{"id":179,"date":"2026-10-23T18:57:01","date_gmt":"2026-10-23T18:57:01","guid":{"rendered":"https:\/\/quantumleapstar.com\/?p=179"},"modified":"2026-10-23T18:57:01","modified_gmt":"2026-10-23T18:57:01","slug":"lesson-45-areas-and-perimeters","status":"publish","type":"post","link":"https:\/\/quantumleapstar.com\/?p=179","title":{"rendered":"Lesson 45: Areas and Perimeters"},"content":{"rendered":"<p><strong>Lesson 45 \u00b7 GCSE Mathematics \u00b7 Higher focus<\/strong><\/p>\n<p>Area measures the space inside a two-dimensional shape; perimeter measures the distance around its boundary. This lesson develops both ideas through compound shapes, circles, sectors, segments and algebra, with worked examples and a complete downloadable lesson pack.<\/p>\n<h2>Area or perimeter?<\/h2>\n<p>Use length units such as cm or m for perimeter, and square units such as cm\u00b2 or m\u00b2 for area. A rectangle measuring 12 cm by 6 cm has perimeter 2(12 + 6) = <strong>36 cm<\/strong> and area 12 \u00d7 6 = <strong>72 cm\u00b2<\/strong>.<\/p>\n<p>Make units match before calculating. For a 1.4 m by 60 cm rectangle, convert 60 cm to 0.60 m: area = 1.4 \u00d7 0.60 = <strong>0.84 m\u00b2<\/strong>, and perimeter = 2(1.4 + 0.60) = <strong>4 m<\/strong>. Remember: 1 cm\u00b2 = 100 mm\u00b2 and 1 m\u00b2 = 10,000 cm\u00b2.<\/p>\n<h2>Essential area formulas<\/h2>\n<ul>\n<li>Rectangle: A = length \u00d7 width.<\/li>\n<li>Square: A = s\u00b2; perimeter = 4s.<\/li>\n<li>Triangle: A = \u00bdbh.<\/li>\n<li>Parallelogram: A = bh.<\/li>\n<li>Trapezium: A = \u00bd(a + b)h, where a and b are the parallel sides.<\/li>\n<li>Kite or rhombus with perpendicular diagonals: A = \u00bdd\u2081d\u2082.<\/li>\n<\/ul>\n<p>In these formulas, height means <strong>perpendicular height<\/strong>, not a sloping side. A triangle with base 10 cm and perpendicular height 7 cm has area \u00bd \u00d7 10 \u00d7 7 = <strong>35 cm\u00b2<\/strong>. A trapezium with parallel sides 8 cm and 14 cm and height 5 cm has area \u00bd \u00d7 (8 + 14) \u00d7 5 = <strong>55 cm\u00b2<\/strong>.<\/p>\n<h2>Compound shapes: split or subtract<\/h2>\n<p>For an L-shaped region, suppose the full width is 12 cm, full height 9 cm, top width 7 cm and lower right vertical side 4 cm. The missing corner measures (12 \u2212 7) cm by (9 \u2212 4) cm, or 5 cm by 5 cm.<\/p>\n<p><strong>Area = 12 \u00d7 9 \u2212 5 \u00d7 5 = 83 cm\u00b2.<\/strong><br \/>Trace the six outside edges for perimeter: 12 + 4 + 5 + 5 + 7 + 9 = <strong>42 cm<\/strong>. Internal dividing lines are not part of the perimeter. If the question asks for the total boundary of a region with a hole, include the hole\u2019s boundary too.<\/p>\n<h2>Circles and semicircles<\/h2>\n<p>For radius r and diameter d = 2r, circle area is <strong>\u03c0r\u00b2<\/strong> and circumference is <strong>2\u03c0r = \u03c0d<\/strong>. Keep \u03c0 in exact answers and round only at the end.<\/p>\n<p>A semicircle has area <strong>\u00bd\u03c0r\u00b2<\/strong>, but its perimeter is <strong>\u03c0r + 2r<\/strong>: half the circumference plus the straight diameter. For diameter 10 cm, r = 5 cm, so area = <strong>12.5\u03c0 cm\u00b2<\/strong> and perimeter = <strong>5\u03c0 + 10 cm<\/strong>.<\/p>\n<p>If a semicircle is attached to a 6 cm side of a 10 cm by 6 cm rectangle, its radius is 3 cm. Combined area = <strong>60 + 4.5\u03c0 cm\u00b2<\/strong>; outside perimeter = <strong>26 + 3\u03c0 cm<\/strong>. The shared diameter is internal.<\/p>\n<h2>Sectors and arc lengths<\/h2>\n<p>A sector with central angle \u03b8 degrees is \u03b8\/360 of a full circle:<\/p>\n<ul>\n<li>Sector area = (\u03b8\/360) \u00d7 \u03c0r\u00b2.<\/li>\n<li>Arc length = (\u03b8\/360) \u00d7 2\u03c0r.<\/li>\n<li>Sector perimeter = arc length + 2r.<\/li>\n<\/ul>\n<p>For radius 9 cm and angle 120\u00b0, the fraction is \u2153. Area = <strong>27\u03c0 cm\u00b2<\/strong>, arc length = <strong>6\u03c0 cm<\/strong> and perimeter = <strong>6\u03c0 + 18 cm<\/strong>.<\/p>\n<p>Reverse problems use the same formula. A sector of radius 12 cm and area 30\u03c0 cm\u00b2 satisfies (\u03b8\/360) \u00d7 144\u03c0 = 30\u03c0, giving <strong>\u03b8 = 75\u00b0<\/strong>. For a major sector, use the larger angle: if the minor angle is 80\u00b0, the major angle is 280\u00b0.<\/p>\n<h2>Higher extension: segments<\/h2>\n<p>A segment lies between a chord and its arc. For a minor central angle \u03b8, subtract the triangle from the sector:<\/p>\n<p><strong>Minor segment area = (\u03b8\/360)\u03c0r\u00b2 \u2212 \u00bdr\u00b2 sin \u03b8.<\/strong><\/p>\n<p>Use degree mode. With r = 8 cm and \u03b8 = 90\u00b0, sector area = 16\u03c0 cm\u00b2 and triangle area = 32 cm\u00b2. The minor segment is <strong>16\u03c0 \u2212 32 cm\u00b2<\/strong>, approximately <strong>18.27 cm\u00b2<\/strong>. The major segment is the rest of the circle: <strong>48\u03c0 + 32 cm\u00b2<\/strong>.<\/p>\n<p>Segment perimeter is arc length plus chord length, not two radii. In this example, the chord is 8\u221a2 cm, giving minor segment perimeter <strong>4\u03c0 + 8\u221a2 cm<\/strong>.<\/p>\n<h2>Algebra and reverse problems<\/h2>\n<p>A trapezium with parallel sides 9 cm and 15 cm and area 84 cm\u00b2 satisfies 84 = \u00bd(9 + 15)h = 12h, so <strong>h = 7 cm<\/strong>.<\/p>\n<p>For a rectangle of width x cm, length (x + 5) cm and area 84 cm\u00b2: x(x + 5) = 84, so x\u00b2 + 5x \u2212 84 = 0 and (x + 12)(x \u2212 7) = 0. Reject the negative length. The dimensions are <strong>7 cm by 12 cm<\/strong>, and perimeter is <strong>38 cm<\/strong>.<\/p>\n<p>A ring between concentric circles of radii 10 cm and 6 cm has area \u03c0(10\u00b2 \u2212 6\u00b2) = <strong>64\u03c0 cm\u00b2<\/strong>. Its total boundary length includes both circles: 20\u03c0 + 12\u03c0 = <strong>32\u03c0 cm<\/strong>.<\/p>\n<h2>Check your understanding<\/h2>\n<ol>\n<li>A parallelogram has base 13 cm, perpendicular height 6 cm and sloping side 8 cm. Find its area and perimeter.<\/li>\n<li>A trapezium has parallel sides 7 cm and 17 cm and area 96 cm\u00b2. Find its height.<\/li>\n<li>A semicircle has diameter 14 cm. Find its exact area and perimeter.<\/li>\n<li>A sector has radius 12 cm and angle 150\u00b0. Find its exact area, arc length and perimeter.<\/li>\n<\/ol>\n<details>\n<summary>Show practice answers<\/summary>\n<ol>\n<li>Area = 78 cm\u00b2; perimeter = 42 cm.<\/li>\n<li>Height = 8 cm.<\/li>\n<li>Area = 24.5\u03c0 cm\u00b2; perimeter = 7\u03c0 + 14 cm.<\/li>\n<li>Area = 60\u03c0 cm\u00b2; arc length = 10\u03c0 cm; perimeter = 10\u03c0 + 24 cm.<\/li>\n<\/ol>\n<\/details>\n<h2>Download the Lesson 45 pack<\/h2>\n<p>Use the illustrated notes first, attempt the questions independently, then check your method against the worked answers.<\/p>\n<ul>\n<li><a href=\"https:\/\/quantumleapstar.com\/wp-content\/uploads\/2026\/10\/notes.pdf\">Illustrated lesson notes and practice solutions (PDF)<\/a><\/li>\n<li><a href=\"https:\/\/quantumleapstar.com\/wp-content\/uploads\/2026\/10\/QLS_45_Question_Paper_27_Questions.pdf\">27 Higher Tier past-paper questions \u2014 103 marks (PDF)<\/a><\/li>\n<li><a href=\"https:\/\/quantumleapstar.com\/wp-content\/uploads\/2026\/10\/QLS_45_Worked_Answers_27_Questions.pdf\">Separate fully worked answers (PDF)<\/a><\/li>\n<li><a href=\"https:\/\/quantumleapstar.com\/wp-content\/uploads\/2026\/10\/QLS_45_Areas_and_Perimeters_Presentation.pptx\">16-slide teaching presentation (PowerPoint)<\/a><\/li>\n<\/ul>\n<p>The past-paper set includes source references and original examination diagrams. Its coverage page identifies the selected papers; it is not an exhaustive archive of every board and session. The teaching examples and practice above are original QLS material.<\/p>\n<p><strong>Before you finish:<\/strong> check units, use perpendicular heights, distinguish radius from diameter, include every required boundary, reject negative lengths and round only at the end.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Lesson 45 \u00b7 GCSE Mathematics \u00b7 Higher focus Area measures the space inside a two-dimensional shape; perimeter measures the distance around its boundary. This lesson develops both ideas through compound<\/p>\n<p><a href=\"https:\/\/quantumleapstar.com\/?p=179\" class=\"more-link\">Continue reading<span class=\"screen-reader-text\">Lesson 45: Areas and Perimeters<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12],"tags":[],"class_list":["post-179","post","type-post","status-publish","format-standard","hentry","category-gcse-mathematics"],"_links":{"self":[{"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=\/wp\/v2\/posts\/179","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=179"}],"version-history":[{"count":1,"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=\/wp\/v2\/posts\/179\/revisions"}],"predecessor-version":[{"id":180,"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=\/wp\/v2\/posts\/179\/revisions\/180"}],"wp:attachment":[{"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=179"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=179"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/quantumleapstar.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=179"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}