This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1908 Excerpt: ...Draw line AE II DF, meeting CB prolonged at E. 2. Then, rt. A ABE and DCF are equal. ( 61, 104) 3. If from figure ADCE we take A ABE, there remains O AC; if we take A DCF, there remains rect. AF. 4. Then, area ABCD = area AEFD; whence equation (1). Prop. IV. Theorem 283. It follows from 282 that: 1. Two parallelograms ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1908 Excerpt: ...Draw line AE II DF, meeting CB prolonged at E. 2. Then, rt. A ABE and DCF are equal. ( 61, 104) 3. If from figure ADCE we take A ABE, there remains O AC; if we take A DCF, there remains rect. AF. 4. Then, area ABCD = area AEFD; whence equation (1). Prop. IV. Theorem 283. It follows from 282 that: 1. Two parallelograms having equal bases and equal altitudes are equivalent ( 279). 2. Two parallelograms having equal altitudes are to each other as their bases. 3. Two parallelograms having equal bases are to each other as their altitudes. 4. Any two parallelograms are to each other as the products of their bases by their altitudes. Ex. 4. The area of a parallelogram is 288, the base Is twice the altitude. Find the dimensions. Construct the parallelogram. Can more than one such parallelogram be drawn? How many and what parts are necessary for a definite figure? Why? Ex. 5. Find the ratio of the area of a rhombus to the product of its diagonals. Prop. V. Theorem 284. The area o f a triangle is equal to one-half the product of its base and altitude. Draw A ABC having base BC (6); draw AE BC, meeting BC, or BC extended, at E. We then have: Given b the base, and a the altitude of A ABC. To Prove area ABC =-J a xb. (Draw line AD II BC, and line CD II AB. By 105 AC divides O ABCD into two equal A.) 285. It follows from 284 that: 1. Two triangles having equal bases and equal altitudes are equivalent. 2. Two triangles having equal altitudes are to each other as their bases. 3. Two triangles having equal bases are to each other as their altitudes. 4. Any two triangles are to each other as the products of their bases by their altitudes. 5. The area of any triangle is one-half that of a parallelogram having the same base and altit...
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Add this copy of New Plane And Solid Geometry to cart. $50.00, new condition, Sold by Booksplease rated 4.0 out of 5 stars, ships from Southport, MERSEYSIDE, UNITED KINGDOM, published 2010 by Kessinger Publishing.
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New. Sewn binding. Cloth over boards. 328 p. In Stock. 100% Money Back Guarantee. Brand New, Perfect Condition, allow 4-14 business days for standard shipping. To Alaska, Hawaii, U.S. protectorate, P.O. box, and APO/FPO addresses allow 4-28 business days for Standard shipping. No expedited shipping. All orders placed with expedited shipping will be cancelled. Over 3, 000, 000 happy customers.
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New. Sewn binding. Cloth over boards. 328 p. In Stock. 100% Money Back Guarantee. Brand New, Perfect Condition, allow 4-14 business days for standard shipping. To Alaska, Hawaii, U.S. protectorate, P.O. box, and APO/FPO addresses allow 4-28 business days for Standard shipping. No expedited shipping. All orders placed with expedited shipping will be cancelled. Over 3, 000, 000 happy customers.
