{"id":67419,"date":"2023-04-03T06:54:00","date_gmt":"2023-04-03T01:24:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67419"},"modified":"2023-12-19T11:07:14","modified_gmt":"2023-12-19T05:37:14","slug":"rs-aggarwal-chapter-15-class-9-maths-exercise-15-1-solutions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-15-class-9-maths-exercise-15-1-solutions\/","title":{"rendered":"RS Aggarwal Chapter 15 Class 9 Maths Exercises 15.1 (ex 15a) Solutions 2024"},"content":{"rendered":"\n<p><strong>RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.1 Solutions:<\/strong> Wherever we look, usually we see solids. So far, in all our studies, we have been dealing with figures that can be easily drawn on our notebooks or blackboards. These are called plane figures. We have understood what rectangles, squares, and circles are, what we mean by their perimeters and areas, and how we can find them. We have learned these in earlier classes. It would be interesting to see what happens if we cut out many of these plane figures of the same shape and size from cardboard sheets and stack them up in a vertical pile. Through this process, we shall obtain some solid figures (briefly called solids) such as a cuboid, a cylinder, etc. Know more about this chapter <a href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-solutions-class-9-maths-chapter-15-volume-and-surface-area-of-solids\/\">here<\/a>.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d505a57930c\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" 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15.1\u00a0Solutions?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-15-class-9-maths-exercise-15-1-solutions\/#can-i-access-the-rs-aggarwal-chapter-15-class-9-maths-exercise-151-solutions-pdf-offline\" title=\"Can I access the RS Aggarwal Chapter 15\u00a0Class 9 Maths Exercise 15.1\u00a0Solutions PDF offline?\">Can I access the RS Aggarwal Chapter 15\u00a0Class 9 Maths Exercise 15.1\u00a0Solutions PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rs-aggarwal-chapter-15-class-9-maths-exercise-151-solutions\"><\/span>Download RS Aggarwal Chapter 15\u00a0Class 9 Maths Exercise 15.1\u00a0Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/EXERCISE-\u2013-15A.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/EXERCISE-\u2013-15A.pdf\">EXERCISE \u2013 15A<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-definition-for-rs-aggarwal-chapter-15-class-9-maths-ex-15a-solutions\"><\/span>Important Definition for RS Aggarwal Chapter 15 Class 9\u00a0Maths Ex 15a\u00a0Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table>\n<tbody>\n<tr>\n<td>\n<p><strong><b>Name of the Solid Figure<\/b><\/strong><\/p>\n<\/td>\n<td>\n<p><strong><b>Formulas<\/b><\/strong><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Cuboid<\/p>\n<\/td>\n<td>\n<p><strong><b>LSA:<\/b><\/strong>\u00a02h(l + b)<br \/><strong><b>TSA:<\/b><\/strong>\u00a02(lb + bh + hl)<br \/><strong><b>Volume:<\/b><\/strong>\u00a0l \u00d7 b \u00d7 h<br \/><br \/>l = length,<br \/>b = breadth,<br \/>h = height<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Cube<\/p>\n<\/td>\n<td>\n<p><strong><b>LSA:<\/b><\/strong>\u00a04a2<br \/><strong><b>TSA:<\/b><\/strong>\u00a06a2<br \/><strong><b>Volume:<\/b><\/strong>\u00a0a3<br \/><br \/>a = sides of a cube<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Right Circular Cylinder<\/p>\n<\/td>\n<td>\n<p><strong><b>LSA:<\/b><\/strong>\u00a02(\u03c0 \u00d7 r \u00d7 h)<br \/><strong><b>TSA:<\/b><\/strong>\u00a02\u03c0r (r + h)<br \/><strong><b>Volume:<\/b><\/strong>\u00a0\u03c0 \u00d7 r2\u00a0\u00d7 h<br \/><br \/>r = radius,<br \/>h = height<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Right Pyramid<\/p>\n<\/td>\n<td>\n<p><strong><b>LSA:<\/b><\/strong>\u00a0\u00bd \u00d7 p \u00d7 l<br \/><strong><b>TSA:<\/b><\/strong>\u00a0LSA + Area of the base<br \/><strong><b>Volume:<\/b><\/strong>\u00a0\u2153 \u00d7 Area of the base \u00d7 h<br \/><br \/>p = perimeter of the base,<br \/>l = slant height, h = height<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Prism<\/p>\n<\/td>\n<td>\n<p><strong><b>LSA:<\/b><\/strong>\u00a0p \u00d7 h<br \/><strong><b>TSA:<\/b><\/strong>\u00a0LSA \u00d7 2B<br \/><strong><b>Volume:<\/b><\/strong>\u00a0B \u00d7 h<br \/><br \/>p = perimeter of the base,<br \/>B = area of base, h = height<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Right Circular Cone<\/p>\n<\/td>\n<td>\n<p><strong><b>LSA:<\/b><\/strong>\u00a0\u03c0rl<br \/><strong><b>TSA:<\/b><\/strong>\u00a0\u03c0 \u00d7 r \u00d7 (r + l)<br \/><strong><b>Volume:\u00a0<\/b><\/strong>\u2153 \u00d7 (\u03c0r2h)<br \/><br \/>r = radius,<br \/>l = slant height,<br \/>h = height<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Hemisphere<\/p>\n<\/td>\n<td>\n<p><strong><b>LSA:<\/b><\/strong>\u00a02 \u00d7 \u03c0 \u00d7 r2<br \/><strong><b>TSA:<\/b><\/strong>\u00a03 \u00d7 \u03c0 \u00d7 r2<br \/><strong><b>Volume:<\/b><\/strong>\u00a0\u2154 \u00d7 (\u03c0r3)<br \/><br \/>r = radius<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Sphere<\/p>\n<\/td>\n<td>\n<p><strong><b>LSA:<\/b><\/strong>\u00a04 \u00d7 \u03c0 \u00d7 r2<br \/><strong><b>TSA:<\/b><\/strong>\u00a04 \u00d7 \u03c0 \u00d7 r2<br \/><strong><b>Volume:<\/b><\/strong>\u00a04\/3 \u00d7 (\u03c0r3)<br \/><br \/>r = radius<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-the-rs-aggarwal-chapter-15-class-9-maths-exercise-151-solutions\"><\/span><strong>Access The RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.1 Solutions<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>(1)<\/strong>\u00a0Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:<\/p>\n<p><strong>(i)<\/strong> Length = 12 cm, breadth = 8 cm and height = 4.5 cm<\/p>\n<p>Volume of cuboid = (lbh) cm<sup>3<\/sup><br \/>=12\u00d78\u00d74.5 cm<sup>3<\/sup><br \/>=432 cm<sup>3<\/sup><\/p>\n<p>Lateral surface area =[2(l+b)\u00d7h] cm<sup>2<\/sup><br \/>=[2(12+8)\u00d74.5]<br \/>=40\u00d74.5<br \/>=180 cm<sup>2<\/sup><\/p>\n<p>Total surface area =2(lb+bh+hl) cm<sup>2<\/sup><br \/>=2{12\u00d78+8\u00d74.5+4.5\u00d712} cm<sup>2<\/sup><br \/>=2(96+36+54) cm<sup>2<\/sup><br \/>=372 cm<sup>2<\/sup><\/p>\n<p>Hence volume of cuboid is 432 cm<sup>3<\/sup>, lateral surface area is 180 cm<sup>2<\/sup>\u00a0and total surface area is 372 cm<sup>2<\/sup><\/p>\n<p><strong>(ii)<\/strong>\u00a0Length = 26 m, breadth = 14 m and height = 6.5m<\/p>\n<p>Volume of cuboid = (lbh) m<sup>3<\/sup><br \/>=26\u00d714\u00d76.5 m<sup>3<\/sup><br \/>=2366 m<sup>3<\/sup><\/p>\n<p>Lateral surface area =[2(l+b)\u00d7h] m<sup>2<\/sup><br \/>=[2(26+14)\u00d76.5]<br \/>=80\u00d76.5<br \/>=520 m<sup>2<\/sup><\/p>\n<p>Total surface area =2(lb+bh+hl) m<sup>2<\/sup><br \/>= 2{26\u00d714+14\u00d76.5+6.5\u00d726} m<sup>2<\/sup><br \/>= 2(364+91+169) m<sup>2<\/sup><br \/>= 1248 m<sup>2<\/sup><\/p>\n<p>Hence volume of cuboid is 2366 m<sup>3<\/sup>, lateral surface area is 520 m<sup>2<\/sup>\u00a0and total surface area is 1248 m<sup>2<\/sup><\/p>\n<p><strong>(iii)\u00a0<\/strong>Length = 15 m, breadth = 6 m and height = 5 dm<br \/>5 dm = 50 cm = 0.5 m<br \/>Volume of cuboid = (lbh) m<sup>3<\/sup><br \/>=15\u00d76\u00d70.50 cm<sup>3<\/sup><br \/>=45 m<sup>3<\/sup><\/p>\n<p>Lateral surface area =[2(l+b)\u00d7h] m<sup>2<\/sup><br \/>=[2(15+6)\u00d70.50]<br \/>=42\u00d70.50<br \/>=21 cm<sup>2<\/sup><\/p>\n<p>Total surface area =2(lb+bh+hl) m<sup>2<\/sup><br \/>= 2{15\u00d76+6\u00d70.50+0.50\u00d715} m<sup>2<\/sup><br \/>= 2(90+3+7.5) m<sup>2<\/sup><br \/>= 201 m<sup>2<\/sup><\/p>\n<p>Hence the volume of the cuboid is 45 m<sup>3<\/sup>, the lateral surface area is 21 m<sup>2<\/sup> and the total surface area is 201m<sup>2<\/sup><\/p>\n<p><strong>(iv)<\/strong>\u00a0Length = 24m, breadth = 25 cm and height = 6m<br \/>breadth = 25 cm = 0.25 m<br \/>Volume of cuboid = (lbh) m<sup>3<\/sup><br \/>=24\u00d70.25\u00d76 m<sup>3<\/sup><br \/>=36 m<sup>3<\/sup><\/p>\n<p>Lateral surface area =[2(l+b)\u00d7h] m<sup>2<\/sup><br \/>=[2(24+0.25)\u00d76]<br \/>=48.5\u00d76<br \/>=291 m<sup>2<\/sup><\/p>\n<p>Total surface area =2(lb+bh+hl) m<sup>2<\/sup><br \/>= 2{24\u00d70.25+0.25\u00d76+6\u00d724} m<sup>2<\/sup><br \/>= 2(6+1.5+144) m<sup>2<\/sup><br \/>=303 m<sup>2<\/sup><\/p>\n<p><strong>(2)\u00a0<\/strong>A match box measures 4 cm\u00d72.5 cm\u00d71.5 cm. What is the volume of a packet containing 12 such matchbox?<\/p>\n<p>Volume of match box = 4\u00d72.5 \u00d71.5=15 cm<sup>3<\/sup><br \/>Volume of packet containing 12 match box = 15\u00d712=180 cm<sup>3<\/sup><\/p>\n<p><strong>(3)<\/strong>\u00a0A cuboid water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold?<\/p>\n<p>Volume of tank = 6\u00d75 \u00d74.5<br \/>=135 m<sup>3<\/sup><br \/>Water in litre = 135\u00d71000<br \/>=135000 litres [\u2235 1 m<sup>3<\/sup>\u00a0=1000 litres]<\/p>\n<p><strong>(4)<\/strong> The capacity of the Cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10m and 2.5 m. (Given 1000 litres = 1m<sup>3<\/sup>)<\/p>\n<p>Let the breadth of the tank is x, then<br \/>Volume of tank =10\u00d7x\u00d72.5<br \/>=25x m<sup>3<\/sup><br \/>Water in litre =25x\u00d71000<br \/>=25000x litres<br \/>According to the question,<br \/>\u21d2 25000x=50000<br \/>\u21d2 x = 2 m<\/p>\n<p><strong>(5)<\/strong> A godown measures 40m\u00d725m\u00d715m . Find the maximum number of wooden crates, each measuring1.5m\u00d71.25m\u00d70.5m, that can be stored in the godown.<\/p>\n<p>Volume of godown =40\u00d725\u00d715<br \/>=15000 m<sup>3<\/sup><br \/>The volume of one crate = 1.5\u00d71.25\u00d70.5<br \/>= 0.9375 m<sup>3<\/sup><br \/>Number of crates =15000\/0.9375<br \/>= 16000<br \/>Hence 16000 crates can be stored in the godown.<\/p>\n<p><strong>(6)<\/strong>\u00a0How many planks of dimension (5m\u00d725cm\u00d710cm) can be stored in a pit which is 20m long, 6m wide and 80 cm deep?<\/p>\n<p>Volume of pit = 20 m\u00d76 m\u00d780 cm<br \/>=2000 cm\u00d7600 cm\u00d780 cm<br \/>=96000000 cm<sup>3<\/sup><br \/>The volume of one plank = 5m\u00d725cm\u00d710cm<br \/>=500cm\u00d725cm\u00d710cm<br \/>=125000 cm<sup>3<\/sup><br \/>Number of planks = 96000000\/125000<br \/>= 768<\/p>\n<p>Hence total 768 planks can be stored.<\/p>\n<p><strong>(7)<\/strong>\u00a0How many bricks will be required to construct a wall 8 m long, 6 m high, and 22.5 cm thick if each brick measures(25cm\u00d711.25cm\u00d76cm)?<\/p>\n<p>Volume of wall = 8m \u00d7 6m \u00d7 22.5cm<br \/>= 800 cm\u00d7600 cm\u00d722.5 cm<br \/>=10800000 cm<sup>3<\/sup><br \/>Volume of one brick = 25cm \u00d7 11.25cm \u00d7 6cm<br \/>=1687.5 cm<sup>3<\/sup><br \/>Number of bricks = 6400<\/p>\n<p>Hence, 6400 bricks are required.<\/p>\n<p><strong>(8)<\/strong>\u00a0Find the capacity of a closed rectangular cistern whose length is 8m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.<\/p>\n<p>Volume of cistern = 8 m\u00d76 m\u00d72.5 m<br \/>=120 m<sup>3<\/sup><br \/>Total surface area =2(lb+bh+hl)<br \/>=2(8\u00d76+6\u00d72.5+2.5\u00d78)<br \/>=2(48+15+20)<br \/>=2(83)<br \/>=166 m<sup>2<\/sup><\/p>\n<p>Hence, Volume is 120 m<sup>3<\/sup>\u00a0and 166 m<sup>2<\/sup>\u00a0iron sheet is required.<\/p>\n<p><strong>(9)<\/strong>\u00a0The dimensions of the room are(9m\u00d78m\u00d76.5m). It has one door of dimensions (2m\u00d71.5m) and two windows, each of dimensions(1.5m\u00d71m). Find the cost of white washing the walls at Rs. 25 per square metre.<\/p>\n<p>Area of room\u2019s wall =[2(l+b)\u00d7h]<br \/>=[2(9m+8m)\u00d76.5m]<br \/>=221 m<sup>2<\/sup><\/p>\n<p>And, Area of one door + two windows<br \/>= (2m\u00d71.5m)+2(1.5m\u00d71m)<br \/>=(3+3)m<sup>2<\/sup><br \/>Net area for white washing = 221-6<br \/>=215 m<sup>2<\/sup><br \/>Cost of white washing = 215\u00d725<br \/>= 5375<\/p>\n<p>Hence the cost of white washing is Rs. 5375<\/p>\n<p><strong>(10)\u00a0<\/strong>A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring(22 cm\u00d712.5 cm\u00d77.5 cm) . If 1\/12 of the total volume of wall consists of mortar, now many bricks are there in the wall?<\/p>\n<p>Volume of wall = (15 m\u00d730 cm\u00d74 m)<br \/>=(1500 cm\u00d730 cm\u00d7400 cm)<br \/>=18000000 cm<sup>3<\/sup><br \/>Volume of mortar=1\/12\u00d718000000<br \/>=1500000 cm<sup>3<\/sup><br \/>Volume of wall without mortar = (18000000-1500000) cm<sup>3<\/sup><br \/>=16500000 cm<sup>3<\/sup><br \/>Volume of one brick = (22 cm\u00d712.5 cm\u00d77.5 cm)<br \/>=2062.5 cm<sup>3<\/sup><br \/>Number of bricks = 16500000\/2062.5<br \/>= 8000<\/p>\n<p>Hence 8000 bricks are in the wall.<\/p>\n<p><strong>(11)<\/strong>\u00a0How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm<sup>3<\/sup>\u00a0of iron weight 15 g, find the weight of the empty box in kilograms.<\/p>\n<p>External length = 36 cm<br \/>External breadth = 25 cm<br \/>External height = 16.5 cm<br \/>Internal length = 36 \u2013 2(1.5) = 33 cm<br \/>Internal breadth = 25 \u2013 2(1.5) = 22 cm<br \/>Internal height = 16.5 \u2013 1.5 = 15 cm<br \/>Internal volume of the box = (33\u00d722\u00d715)<br \/>=10890 cm<sup>3<\/sup><br \/>Volume of iron sheet = 14850-10890<br \/>=3960 cm<sup>3<\/sup><br \/>Weight of sheet = 3960\u00d715g<br \/>= 59.4 kg<\/p>\n<p>Hence 3960 cm<sup>3<\/sup>\u00a0iron sheet is required and weight of sheet is 59.4 kg<\/p>\n<p><strong>(12)\u00a0<\/strong>A box made of sheet metal costs Rs. 6480 at Rs. 120 per square metre. If the box is 5 m long and 3 m wide, find its height.<\/p>\n<p>Let height = x metre<br \/>Total surface area of box =2(lb+bh+hl)<br \/>=2(5\u00d73+3\u00d7x+x\u00d75)=2(15+8x)<br \/>Cost of metal at Rs. 120 =2(15+8x)\u00d7120<br \/>=3600+1920x<br \/>According to question, =3600+1920x = 6480<br \/>1920x =6480-3600 =2880<br \/>x=2880\/1920 = 1.5 m<\/p>\n<p>Hence height of box is 1.5 m<\/p>\n<p><strong>(13)<\/strong>\u00a0The volume of cuboid is 1536 m<sup>3<\/sup>\u00a0. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.<\/p>\n<p>Length = 16 m, breadth = 3x and height = 2x<br \/>Volume of cuboid = lbh m<sup>3<\/sup><br \/>=16\u00d73x\u00d72x<br \/>= 96 x<sup>2<\/sup>m<sup>3<\/sup><br \/>By question,=96 x<sup>2<\/sup>m<sup>3<\/sup><br \/>=1536 m<sup>3<\/sup><br \/>\u21d2 x<sup>2<\/sup>\u00a0=1536\/96 = 16<br \/>\u21d2 x = 4<\/p>\n<p>Therefore breadth =3\u00d74=12 m and height =2\u00d74=8 m<\/p>\n<p><strong>(14)\u00a0<\/strong>How many persons can be accommodated in dining hall of dimensions (20 m\u00d716 m\u00d74.5 m), assuming that each person requires 5 cubic metres of air?<\/p>\n<p>Volume of dining hall = 20 m\u00d716 m\u00d74.5 m<br \/>=1440 m<sup>3<\/sup><br \/>Each person required 5 cubic metre air, therefore<br \/>Number of person in hall = 1440\/5 = 288<\/p>\n<p>Hence, 288 person can accommodate in dining hall.<\/p>\n<p><strong>(15)<\/strong>\u00a0A class room is 10 m long, 6.4 m wide and 5 m high. If each student be given 1.6 m2 of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?<\/p>\n<p>Area of the class = 10 m\u00d76.4 m<br \/>=64 m<sup>2<\/sup><br \/>Each student needs 1.6 m<sup>2<\/sup>\u00a0area, therefore<br \/>Number of student in class = 64\/1.6 = 40<br \/>Air for each student = Floor area\u00d7height<br \/>= 1.6\u00d75 = 8 m<sup>3<\/sup><\/p>\n<p>Hence, 40 Student can accommodate in class and each student get 8m<sup>3<\/sup>\u00a0air.<\/p>\n<p><strong>(16)<\/strong>\u00a0The surface area of cuboid is 758 cm2. Its length and breadth are 14 cm and 11 cm respectively. Find its height.<\/p>\n<p>Let height = x cm<br \/>Surface are of cuboid =2(lb+bh+hl)<br \/>=2(14\u00d711+11\u00d7x+x\u00d714)<br \/>=2(154+25x)<br \/>=(308+50x) cm2<br \/>According to question, 308+50x =758<br \/>50x =758-308<br \/>x = 450<br \/>x = 9 cm<\/p>\n<p>Hence height of cuboid is 9 cm.<\/p>\n<p><strong>(17)\u00a0<\/strong>In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.<br \/>1 hectare = 10000 m<sup>2<\/sup><\/p>\n<p>Area of ground = 2 hectare = 20000 m<sup>2<\/sup><br \/>Volume of water = 20000 m<sup>2<\/sup>\u00a0\u00d75 cm<br \/>= 20000 m<sup>2<\/sup>\u00a0\u00d70.05 m<br \/>=1000 m<sup>3<\/sup><\/p>\n<p>Hence, 1000 m<sup>3<\/sup>\u00a0of rain falls.<\/p>\n<p><strong>(18)<\/strong>\u00a0Find the volume, the lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures 9 m. (Take \u221a3=1.73)<\/p>\n<p>Volume of cube = (side)<sup>3<\/sup>\u00a0= 9<sup>3<\/sup><br \/>=729 m<sup>3<\/sup><br \/>Total surface area = 6 \u00d7(side)<sup>3<\/sup><br \/>= 6\u00d79<sup>2<\/sup><br \/>= 6\u00d781<br \/>= 486 m<sup>2<\/sup><br \/>Diagonal = \u221a3\u00d7 (side)<br \/>= 1.73\u00d79<br \/>= 15.57 m<\/p>\n<p>Hence volume of cube is 729 m<sup>3<\/sup>\u00a0,total surface area is 486 m<sup>2<\/sup>\u00a0and diagonal is 15.57m.<\/p>\n<p><strong>(19)\u00a0<\/strong>The total surface area of a cube is 1176 cm<sup>2<\/sup>\u00a0. Find its volume.<\/p>\n<p>Let side is x cm.<br \/>Total surface area of cube = 6 \u00d7(x)<sup>3<\/sup>\u00a0= 1176<br \/>(x)<sup>2<\/sup>\u00a0= 1176\/6 = 196 \u21d2 x = 14<br \/>Volume = (14)<sup>3<\/sup>\u00a0= 2744 cm<sup>3<\/sup><\/p>\n<p>Hence volume is 2744 cm<sup>3<\/sup>.<\/p>\n<p><strong>(20)\u00a0<\/strong>The lateral surface area of a cube is 900 cm<sup>2<\/sup>\u00a0. Find its volume.<\/p>\n<p>Let side of cube is x cm.<br \/>Lateral surface area =[2(l+b)\u00d7h] cm<sup>2<\/sup><br \/>=[2(x+x)\u00d7x]=900<br \/>= 4x<sup>2<\/sup>\u00a0= 900<br \/>= x<sup>2<\/sup>\u00a0= 225<br \/>\u21d2 x = 15<\/p>\n<p>Volume = (15)<sup>3<\/sup>\u00a0= 3375 cm<sup>3<\/sup><\/p>\n<p><strong>(21)\u00a0<\/strong>The volume of a cube is 512 cm<sup>3<\/sup>\u00a0. Find its surface area.<\/p>\n<p>Let side of cube is x cm<br \/>Volume = (x)<sup>3<\/sup><br \/>= 512 cm<sup>3<\/sup><br \/>\u21d2 x = 8<br \/>Surface area = 6\u00d7 (x)<sup>2<\/sup><br \/>= 6\u00d764<br \/>\u21d2 384 cm<sup>2<\/sup><\/p>\n<p>Hence surface area is 384 cm<sup>2<\/sup>\u00a0.<\/p>\n<p><strong>(22)\u00a0<\/strong>Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.<\/p>\n<p>Volume of first cube =3<sup>3<\/sup><br \/>= 27 cm<sup>3<\/sup><br \/>Volume of second cube = 4<sup>3<\/sup><br \/>= 64 cm<sup>3<\/sup><br \/>Volume of third cube =5<sup>3<\/sup><br \/>= 125 cm<sup>3<\/sup><br \/>Total volume = 27 + 64 + 125<br \/>= 216<br \/>Volume of large cube = (x)<sup>3<\/sup>\u00a0= 216<br \/>x = 6 cm<\/p>\n<p>Lateral surface area = [2(x+x)\u00d7x]<br \/>= 4x<sup>2<\/sup><br \/>= 4\u00d736<br \/>= 144 cm<sup>2<\/sup><\/p>\n<p>Hence lateral surface area of the new cube is 144 cm<sup>2<\/sup>\u00a0.<\/p>\n<figure class=\"wp-block-image size-large\"><img class=\"wp-image-2806\" src=\"https:\/\/www.msipg.com\/wp-content\/uploads\/2020\/01\/Capture-1-22.png\" sizes=\"(max-width: 425px) 100vw, 425px\" srcset=\"https:\/\/www.msipg.com\/wp-content\/uploads\/2020\/01\/Capture-1-22.png 425w, https:\/\/www.msipg.com\/wp-content\/uploads\/2020\/01\/Capture-1-22-214x300.png 214w\" alt=\"RS Aggarwal Class 9 Chapter 15 Exercise 15A Solution\" width=\"425\" height=\"597\" \/><\/figure>\n<figure class=\"wp-block-image size-large\"><img class=\"wp-image-2807\" src=\"https:\/\/www.msipg.com\/wp-content\/uploads\/2020\/01\/Capture-2-21.png\" sizes=\"(max-width: 402px) 100vw, 402px\" srcset=\"https:\/\/www.msipg.com\/wp-content\/uploads\/2020\/01\/Capture-2-21.png 402w, https:\/\/www.msipg.com\/wp-content\/uploads\/2020\/01\/Capture-2-21-189x300.png 189w\" alt=\"RS Aggarwal Class 9 Chapter 15 Exercise 15A Solution\" width=\"402\" height=\"639\" \/><\/figure>\n<p><strong>(27)\u00a0<\/strong>Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hour. How much area will it irrigate in 30 minute, if 9 cm of standing water is desired?<\/p>\n<p>Distance covered by water in 30 minute<br \/>=speed of water\u00d7time<br \/>=20\u00d71\/2 h<br \/>=10km<br \/>Volume of water which flown in 30 min<br \/>=30 dm \u00d712 dm\u00d710km<br \/>=3 m\u00d71.2 m\u00d710000 m<br \/>=36000 m<sup>3<\/sup><br \/>Let irrigation area is x m<sup>2<\/sup>\u00a0, then<br \/>x m<sup>2<\/sup>\u00a0\u00d79 cm = 36000 m<sup>3<\/sup><br \/>x m<sup>2<\/sup>\u00a0\u00d70.09 m = 36000 m<sup>3<\/sup><br \/>x = 400000<\/p>\n<p>Hence irrigation area is 400000 m<sup>2<\/sup><\/p>\n<p><strong>(28)\u00a0<\/strong>A solid metallic cuboid of dimensions (9 m\u00d78 m\u00d72 m) is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.<\/p>\n<p>Volume of cuboid = (9 m\u00d78 m\u00d72 m)<br \/>=144 m<sup>3<\/sup><br \/>Volume of cube which side is 2 m<br \/>= 2<sup>3<\/sup><br \/>=8m<sup>3<\/sup><br \/>Number of cubes = 144\/8 =18<\/p>\n<p>Hence 18 cubes formed.<\/p>\n<p>Know more at the <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">official website<\/a>.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rs-aggarwal-chapter-15-class-9-maths-exercise-151-solutions\"><\/span>FAQs on RS Aggarwal Chapter 15\u00a0Class 9 Maths Exercise 15.1\u00a0Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1680499710239\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rs-aggarwal-chapter-15-class-9-maths-exercise-151-solutions\"><\/span>From where can I download the PDF of RS Aggarwal Chapter 15\u00a0Class 9 Maths Exercise 15.1\u00a0Solutions?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1680499738420\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rs-aggarwal-chapter-15-class-9-maths-exercise-151-solutions\"><\/span>How much does it cost to download the PDF of RS Aggarwal Chapter 15\u00a0Class 9 Maths Exercise 15.1\u00a0Solutions?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1680499752787\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rs-aggarwal-chapter-15-class-9-maths-exercise-151-solutions-pdf-offline\"><\/span>Can I access the RS Aggarwal Chapter 15\u00a0Class 9 Maths Exercise 15.1\u00a0Solutions PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.1 Solutions: Wherever we look, usually we see solids. So far, in all our studies, we have been dealing with figures that can be easily drawn on our notebooks or blackboards. These are called plane figures. We have understood what rectangles, squares, and circles are, what we &#8230; <a title=\"RS Aggarwal Chapter 15 Class 9 Maths Exercises 15.1 (ex 15a) Solutions 2024\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-15-class-9-maths-exercise-15-1-solutions\/\" aria-label=\"More on RS Aggarwal Chapter 15 Class 9 Maths Exercises 15.1 (ex 15a) Solutions 2024\">Read more<\/a><\/p>\n","protected":false},"author":236,"featured_media":67427,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73412,73395,2985,73410],"tags":[3081,3086,4711],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67419"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/236"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=67419"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67419\/revisions"}],"predecessor-version":[{"id":524559,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67419\/revisions\/524559"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/67427"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67419"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67419"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67419"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}