{"id":89598,"date":"2025-06-01T10:07:52","date_gmt":"2025-06-01T10:07:52","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89598"},"modified":"2025-06-01T10:07:52","modified_gmt":"2025-06-01T10:07:52","slug":"the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/","title":{"rendered":"The sum of the base and altitude of a triangle is 30 cm. What is the m"},"content":{"rendered":"

The sum of the base and altitude of a triangle is 30 cm. What is the maximum possible area of such a triangle?<\/p>\n

[amp_mcq option1=”100 cm$^2$” option2=”110 cm$^2$” option3=”112.5 cm$^2$” option4=”120 cm$^2$” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2013<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nLet the base of the triangle be $b$ and the altitude be $h$. We are given that the sum of the base and altitude is 30 cm, i.e., $b + h = 30$.
\nThe area of a triangle is given by $A = \\frac{1}{2} \\times \\text{base} \\times \\text{altitude} = \\frac{1}{2}bh$.
\nFrom the given condition, $h = 30 – b$. Substitute this into the area formula:
\n$A = \\frac{1}{2} b (30 – b) = \\frac{1}{2} (30b – b^2) = 15b – \\frac{1}{2}b^2$.
\nThis is a quadratic function of $b$ in the form $Ab^2 + Bb + C$, where $A = -\\frac{1}{2}$, $B = 15$, and $C = 0$. Since the coefficient of $b^2$ is negative, the parabola opens downwards, and its maximum value occurs at the vertex.
\nThe vertex of a parabola $ax^2 + bx + c$ is at $x = -\\frac{B}{2A}$. In this case, $b = -\\frac{15}{2(-\\frac{1}{2})} = -\\frac{15}{-1} = 15$.
\nThe maximum area occurs when the base $b = 15$ cm.
\nWhen $b = 15$, the altitude $h = 30 – b = 30 – 15 = 15$ cm.
\nThe maximum possible area is $A = \\frac{1}{2} \\times 15 \\times 15 = \\frac{1}{2} \\times 225 = 112.5$ cm$^2$.
\n<\/section>\n
\nFor a fixed sum of two non-negative variables, their product is maximized when the two variables are equal. In this case, the area $\\frac{1}{2}bh$ is maximized when the product $bh$ is maximized, given $b+h=30$. The product $bh$ is maximized when $b=h$.
\n<\/section>\n
\nThis result can also be obtained using the AM-GM inequality. For non-negative numbers $b$ and $h$, the arithmetic mean is always greater than or equal to the geometric mean: $\\frac{b+h}{2} \\ge \\sqrt{bh}$.
\nGiven $b+h=30$, we have $\\frac{30}{2} \\ge \\sqrt{bh} \\implies 15 \\ge \\sqrt{bh}$. Squaring both sides gives $225 \\ge bh$. The maximum value of $bh$ is 225.
\nThe area $A = \\frac{1}{2}bh$, so the maximum area is $\\frac{1}{2} \\times 225 = 112.5$. Equality in AM-GM holds when $b=h$, confirming the previous result.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The sum of the base and altitude of a triangle is 30 cm. What is the maximum possible area of such a triangle? [amp_mcq option1=”100 cm$^2$” option2=”110 cm$^2$” option3=”112.5 cm$^2$” option4=”120 cm$^2$” correct=”option3″] This question was previously asked in UPSC CAPF – 2013 Download PDFAttempt Online Let the base of the triangle be $b$ and … <\/p>\n

Detailed SolutionThe sum of the base and altitude of a triangle is 30 cm. What is the m<\/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":[1467,1102],"class_list":["post-89598","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1467","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nThe sum of the base and altitude of a triangle is 30 cm. What is the m<\/title>\n<meta name=\"description\" content=\"Let the base of the triangle be $b$ and the altitude be $h$. We are given that the sum of the base and altitude is 30 cm, i.e., $b + h = 30$. The area of a triangle is given by $A = frac{1}{2} times text{base} times text{altitude} = frac{1}{2}bh$. From the given condition, $h = 30 - b$. Substitute this into the area formula: $A = frac{1}{2} b (30 - b) = frac{1}{2} (30b - b^2) = 15b - frac{1}{2}b^2$. This is a quadratic function of $b$ in the form $Ab^2 + Bb + C$, where $A = -frac{1}{2}$, $B = 15$, and $C = 0$. Since the coefficient of $b^2$ is negative, the parabola opens downwards, and its maximum value occurs at the vertex. The vertex of a parabola $ax^2 + bx + c$ is at $x = -frac{B}{2A}$. In this case, $b = -frac{15}{2(-frac{1}{2})} = -frac{15}{-1} = 15$. The maximum area occurs when the base $b = 15$ cm. When $b = 15$, the altitude $h = 30 - b = 30 - 15 = 15$ cm. The maximum possible area is $A = frac{1}{2} times 15 times 15 = frac{1}{2} times 225 = 112.5$ cm$^2$. For a fixed sum of two non-negative variables, their product is maximized when the two variables are equal. In this case, the area $frac{1}{2}bh$ is maximized when the product $bh$ is maximized, given $b+h=30$. The product $bh$ is maximized when $b=h$.\" \/>\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\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The sum of the base and altitude of a triangle is 30 cm. What is the m\" \/>\n<meta property=\"og:description\" content=\"Let the base of the triangle be $b$ and the altitude be $h$. We are given that the sum of the base and altitude is 30 cm, i.e., $b + h = 30$. The area of a triangle is given by $A = frac{1}{2} times text{base} times text{altitude} = frac{1}{2}bh$. From the given condition, $h = 30 - b$. Substitute this into the area formula: $A = frac{1}{2} b (30 - b) = frac{1}{2} (30b - b^2) = 15b - frac{1}{2}b^2$. This is a quadratic function of $b$ in the form $Ab^2 + Bb + C$, where $A = -frac{1}{2}$, $B = 15$, and $C = 0$. Since the coefficient of $b^2$ is negative, the parabola opens downwards, and its maximum value occurs at the vertex. The vertex of a parabola $ax^2 + bx + c$ is at $x = -frac{B}{2A}$. In this case, $b = -frac{15}{2(-frac{1}{2})} = -frac{15}{-1} = 15$. The maximum area occurs when the base $b = 15$ cm. When $b = 15$, the altitude $h = 30 - b = 30 - 15 = 15$ cm. The maximum possible area is $A = frac{1}{2} times 15 times 15 = frac{1}{2} times 225 = 112.5$ cm$^2$. For a fixed sum of two non-negative variables, their product is maximized when the two variables are equal. In this case, the area $frac{1}{2}bh$ is maximized when the product $bh$ is maximized, given $b+h=30$. The product $bh$ is maximized when $b=h$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:07:52+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=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"The sum of the base and altitude of a triangle is 30 cm. What is the m","description":"Let the base of the triangle be $b$ and the altitude be $h$. We are given that the sum of the base and altitude is 30 cm, i.e., $b + h = 30$. The area of a triangle is given by $A = frac{1}{2} times text{base} times text{altitude} = frac{1}{2}bh$. From the given condition, $h = 30 - b$. Substitute this into the area formula: $A = frac{1}{2} b (30 - b) = frac{1}{2} (30b - b^2) = 15b - frac{1}{2}b^2$. This is a quadratic function of $b$ in the form $Ab^2 + Bb + C$, where $A = -frac{1}{2}$, $B = 15$, and $C = 0$. Since the coefficient of $b^2$ is negative, the parabola opens downwards, and its maximum value occurs at the vertex. The vertex of a parabola $ax^2 + bx + c$ is at $x = -frac{B}{2A}$. In this case, $b = -frac{15}{2(-frac{1}{2})} = -frac{15}{-1} = 15$. The maximum area occurs when the base $b = 15$ cm. When $b = 15$, the altitude $h = 30 - b = 30 - 15 = 15$ cm. The maximum possible area is $A = frac{1}{2} times 15 times 15 = frac{1}{2} times 225 = 112.5$ cm$^2$. For a fixed sum of two non-negative variables, their product is maximized when the two variables are equal. In this case, the area $frac{1}{2}bh$ is maximized when the product $bh$ is maximized, given $b+h=30$. The product $bh$ is maximized when $b=h$.","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\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/","og_locale":"en_US","og_type":"article","og_title":"The sum of the base and altitude of a triangle is 30 cm. What is the m","og_description":"Let the base of the triangle be $b$ and the altitude be $h$. We are given that the sum of the base and altitude is 30 cm, i.e., $b + h = 30$. The area of a triangle is given by $A = frac{1}{2} times text{base} times text{altitude} = frac{1}{2}bh$. From the given condition, $h = 30 - b$. Substitute this into the area formula: $A = frac{1}{2} b (30 - b) = frac{1}{2} (30b - b^2) = 15b - frac{1}{2}b^2$. This is a quadratic function of $b$ in the form $Ab^2 + Bb + C$, where $A = -frac{1}{2}$, $B = 15$, and $C = 0$. Since the coefficient of $b^2$ is negative, the parabola opens downwards, and its maximum value occurs at the vertex. The vertex of a parabola $ax^2 + bx + c$ is at $x = -frac{B}{2A}$. In this case, $b = -frac{15}{2(-frac{1}{2})} = -frac{15}{-1} = 15$. The maximum area occurs when the base $b = 15$ cm. When $b = 15$, the altitude $h = 30 - b = 30 - 15 = 15$ cm. The maximum possible area is $A = frac{1}{2} times 15 times 15 = frac{1}{2} times 225 = 112.5$ cm$^2$. For a fixed sum of two non-negative variables, their product is maximized when the two variables are equal. In this case, the area $frac{1}{2}bh$ is maximized when the product $bh$ is maximized, given $b+h=30$. The product $bh$ is maximized when $b=h$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:07:52+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/","name":"The sum of the base and altitude of a triangle is 30 cm. What is the m","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:07:52+00:00","dateModified":"2025-06-01T10:07:52+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"Let the base of the triangle be $b$ and the altitude be $h$. We are given that the sum of the base and altitude is 30 cm, i.e., $b + h = 30$. The area of a triangle is given by $A = \\frac{1}{2} \\times \\text{base} \\times \\text{altitude} = \\frac{1}{2}bh$. From the given condition, $h = 30 - b$. Substitute this into the area formula: $A = \\frac{1}{2} b (30 - b) = \\frac{1}{2} (30b - b^2) = 15b - \\frac{1}{2}b^2$. This is a quadratic function of $b$ in the form $Ab^2 + Bb + C$, where $A = -\\frac{1}{2}$, $B = 15$, and $C = 0$. Since the coefficient of $b^2$ is negative, the parabola opens downwards, and its maximum value occurs at the vertex. The vertex of a parabola $ax^2 + bx + c$ is at $x = -\\frac{B}{2A}$. In this case, $b = -\\frac{15}{2(-\\frac{1}{2})} = -\\frac{15}{-1} = 15$. The maximum area occurs when the base $b = 15$ cm. When $b = 15$, the altitude $h = 30 - b = 30 - 15 = 15$ cm. The maximum possible area is $A = \\frac{1}{2} \\times 15 \\times 15 = \\frac{1}{2} \\times 225 = 112.5$ cm$^2$. For a fixed sum of two non-negative variables, their product is maximized when the two variables are equal. In this case, the area $\\frac{1}{2}bh$ is maximized when the product $bh$ is maximized, given $b+h=30$. The product $bh$ is maximized when $b=h$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-sum-of-the-base-and-altitude-of-a-triangle-is-30-cm-what-is-the-m\/#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":"The sum of the base and altitude of a triangle is 30 cm. 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