 | Silas Ellsworth Coleman - Arithmetic - 1897 - 178 pages
...the base. The altitude is the length of the perpendicular from the base to the opposite vertex. The area of a triangle is equal to one-half the product of its base and altitude. EXPLANATION. Add to the given triangle an equal triangle inverted. The two together form a parallelogram... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1897 - 374 pages
...parallelograms having equal altitudes are to each other as their bases. " PROPOSITION VI. THEOREM 370. The area of a triangle is equal to one-half the product of its base and altitude. b B GIVEN the triangle ABC with base b and altitude a. To PROVE area ABC = i a X b. From C draw CX... | |
 | George Washington Hull - Geometry - 1897 - 408 pages
...231. The area of a triangle is equal to one half the product of its base and altitude. b Given—ABC a triangle having its altitude equal to a and its base equal to b. To Prove—Area ABC= J a X b. Dem.—Construct the parallelogram ABEC. Now, AABC = I CD'ABEC. § 220 But... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1897 - 376 pages
...+ bf) a. § 370 Draw the diagonal AC. Then area triangle ADC—\ ah, area triangle ABC=$ ab' . [The area of a triangle is equal to one-half the product of its base and altitude.] Adding, area trapezoid ABCD=$ab+$ab'. Ax. II a. QED * The ancient Egyptians attempted to find the area... | |
 | Mathematics - 1898 - 228 pages
...proportional between AD and AE. 5. Quote a theorem, a definition and an axiom used in proving that the area of a triangle is equal to one-half the product of its base and altitude. Write nothing else. JUNE, 1901. (B) 1. Find the longest and shortest lines that can be drawn from a... | |
 | Henry W. Keigwin - Geometry - 1898 - 250 pages
...rectangle is equal to the product of its base and altitude. (n =ab) In the same way § 306 gives: The area of a triangle is equal to one-half the product of its base and altitude. ab \ § 307 is, when restated : The area of a trapezoid is equal to the product of its mid-parallel... | |
 | George Albert Wentworth - Geometry - 1898 - 462 pages
...are to each other as the products of their bases by their altitudes. PROPOSITION V. THEOREM. 368. The area of a triangle is equal to one-half the product of its base by its altitude. H BD Let ABC be a triangle, AB its base, and DC its altitude. To prove the area of... | |
 | Webster Wells - Geometry - 1898 - 264 pages
...The base of an isosceles triangle is 56, and each of the equal sides is 53 ; find its area. 30. The area of a triangle is equal to one-half the product of its perimeter by the radius of the inscribed circle. D B 31. The area of an isosceles right triangle is... | |
 | James Howard Gore - Geometry - 1898 - 232 pages
...a triangle are 18 and 12, what is the length of the side of an equivalent square? 3. Show that the area of a triangle is equal to onehalf the product of its perimeter by the radius of the inscribed circle. PROPOSITION VI. THEOREM. 258. The area of a trapezoid... | |
 | William Taylor Campbell - Geometry - 1899 - 264 pages
...figure belongs. (f) The area of your plan. (g) The real area of the ground. 3. Area of a Triangle. The area of a triangle is equal to one-half the product of its base and altitude. C The base is any one of its sides. The altitude is the perpendicular distance from the base to the... | |
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The area of a triangle is equal to one-half the product of its base and altitude. 
