Measuring Space: Perimeter and Area Worksheet

(b) Doubles

1. Circumference C = 2πr.
2. Replacing r by 2r multiplies C by 2.

(b) Cyclic quadrilateral

1. The formula √((s − a)(s − b)(s − c)(s − d)) applies to a cyclic quadrilateral.
2. The cyclicity condition is essential.

(a) 3π cm²

1. The sector is 120/360 = 1/3 of the circle.
2. Area = (1/3) × π × 3² = 3π cm².

84 cm².

1. The semiperimeter is s = (13 + 14 + 15)/2 = 21 cm.
2. Area = √(s(s − a)(s − b)(s − c)) = √(21 × 8 × 7 × 6).
3. The product is 7056 and its square root is 84 cm².

48 cm².

1. Semiperimeter s = 16; Heron gives √(16 × 6 × 6 × 4) = 48.
2. The altitude bisects the 12 cm base into two 6 cm parts.
3. Pythagoras gives height √(10² − 6²) = 8 cm; area = (1/2) × 12 × 8 = 48 cm².

π is irrational; the arc length is approximately 22 cm.

1. For any circle C/d = π; π cannot equal any rational fraction exactly.
2. Historical approximations include Archimedes’ bounds 223/71 < π < 22/7, Aryabhata’s 3.1416, and Zu Chongzhi’s 355/113.
3. An arc of central angle θ has length (θ/360°) × 2πr.
4. For θ = 90° and r = 14, using 22/7 gives (1/4) × 2 × (22/7) × 14 = 22 cm. A right-angle arc is one quarter of the circumference.

Area 42 cm²; each half has area 21 cm².

1. Triangle area is half the area of a parallelogram with the same base and perpendicular height: A = bh/2.
2. For the given values, A = 12 × 7/2 = 42 cm².
3. A median bisects the base. The two smaller triangles have equal bases and the same altitude, so their areas are equal, each 21 cm².
4. A rectangle has area length × width. A parallelogram can be cut and rearranged into a rectangle, giving base × perpendicular height.

Circle area πr²; sector area 6π cm².

1. Cut the circle into many narrow equal sectors and alternate them to approximate a rectangle/parallelogram.
2. As the sectors become thinner, the limiting base is half the circumference, πr, and height is r.
3. Thus the limiting area is πr × r = πr².
4. A 60° sector is 1/6 of the circle, so its area is (1/6) × π × 6² = 6π cm².

The square side is √(ab), so its area is ab.

1. On a line mark X, Y, Z with XY = a and YZ = b. Draw a semicircle with diameter XZ.
2. At Y draw a perpendicular meeting the semicircle at P. Let O be the midpoint of XZ.
3. OP = (a + b)/2 and OY = |a − b|/2. Pythagoras gives YP² = OP² − OY² = ab.
4. Construct a square on side YP. Its area equals the rectangle’s. This is an equivalent geometric justification of squaring, not a claim that every historical construction uses this exact semicircle diagram.

Area 6√10 cm²; the limiting formula is √(s(s − a)(s − b)(s − c)).

1. For the quadrilateral s = (3 + 4 + 5 + 6)/2 = 9.
2. Area = √((9 − 3)(9 − 4)(9 − 5)(9 − 6)) = √360 = 6√10 cm².
3. If d tends to 0, the semiperimeter tends to (a + b + c)/2 and the factor s − d tends to s.
4. The remaining three factors give Heron’s formula. A zero-length side itself is a degenerate boundary case, not a genuine quadrilateral.

Minor segment area 25π − 50 cm²; arc length 5π cm.

1. The 90° sector area is π × 10²/4 = 25π cm².
2. The two radii form a right triangle of area (1/2) × 10 × 10 = 50 cm².
3. Subtract the triangle from the sector to get the minor segment: 25π − 50 cm².
4. The arc is one quarter of the circumference: (1/4) × 2π × 10 = 5π cm.

Outer perimeter 48 m; path area 44 m².

1. Add twice the width to each dimension: outer dimensions 14 m × 10 m.
2. Outer perimeter = 2(14 + 10) = 48 m.
3. Subtract areas: path area = 14 × 10 − 12 × 8 = 140 − 96 = 44 m².
4. The algorithm forms the complete outer rectangle and subtracts the inner one once; this automatically includes each corner exactly once.

Area 60 cm².

1. A rectangle has supplementary opposite angles, so it is cyclic.
2. The four sides are 5,12,5,12 and semiperimeter s = 17.
3. Brahmagupta gives √((17 − 5)(17 − 12)(17 − 5)(17 − 12)) = √3600 = 60.
4. This agrees with length × width = 12 × 5 = 60 cm²; cyclicity justifies the formula.