8th Class Maths 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers<\/h3>\n
Do this<\/span><\/p>\n Question 1. ii) l = 6 cm, b = 4 cm, h = 2 cm. Question 2. Question 3. 2) l = 32 cm, b = 2 cm, h = 1 cm. 3) l = 16 cm, b = 4 cm, h = 1 cm. No, the volume of a cuboid is not same as the surface area of a cuboid.<\/p>\n Try These<\/span><\/p>\n Question 1. Question 2. Question 3. Question 4. Think, Discuss and Write<\/span><\/p>\n Question 1. Question 2. Question 3. AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions AP State Syllabus\u00a0AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers. 8th Class Maths 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers Do this Question 1. Find the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/7973"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=7973"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/7973\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=7973"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=7973"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=7973"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nFind the total surface area of the following cuboid.\u00a0 \u00a0 \u00a0 [Page No. 298]
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-1.png\")
\nAnswer:
\ni) l = 4 cm, b = 4 cm, h = 10 cm.
\nThe total surface area of a cuboid = 2 (lb + bh + lh)
\n= 2 (4 \u00d7 4 + 4 \u00d7 10 + 4 \u00d7 10) = 2(16 + 40 + 40)
\n= 2 \u00d7 96
\n= 192 Sq. cms.<\/p>\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\")
<\/p>\n
\nThe total surface area of a cuboid = 2 (lb + bh + lh)
\n= 2(6 \u00d7 4 + 4 \u00d7 2 + 6 \u00d7 2)
\n= 2 (24 + 8 + 12)
\n= 2 \u00d7 44
\n= 88 sq. cms.<\/p>\n
\nLet us find the volume of a cuboid whose length, breadth and height are 6 cm, 4 cm and 5 cm respectively.\u00a0 \u00a0 \u00a0 [Page No. 287]
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-2.png\")
\nLet place 1 cubic centimeter blocks along the length of the cuboid. How many blocks can we place along the length? 6 blocks, as the length of the cuboid is 6 cm.
\nHow many blocks can we place along its breadth? 4 blocks, as the breadth of the cuboid is 4 cm. So there are 6 \u00d7 4 blocks can be placed in a layer.
\nHow many layers of blocks can be placed in the cuboid? 5 layers, as the height of the cuboid is 5 cm. Each layer has 6 \u00d7 4 blocks. So, all the 5 layers will have 6 \u00d7 4 \u00d7 5 blocks i.e. length \u00d7 breadth \u00d7 height.
\nThis discussion leads us to the formula for the volume of a cuboid.
\nVolume of a cuboid = length \u00d7 breadth \u00d7 height\u00a0 \u00a0 \u00a0 [Page No. 305]
\nAnswer:
\nThe dimensions of a cuboid are 6 cm, 4 cm, 5 cm respectively.
\n\u2234 Volume (V) = lbh
\n= 6 \u00d7 4 \u00d7 5.
\n= 120 cm3<\/sup><\/p>\n<\/p>\n
\nArrange 64 unit cubes in as many ways as you can to form a cuboid. Find the surface area of each arrangement. Can solid cuboid of same volume have same surface area? [Page No. 306]
\nAnswer:
\nNo. of cuboids are formed using 64 unit cubes
\n64 = 1 \u00d7 64 ……. (1)
\n= 2 \u00d7 32 …….. (2)
\n= 4 \u00d7 16 …….. (3)
\n1) l = 64 cm, b = 1 cm, h = 1 cm.
\nThe total surface area of a cuboid, A = 2 (lb + bh + lh)
\n= 2 (64 \u00d7 1 + 1 \u00d7 1 + 1 \u00d7 64)
\n= 2 (64 + 1 + 64)
\n= 2 \u00d7 129
\n= 258 Sq. cm.<\/p>\n
\nA = 2 (lb + bh + lh)
\n= 2 (32 \u00d7 2 + 2 \u00d7 1 + 32 \u00d7 1)
\n= 2 (64 + 2 + 32)
\n= 2 \u00d7 98 = 196 Sq. cm.<\/p>\n
\nA = 2 (lb + bh + lh)
\n= 2 (16 \u00d7 4 + 4 \u00d7 1 + 16 \u00d7 1)
\n= 2 (64 + 4 + 16)
\n= 2 \u00d7 84 = 168 Sq. cm.<\/p>\n<\/p>\n
\nFind the surface area of cube \u2018A\u2019 and lateral surface area of cube \u2018B\u2019.\u00a0 \u00a0 \u00a0 [Page No. 300]
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-3.png\")
\nAnswer:
\na = 10 cm.
\nThe total surface area of a figure ‘A’ = 6a2<\/sup>
\n= 6 \u00d7 (10)2<\/sup>
\n= 6 \u00d7 100 = 600 Sq. cm.
\nLateral surface area of a figure ‘B’ – 4a2<\/sup>
\n= 4 \u00d7 (8)2<\/sup> [\u2235 a = 8 cm.]
\n= 4 \u00d7 64 = 256 Sq. cm.<\/p>\n
\nTwo cubes each with side \u2018b\u2019 are joined to form a cuboid as shown in the given fig. What is the total surface area of this cuboid?\u00a0 \u00a0 \u00a0 [Page No. 300]
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-4.png\")
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-5.png\")
\nTotal surface area of a cuboid = 2 (lb + bh + lh)
\n= 2 (2b \u00d7 b + b \u00d7 b + 2b \u00d7 b)
\n= 2 (2b2<\/sup> + b2<\/sup> + 2b2<\/sup>)
\n= 2(5b2<\/sup>) = 10b2<\/sup> Sq. cm.<\/p>\n<\/p>\n
\nHow will you arrange 12 cubes of equal lengths to form a cuboid of smallest surface area?\u00a0 \u00a0 \u00a0 [Page No. 300]
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-6.png\")
\nAnswer:
\nWe can’t obtain the least total surface area by arranging 12 cubes by side by side.
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-7.png\")
\n\u2234 A = 2 (lb + bh + lh)
\n= 2 (12 \u00d7 1 + 1 \u00d7 1 + 12 \u00d7 1)
\n= 2 (12 + 1 + 12)
\n= 2 \u00d7 25 = 50 Sq. cm.
\nWe can obtain the least total surface area by arranging 3 cubes by 4 cubes.
\n\u2234 A = 2 (lb + bh + lh)
\n= 2 (3 \u00d7 1 + 1 \u00d7 4 + 3 \u00d7 4) (\u2235 l = 3; b = 1; h = 4)
\n= 2 (3 + 4 + 12)
\n= 2 \u00d7 19
\n= 38 Sq. cm.<\/p>\n
\nThe surface area of a cube of 4 \u00d7 4 \u00d7 4 dimensions is painted. The cube is cut into 64 equal cubes. How many cubes have
\n(a) 1 face painted? (b) 2 faces painted? (c) 3 faces painted? (d) no face painted?\u00a0 \u00a0 \u00a0 \u00a0[Page No. 300]
\nAnswer:
\nIf the 4 \u00d7 4 \u00d7 4 cube is divided into 64 equal cubes then the length of its each side = 1 unit.
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-8.png\")
\n[\u2235 \\(\\frac{4 \\times 4 \\times 4}{64}\\) = 1]
\na) No.of cubes (a = 4) have painted 1 face = 6(a – 2)2<\/sup> = 6(4 – 2)2<\/sup> = 6 \u00d7 4 = 24
\nb) No.of cubes have painted 2 faces = 12(a – 2) = 12(4 – 2) = 24
\nc) No.of cubes have painted 3 faces = 4 \u00d7 a = 4 \u00d7 2 = 8
\nd) No.of cubes have painted no faces = (a – 2)3<\/sup> = (4 – 2)3<\/sup> = (2)3<\/sup> = 8<\/p>\n<\/p>\n
\nCan we say that the total surface area of cuboid = lateral surface area + 2 \u00d7 area of base.\u00a0 \u00a0 \u00a0 [Page No. 299]
\nAnswer:
\nTotal surface area of a cuboid = L.S.A + 2 \u00d7 Area of base
\n= 2h (l + b) + 2 \u00d7 lb
\n= 2lh + 2bh + 2lb
\n= 2 (lb + bh + lh)
\nWe can conclude that total surface area of a cuboid = L.S.A + 2 \u00d7 Area of base<\/p>\n
\nIf we change the position of cuboid from Fig. (i) to Fig. (ii) do the lateral surface areas become equal?\u00a0 \u00a0 \u00a0[Page No. 299]
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-9.png\")
\nAnswer:
\nThere will be no change in the L.S.A of a cuboid if its positions are changed.<\/p>\n<\/p>\n
\nDraw a figure of cuboid whose dimensions are l, b, h are equal. Derive the formula for LSA and TSA.\u00a0 \u00a0 \u00a0 \u00a0[Page No. 299]
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-Board-8th-Class-Maths-Solutions-Chapter-14-Surface-Areas-and-Volume-Cube-Cuboid-InText-Questions-10.png\")
\nLateral surface area of a cuboid
\n= 4 \u00d7 (areas of 4 faces)
\n= 2 (l \u00d7 h) + 2 \u00d7 (b \u00d7 h) (1 + 2 + 3 + 4 faces)
\n= 2h(l + b) sq.units (1 = 3, 4 = 2)
\n\u2234 Total surface area of a cuboid
\n= 4 \u00d7 (Area of 4 faces) + (Areas of upper & lower faces)
\n= 2h (l + b) + 2 (lb)
\n= 2lh + 2bh + 27b
\n= 2 (lb + bh + lh) sq.units.<\/p>\n","protected":false},"excerpt":{"rendered":"