \( 0.82P \). Parthi pays \$166.05 for 5 frames:
\[ 5 \times 0.82P = 166.05 \implies P = \frac{166.05}{5 \times 0.82} = 40.50 \text{ dollars}. \]
(d) Ans: \$2.23
Convert 48.20 euros to dollars: \( 48.20 \div 0.88 = 54.7727… \) dollars. The difference between \$57 and \$54.77 is:
\[ 57 – 54.7727… \approx 2.23 \text{ dollars}. \]
(a) A rectangle measures 8.5 cm by 10.7 cm, both correct to 1 decimal place.
Calculate the upper bound of the perimeter of the rectangle.
(b) 
ABDF is a parallelogram and BCDE is a straight line.
AF = 12 cm, AB = 9 cm, angle CFD = 40° and angle FDE = 80°.
(i) Calculate the height, h, of the parallelogram.
(ii) Explain why triangle CDF is isosceles.
(iii) Calculate the area of the trapezium ABCF.
(c) 
A, B, C and D are points on the circle, centre O.
Angle ABD = 21° and CD = 12 cm.
Calculate the area of the circle.
(d) 
The diagram shows a square with side length 8 cm and a sector of a circle with radius 9.5 cm and sector angle x°.
The perimeter of the square is equal to the perimeter of the sector.
Calculate the value of x.
▶️ Answer/Explanation
(a) Ans: 38.6 cm
Upper bounds: length = 10.75 cm, width = 8.55 cm.
Perimeter = 2 × (10.75 + 8.55) = 38.6 cm.
(b)(i) Ans: 8.86 cm
Using sine rule: \( h = AF \times \sin(40°) = 12 \times 0.6428 ≈ 8.86 \).
(b)(ii) Ans: Two equal angles (40° each)
∠CDF = 100° ⇒ ∠DCF = 40° (angles in triangle).
Thus, triangle CDF is isosceles.
(b)(iii) Ans: 66.5 cm²
Area of trapezium ABCF = \(\frac{1}{2} \times (AB + CF) \times h = \frac{1}{2} \times (9 + 12) \times 8.86 ≈ 66.5\).
(c) Ans: 130 cm²
Radius \( r = \frac{CD}{2 \sin(21°)} ≈ 17.2 \).
Area = \( \pi r^2 ≈ 130 \, \text{cm}^2 \).
(d) Ans: 78.4°
Perimeter of square = 32 cm.
Sector perimeter: \( 2 \times 9.5 + \frac{x}{360} \times 2 \pi \times 9.5 = 32 \).
Solve for \( x ≈ 78.4° \).