{"id":4357,"date":"2023-02-09T06:58:43","date_gmt":"2023-02-09T06:58:43","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4357"},"modified":"2023-02-09T07:00:48","modified_gmt":"2023-02-09T07:00:48","slug":"ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-5","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-5\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 7 Triangles Ex 7.5"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 7 (Triangles)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 7 Triangles <\/strong>Exercise 7.5 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 7: Triangles<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.4<\/a><\/li>\n<\/ul>\n<div>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 7.5<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1.\u00a0ABC is a triangle. Locate a point in the interior of \u2206ABC which is equidistant from all the vertices of \u2206ABC.<br \/>\nAnswer &#8211; <\/strong>Let us consider a \u2206ABC. <\/span><br \/>\n<span style=\"color: #000000;\">Draw l, the perpendicular bisector of AB.<\/span><br \/>\n<span style=\"color: #000000;\">Draw m, the perpendicular bisector of BC.<\/span><br \/>\n<span style=\"color: #000000;\">Let the two perpendicular bisectors l and m meet at O.<\/span><br \/>\n<span style=\"color: #000000;\">O is the required point which is equidistant from A, B and C.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4422\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"235\" height=\"223\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><em><strong>Note:<\/strong> <\/em>If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.<br \/>\nAnswer &#8211;<\/strong> Let us consider a \u2206ABC.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4423\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Ans2a.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"197\" height=\"158\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Draw m, the bisector of \u2220C.<\/span><br \/>\n<span style=\"color: #000000;\">Let the two bisectors l and m meet at O.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, O is the required point which is equidistant from the sides of \u2206ABC.<\/span><br \/>\n<span style=\"color: #000000;\">Note: If we draw OM \u22a5 BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle. Point O is called incentre of the triangle.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4424\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Ans2b.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"200\" height=\"163\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In a huge park, people are concentrated at three points (see figure)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4425\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"258\" height=\"234\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Q3.png 334w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Q3-300x272.png 300w\" sizes=\"auto, (max-width: 258px) 100vw, 258px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>A: <\/strong>where these are different slides and swings for children.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>B: <\/strong>near which a man-made lake is situated.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>C: <\/strong>which is near to a large parking and exist.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?<br \/>\n[Hint The parlour should be equidistant from A, B and C.]<br \/>\nAnswer &#8211;\u00a0 <\/strong>Let us join A and B, and draw l, the perpendicular bisector of AB.<\/span><br \/>\n<span style=\"color: #000000;\">Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O.<\/span><br \/>\n<span style=\"color: #000000;\">The point O is the required point where the ice cream parlour be set up.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4426\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"172\" height=\"178\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Complete the hexagonal and star shaped Rangolis\u2019 [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4427\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"523\" height=\"343\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Q4.png 523w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Q4-300x197.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Q4-480x315.png 480w\" sizes=\"auto, (max-width: 523px) 100vw, 523px\" \/><br \/>\nAnswer &#8211;\u00a0<\/strong>We first divide the hexagon into six equilateral triangles of side 5cm as follow.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4428\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Ans4i.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"158\" height=\"125\" \/><\/span><br \/>\n<span style=\"color: #000000;\">We take one triangle from six equilateral triangles as shown above and make as many equilateral triangles of one side 1 cm as shown in the figure.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4429\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.5-Ans4ii.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"162\" height=\"141\" \/><\/span><br \/>\n<span style=\"color: #000000;\">The number of equilateral triangles of side 1 cm = 1 + 3 + 5 + 7 + 9 = 25 <\/span><br \/>\n<span style=\"color: #000000;\">So, the total number of triangles in the hexagon = 6 \u00d7 25 = 150<\/span><br \/>\n<span style=\"color: #000000;\">To find the number of triangles in the Fig. (ii), we adopt the same procedure.<\/span><br \/>\n<span style=\"color: #000000;\">So, the number of triangles in the Fig. (ii) = 12 \u00d7 25 = 300 <\/span><br \/>\n<span style=\"color: #000000;\">Hence, Fig. (ii) has more triangles.<\/span><\/p>\n<p style=\"text-align: justify;\">\n<\/div>\n<div><\/div>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 7 (Triangles)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 7 Triangles Exercise 7.5 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 7: Triangles NCERT Solution Class 9 Maths Ex &#8211; 7.1 NCERT Solution Class 9 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[713,702,186,714,701,5],"class_list":["post-4357","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-7-triangles-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-7-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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