{"id":89218,"date":"2025-06-01T09:59:13","date_gmt":"2025-06-01T09:59:13","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89218"},"modified":"2025-06-01T09:59:13","modified_gmt":"2025-06-01T09:59:13","slug":"a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/","title":{"rendered":"A square is drawn inside the circle as shown in the figure above. If t"},"content":{"rendered":"

A square is drawn inside the circle as shown in the figure above. If the area of the shaded portion is 32\/7 units then the radius of the circle is :<\/p>\n

[amp_mcq option1=”\u221a2 units” option2=”2 units” option3=”3 units” option4=”4 units” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2009<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct option is B) 2 units.
\n<\/section>\n
\nThe problem involves a square inscribed in a circle. Let the radius of the circle be `r`. The diameter of the circle is `2r`. When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Let the side of the square be `s`. The diagonal of the square is `s\u221a2`. Thus, `s\u221a2 = 2r`, which means `s = 2r\/\u221a2 = r\u221a2`. The area of the square is `s\u00b2 = (r\u221a2)\u00b2 = 2r\u00b2`. The area of the circle is `\u03c0r\u00b2`. The shaded portion is the area of the circle minus the area of the square. Given the area of the shaded portion is 32\/7 units, we have `\u03c0r\u00b2 – 2r\u00b2 = 32\/7`. Factoring out `r\u00b2`, we get `r\u00b2(\u03c0 – 2) = 32\/7`. Using the approximation `\u03c0 \u2248 22\/7`, we have `r\u00b2(22\/7 – 2) = 32\/7`. This simplifies to `r\u00b2((22 – 14)\/7) = 32\/7`, which is `r\u00b2(8\/7) = 32\/7`. Multiplying both sides by 7\/8, we get `r\u00b2 = (32\/7) * (7\/8) = 32\/8 = 4`. Taking the square root, `r = \u221a4 = 2` (since radius must be positive). The radius of the circle is 2 units.
\n<\/section>\n
\nFor a square inscribed in a circle of radius `r`, the side length is `r\u221a2` and the area is `2r\u00b2`. The ratio of the area of the inscribed square to the area of the circle is `2r\u00b2 \/ (\u03c0r\u00b2) = 2\/\u03c0`. The shaded area calculation relies on the difference between the circle’s area and the square’s area.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A square is drawn inside the circle as shown in the figure above. If the area of the shaded portion is 32\/7 units then the radius of the circle is : [amp_mcq option1=”\u221a2 units” option2=”2 units” option3=”3 units” option4=”4 units” correct=”option2″] This question was previously asked in UPSC CAPF – 2009 Download PDFAttempt Online The … <\/p>\n

Detailed SolutionA square is drawn inside the circle as shown in the figure above. If t<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1085],"tags":[1462,1102],"class_list":["post-89218","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1462","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nA square is drawn inside the circle as shown in the figure above. If t<\/title>\n<meta name=\"description\" content=\"The correct option is B) 2 units. The problem involves a square inscribed in a circle. Let the radius of the circle be `r`. The diameter of the circle is `2r`. When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Let the side of the square be `s`. The diagonal of the square is `s\u221a2`. Thus, `s\u221a2 = 2r`, which means `s = 2r\/\u221a2 = r\u221a2`. The area of the square is `s\u00b2 = (r\u221a2)\u00b2 = 2r\u00b2`. The area of the circle is `\u03c0r\u00b2`. The shaded portion is the area of the circle minus the area of the square. Given the area of the shaded portion is 32\/7 units, we have `\u03c0r\u00b2 - 2r\u00b2 = 32\/7`. Factoring out `r\u00b2`, we get `r\u00b2(\u03c0 - 2) = 32\/7`. Using the approximation `\u03c0 \u2248 22\/7`, we have `r\u00b2(22\/7 - 2) = 32\/7`. This simplifies to `r\u00b2((22 - 14)\/7) = 32\/7`, which is `r\u00b2(8\/7) = 32\/7`. Multiplying both sides by 7\/8, we get `r\u00b2 = (32\/7) * (7\/8) = 32\/8 = 4`. Taking the square root, `r = \u221a4 = 2` (since radius must be positive). The radius of the circle is 2 units.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A square is drawn inside the circle as shown in the figure above. If t\" \/>\n<meta property=\"og:description\" content=\"The correct option is B) 2 units. The problem involves a square inscribed in a circle. Let the radius of the circle be `r`. The diameter of the circle is `2r`. When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Let the side of the square be `s`. The diagonal of the square is `s\u221a2`. Thus, `s\u221a2 = 2r`, which means `s = 2r\/\u221a2 = r\u221a2`. The area of the square is `s\u00b2 = (r\u221a2)\u00b2 = 2r\u00b2`. The area of the circle is `\u03c0r\u00b2`. The shaded portion is the area of the circle minus the area of the square. Given the area of the shaded portion is 32\/7 units, we have `\u03c0r\u00b2 - 2r\u00b2 = 32\/7`. Factoring out `r\u00b2`, we get `r\u00b2(\u03c0 - 2) = 32\/7`. Using the approximation `\u03c0 \u2248 22\/7`, we have `r\u00b2(22\/7 - 2) = 32\/7`. This simplifies to `r\u00b2((22 - 14)\/7) = 32\/7`, which is `r\u00b2(8\/7) = 32\/7`. Multiplying both sides by 7\/8, we get `r\u00b2 = (32\/7) * (7\/8) = 32\/8 = 4`. Taking the square root, `r = \u221a4 = 2` (since radius must be positive). The radius of the circle is 2 units.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T09:59:13+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A square is drawn inside the circle as shown in the figure above. If t","description":"The correct option is B) 2 units. The problem involves a square inscribed in a circle. Let the radius of the circle be `r`. The diameter of the circle is `2r`. When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Let the side of the square be `s`. The diagonal of the square is `s\u221a2`. Thus, `s\u221a2 = 2r`, which means `s = 2r\/\u221a2 = r\u221a2`. The area of the square is `s\u00b2 = (r\u221a2)\u00b2 = 2r\u00b2`. The area of the circle is `\u03c0r\u00b2`. The shaded portion is the area of the circle minus the area of the square. Given the area of the shaded portion is 32\/7 units, we have `\u03c0r\u00b2 - 2r\u00b2 = 32\/7`. Factoring out `r\u00b2`, we get `r\u00b2(\u03c0 - 2) = 32\/7`. Using the approximation `\u03c0 \u2248 22\/7`, we have `r\u00b2(22\/7 - 2) = 32\/7`. This simplifies to `r\u00b2((22 - 14)\/7) = 32\/7`, which is `r\u00b2(8\/7) = 32\/7`. Multiplying both sides by 7\/8, we get `r\u00b2 = (32\/7) * (7\/8) = 32\/8 = 4`. Taking the square root, `r = \u221a4 = 2` (since radius must be positive). The radius of the circle is 2 units.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/","og_locale":"en_US","og_type":"article","og_title":"A square is drawn inside the circle as shown in the figure above. If t","og_description":"The correct option is B) 2 units. The problem involves a square inscribed in a circle. Let the radius of the circle be `r`. The diameter of the circle is `2r`. When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Let the side of the square be `s`. The diagonal of the square is `s\u221a2`. Thus, `s\u221a2 = 2r`, which means `s = 2r\/\u221a2 = r\u221a2`. The area of the square is `s\u00b2 = (r\u221a2)\u00b2 = 2r\u00b2`. The area of the circle is `\u03c0r\u00b2`. The shaded portion is the area of the circle minus the area of the square. Given the area of the shaded portion is 32\/7 units, we have `\u03c0r\u00b2 - 2r\u00b2 = 32\/7`. Factoring out `r\u00b2`, we get `r\u00b2(\u03c0 - 2) = 32\/7`. Using the approximation `\u03c0 \u2248 22\/7`, we have `r\u00b2(22\/7 - 2) = 32\/7`. This simplifies to `r\u00b2((22 - 14)\/7) = 32\/7`, which is `r\u00b2(8\/7) = 32\/7`. Multiplying both sides by 7\/8, we get `r\u00b2 = (32\/7) * (7\/8) = 32\/8 = 4`. Taking the square root, `r = \u221a4 = 2` (since radius must be positive). The radius of the circle is 2 units.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T09:59:13+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/","name":"A square is drawn inside the circle as shown in the figure above. If t","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T09:59:13+00:00","dateModified":"2025-06-01T09:59:13+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct option is B) 2 units. The problem involves a square inscribed in a circle. Let the radius of the circle be `r`. The diameter of the circle is `2r`. When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Let the side of the square be `s`. The diagonal of the square is `s\u221a2`. Thus, `s\u221a2 = 2r`, which means `s = 2r\/\u221a2 = r\u221a2`. The area of the square is `s\u00b2 = (r\u221a2)\u00b2 = 2r\u00b2`. The area of the circle is `\u03c0r\u00b2`. The shaded portion is the area of the circle minus the area of the square. Given the area of the shaded portion is 32\/7 units, we have `\u03c0r\u00b2 - 2r\u00b2 = 32\/7`. Factoring out `r\u00b2`, we get `r\u00b2(\u03c0 - 2) = 32\/7`. Using the approximation `\u03c0 \u2248 22\/7`, we have `r\u00b2(22\/7 - 2) = 32\/7`. This simplifies to `r\u00b2((22 - 14)\/7) = 32\/7`, which is `r\u00b2(8\/7) = 32\/7`. Multiplying both sides by 7\/8, we get `r\u00b2 = (32\/7) * (7\/8) = 32\/8 = 4`. Taking the square root, `r = \u221a4 = 2` (since radius must be positive). The radius of the circle is 2 units.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-square-is-drawn-inside-the-circle-as-shown-in-the-figure-above-if-t\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"A square is drawn inside the circle as shown in the figure above. If t"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/89218","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=89218"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/89218\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=89218"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=89218"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=89218"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}