{"id":18792,"date":"2018-01-18T11:13:00","date_gmt":"2018-01-18T11:13:00","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=18792"},"modified":"2020-11-26T12:05:36","modified_gmt":"2020-11-26T06:35:36","slug":"math-labs-activity-pythagoras-theorem-method-6","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/math-labs-activity-pythagoras-theorem-method-6\/","title":{"rendered":"Math Labs with Activity – Pythagoras’ theorem (Method 6)"},"content":{"rendered":"
OBJECTIVE<\/strong><\/span><\/p>\n To verify Pythagoras’ theorem (Method 6)<\/p>\n Materials Required<\/strong><\/span><\/p>\n Theory<\/strong> <\/span> Procedure<\/strong> <\/span> The diagram will appear as shown in Figure 14.1. Observations and Calculations<\/strong><\/span> Result<\/strong> <\/span> Math Labs with Activity<\/a>Math Labs<\/a>Science Practical Skills<\/a>Science Labs<\/a><\/p>\n","protected":false},"excerpt":{"rendered":" Math Labs with Activity – Pythagoras theorem (Method 6) OBJECTIVE To verify Pythagoras’ theorem (Method 6) Materials Required A piece of cardboard Two sheets of white paper A drawing sheet A pair of scissors A geometry box A tube of glue Theory Pythagoras’ theorem: In a right-angled triangle, the square of the hypotenuse is equal … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6805],"tags":[],"yoast_head":"\n\n
\nPythagoras’ theorem:<\/strong> In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.<\/p>\n
\nStep 1:<\/strong> Paste a sheet of white paper on the cardboard.
\nOn this paper, draw a right-angled triangle ABC, right angled at C. Let the lengths of the sides AB, BC and CA.be c, a and b units respectively.
\nStep 2:<\/strong> Construct the following three squares:<\/p>\n\n
\n
\nStep 3:<\/strong> On the other sheet of paper, make exact copies of each one of the three squares. Also, make eight exact copies of the \u0394ABC. Shade the squares BFGC and ACHI differently.
\nStep 4:<\/strong> Cut all the three squares and the eight triangles.
\nStep 5:<\/strong> Arrange the square DEBA and the four triangles as shown in Figure 14.2, and paste this arrangement – on the upper half of a drawing sheet.
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\nStep 6:<\/strong> Arrange the remaining two squares (i.e., BFGC and ACHI) and the four triangles as shown in Figure 14.3. Paste this arrangement on the lower half of the drawing sheet.<\/p>\n
\nWe observe that each of the two arrangements (shown in Figures 14.2 and 14.3) forms a square of side (a+b) units. Therefore, both the squares have the equal area.
\nNow, if we remove the four triangles from each one of the two figures, the areas of the remaining figures must still be the same.
\nSo, area of the square in Figure 14.2 = sum of the areas of the squares in Figure 14.3,
\ni.e., c\u00b2 = a\u00b2 + b\u00b2.
\nApplying this result on the right-angled \u0394ABC we conclude that the square of the hypotenuse of the right-angled \u0394ABC is equal to the sum of the squares of the other two sides.<\/p>\n
\nPythagoras’ theorem is verified.<\/p>\n