Area and Perimeter of Triangle Worksheet PDF<\/h2>\n
I. <\/strong>Find the area and perimeter of the triangle whose sides are ii. 7 cm, 9cm, 11cm? Given sides of the triangle=4cm, 6cm, 8cm II. <\/strong>The three sides of a triangle are in the ratio 3:5:7 and the perimeter 345m. Find its area? Given that, III. <\/strong>Find the height of a triangle whose base is 30 cm and whose area is 600 sq cm? Given that, IV. <\/strong>Find the area of an isosceles right-angled triangle of equal sides 12 cm each? We know that area of the triangle=1\/2 \u00d7 b \u00d7 h V. <\/strong>The area of the triangle is equal to that of a square whose each side measures 20 cm. Find the base of the triangle whose corresponding altitude is 24 cm?<\/p>\n Given that, VI. <\/strong>The sides of a triangle are in the ratio 8: 5: 7, and its perimeter is 600 cm. Find the area. Also, find the altitude corresponding to the smallest side?<\/p>\n Given that, VII. <\/strong>The length of one of the diagonals of a field in the form of a quadrilateral is 84 m. The perpendicular distance of the other two vertices from the diagonal is 15 m and 12 m, find the area of the field?<\/p>\n Given that, VIII. <\/strong>The perimeter of a triangle is 68 cm. The first side is 7 cm shorter than the second side. The third side is 5 cm shorter than four times the length of the first side. What is the length of each side?<\/p>\n Given that, IX. <\/strong>If the height of a triangle is tripled and the base is doubled then find its area? Let the height of a triangle = h X. <\/strong>Find the area of a right-angled triangle whose hypotenuse is 5 cm and one of its sides containing the right angle is 3 cm. Find the length of the other side? Given that, XI. <\/strong>A piece of string is 60 cm long. What will be the length of each side of the string is used to form an equilateral triangle and find its area?<\/p>\n The string length is the perimeter of the shape. XII. <\/strong>Find the area of a triangle having \u2026\u2026\u2026\u2026<\/p>\n i. base = 16 cm height = 14 cm <\/p>\n","protected":false},"excerpt":{"rendered":" Test your knowledge on the concept of Area and Perimeter of Triangles in an interesting way by making the most out of our Math Worksheet on Area and Perimeter of Triangle. Find the PDF Formatted Area and Perimeter of Triangle Worksheets for free and resolve all your queries. Area and Perimeter of Triangles Word Problems … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/143357"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=143357"}],"version-history":[{"count":2,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/143357\/revisions"}],"predecessor-version":[{"id":143379,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/143357\/revisions\/143379"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=143357"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=143357"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=143357"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-1.png\")
\ni. 4cm, 6cm, 8cm?<\/p>\n
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-2.png\")
<\/p>\nSolution:<\/strong><\/summary>\n
\nThe perimeter of the triangle=a+b+c where a,b, and c are the sides of the triangle.
\nThe perimeter of the triangle=4 cm + 6 cm + 8 cm
\n=18 cm
\nTherefore, the perimeter of the triangle is 18 cm.
\nii. Given sides of the triangle=7 cm, 9cm, 11cm
\nWe know that the perimeter of the triangle=a+b+c where a,b, and c are the sides of the triangle.
\nThe perimeter of the triangle=7 cm + 9 cm + 11 cm
\n=27 cm
\nTherefore, the perimeter of the triangle is 27 cm.<\/p>\n<\/details>\n
\n
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-3.png\")
<\/p>\nSolution:<\/strong><\/summary>\n
\nThe three sides of a triangle are in the ratio of 3:5:7.
\nPerimeter=345 m
\nLength of first side =3x
\nLength of second side =5x
\nLength of third side =7x
\nThe perimeter of the triangle=side1+side2+side3
\n345 m=3x + 5x + 7x
\n345 m=15x
\nx=345 m\/15=23 m
\nLength of first side=3x=3(23 m)=69 m
\nLength of the second side=5x=5(23 m)=115 m
\nLength of the third side=7x=7(23 m)=161 m
\nArea of the triangle=\\(\\sqrt{ s(s-a)(s-b)(s-c) }\\)
\nS =perimeter of triangle \/2 = 345 m\/2 =172.5 m
\n=\\(\\sqrt{172.5(172.5-69)(172.5-115)(172.5-161) }\\)
\n=\\(\\sqrt{172.5(103.5)(57.5)(11.5)\u00a0 }\\)
\n=\\(\\sqrt{11805792.1875 }\\)
\n=3435.95 sq m
\nTherefore, Area of the triangle is 3435.95 sq m.<\/p>\n<\/details>\n
\n
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-4.png\")
<\/p>\nSolution:<\/strong><\/summary>\n
\nBase=30 cm, Area=600 sq cm
\nWe know that Area= 1\/2 \u00d7 b \u00d7 h
\n600 sq cm=1\/2 \u00d7 30 cm \u00d7 h
\nh=600 \u00d7 2 \/30
\n=1200\/30
\n=40 cm
\nTherefore, the Height is 40 cm.<\/p>\n<\/details>\n
\n
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-5.png\")
<\/p>\nSolution:<\/strong><\/summary>\n
\n=1\/2 \u00d7 12 \u00d7 12
\n=72 sq cm
\nTherefore, the Area of an isosceles right-angled triangle is 72 sq cm.<\/p>\n<\/details>\n
\nSolution:<\/strong><\/summary>\n
\nThe area of the triangle is equal to the area of a square.
\nSide of a square=20 cm
\nArea of a square=side \u00d7 side
\n=20 \u00d7 20=400 sq cm
\nTherefore, the Area of triangle=400 sq cm.
\nwe know that area of triangle=1\/2 \u00d7 b \u00d7 corresponding altitude
\n400=1\/2 \u00d7 b \u00d7 24
\nb=400\/12
\nb=33.33
\nTherefore, the base of the triangle is 33.33 cm.<\/p>\n<\/details>\n
\nSolution:<\/strong><\/summary>\n
\n<\/strong>sides of triangle are in ratio of 8:5:7
\nperimeter of triangle = 600
\n8x + 5x + 7x = 600
\n20x = 600
\nx = 600\/20
\nx = 30
\nTherefore the sides are
\n8x = 8 \u00d7 30 = 240
\n5x = 5 \u00d7 30 = 150
\n7x = 7 \u00d7 30 = 210
\nArea of triangle with sides given we use herons formula
\narea = \\(\\sqrt{s(s-a)(s-b)(s-c)\u00a0 }\\)
\ns = (a+b+c)\/2 where a, b, c are the sides of triangle
\n=(240+150+210)\/2
\n=600\/2=300
\narea=\\(\\sqrt{ 300(300 – 240)(300 – 150)(300 – 210) }\\)
\n=\\(\\sqrt{300(60)(150)(90)}\\)
\n=\\(\\sqrt{243000000}\\)
\n=15588.45 sq cm
\nAltitude corresponding to smallest side is 150
\n15588.45=1\/2 \u00d7 150 \u00d7 h
\nh=31176.9\/150
\nh=207.84 cm
\nHence, height is 207.84 cm.<\/p>\n<\/details>\n
\nSolution:<\/strong><\/summary>\n
\nThe length of diagonal d=84m
\nThe perpendicular distance of the other two vertices from the diagonal h1 = 15 m and
\nh2=12 m
\nWe know that Area of the quadrilateral=1\/2\u00d7d(h1+h2)
\n=1\/2\u00d7 84(15+12)
\n=42(27)
\n=1134 sq m
\nHence, the Area of the field is 1134 sq m.<\/p>\n<\/details>\n
\nSolution:<\/strong><\/summary>\n
\nThe perimeter of a triangle is= 68 cm
\nlet X = first side
\nY = second side
\nZ = third side
\nX = Y – 7 (1)
\nZ = 4X – 5 (2)
\n68 = X + Y + Z (3)
\nuse (1) (2) into (3)
\n68 = ( Y – 7 ) + Y + ( 4X – 5 )
\n68 = Y – 7 + Y + 4X – 5
\n68 + 7 + 5 = 2Y + 4X
\n80 = 2(Y + 2X)
\ny+2x=40
\ny=40-2x–\u00a0(4)
\nsubstitute (4) to (1)
\nX = (40 – 2X) – 7
\nX = 33 – 2X
\n3X = 33
\nX = 11 cm
\nlength of first side
\nuse X in (1)
\n11 = Y – 7 ; Y = 18 cm
\nuse X in (2)
\nZ = 4(11) -5
\nZ =39 cm
\nHence, x=11cm, y=18 cm,z=39 cm.<\/p>\n<\/details>\n
\n
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-6.png\")
<\/p>\nSolution:<\/strong><\/summary>\n
\nThe base of a triangle = b
\n\u2234 The area of a triangle =1\/2bh ……. (1)
\nThe new height of a triangle = 3h
\nThe new base of a triangle = 2b
\nThe area of a new triangle=1\/2(3h)(2b)
\n=6(1\/2hb)
\nUsing equation (1), we get
\n= 6 \u00d7 The area of a triangle
\nThus, the area of the triangle becomes “6 times”.<\/p>\n<\/details>\n
\n
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-7.png\")
<\/p>\nSolution:<\/strong><\/summary>\n
\nHypotenuse AC=5 cm
\none of the side containing the right angle BC=3 cm
\nLet the other side AB be= x cm
\nBy pythagerous theorem,
\nAC2<\/sup>=AB2<\/sup>+BC2<\/sup>
\n(x)2<\/sup>+(3)2<\/sup>= (5)2<\/sup>
\n(x)2<\/sup>+9=25
\n(x)2<\/sup>=25-9
\nx= 4 cm
\nArea of triangle= 1\/2\u00d73\u00d74=6 sq cm.<\/p>\n<\/details>\n
\nSolution:<\/strong><\/summary>\n
\nperimeter=60cm
\nAn equilateral triangle has three equal sides.
\nPerimeter=3\u00d7(side)
\n\u2234side=perimeter\/3
\n\u200b=60\/3
\n=20cm
\nHence, the length of each side of the string is used to form an equilateral triangle will be 20 cm.
\nArea=\\(\\sqrt{ 3}\\)a2<\/sup>\/4
\n=\\(\\sqrt{ 3}\\) (20)2\/4
\n=\\(\\sqrt{ 3}\\) 400\/4
\n=100\\(\\sqrt{ 3}\\) sq cm
\nTherefore, The area of an equilateral triangle is 100\\(\\sqrt{ 3}\\) sq cm.<\/p>\n<\/details>\n
\n
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-8.png\")
\nii. base = 7.5 m height = 8 cm
\n![\"Worksheet](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Worksheet-on-Area-and-Perimeter-of-Triangle-9.png\")
\nSolution:
\n<\/strong>i. Given that,
\nbase = 16 cm height = 14 cm
\nWe know that area of the triangle=1\/2bh.
\nA=1\/2(16)(14)
\n=112
\nTherefore, the Area of the triangle is 112 sq cm.
\nii. Given that,
\nbase = 7.5 m height = 8 cm
\nwe know that 1m=100 cm.
\n7.5 m=7.5 \u00d7 100=750 cm.
\nWe know that area of the triangle=1\/2bh.
\nA=1\/2(750)(8)
\n=3000
\nTherefore, the Area of the triangle is 3000 sq cm.<\/p>\n