 | George Roberts Perkins - Geometry - 1860 - 472 pages
...And as we have proved (B. L, T. XXXIII.) that EF = ^t00, we have ABDC - IK x EF. THEOREM XXIV. The area, of a triangle is equal to half the product of its perimeter by the radius of its inscribed circle. Let K be the centre of the inscribed circle. Draw KD, KE, KF... | |
 | Evan Wilhelm Evans - Geometry - 1862 - 116 pages
...Theo. XIII) on the same base AB and having the same altitude CE (Def. 4, Sec. V), it follows that the area of a triangle is equal to half the product of its base by its altitude. THEOREM XVIII. The area of a trapezoid is equal to half the product of the sum... | |
 | William Thynne Lynn - Physics - 1863 - 136 pages
...in one second a velocity represented by DE, will be measured by the triangle ABC itself. But as the area of a triangle is equal to half the product of its base and altitude, this will be £ AB x BC ; or s, the space, will be =%vt; or, since v=ft, we have... | |
 | Edward Brooks - Geometry - 1868 - 284 pages
...bases; and parallelograms having equal bases are to each other as their altitudes. THEOREM III. The area of a triangle is equal to half the product of its base and altitude. T1etABCbe a triangle, AB its base, and CD its altitude; then will its area be equal... | |
 | William Chauvenet - Geometry - 1871 - 380 pages
...to each other as the products of their bases by their altitudes. PROPOSITION V.— THEOREM. 13. The area of a triangle is equal to half the product of its base and altitude. Let ABCbe a triangle, k the numerical measure of its base BC, h that of its altitude... | |
 | André Darré - 1872 - 226 pages
...to the sides, the two parallelograms through which the diagonal does not pass are equivalent. 5. The area of a triangle is equal to half the product of...perimeter and the radius of the inscribed circle. 6. Two sides of a triangle being given, the area is greatest when the angle which they form is a right... | |
 | William Chauvenet - Mathematics - 1872 - 382 pages
...to each other as the products of their bases by their altitudes. PROPOSITION V.— THEOREM. 13. The area of a triangle is equal to half the product of its base and altitude. Let ABCbe a triangle, k the numerical measure of its base BC, h that of its altitude... | |
 | William Frothingham Bradbury - Geometry - 1872 - 262 pages
...of the rectangle = ADXDF (7) ; therefore the area of the parallelogram = ADXD F. THEOREM IV. 11i The area of a triangle is equal to half the product of 'its , lase and altitude. ' Let BD be the altitude of the triangle ABC ; then the area of ABC = lACX BD.... | |
 | Charles Davies - Geometry - 1872 - 464 pages
...If their bases are «qual, they are to each other as their altitudes. PROPOSITION VI. THEOREM. The area of a triangle is equal to half the product of its base and altitude. Let ABC be a triangle, BC its base, and AD is Mtitude : then will the area of the... | |
 | William Frothingham Bradbury - Geometry - 1873 - 132 pages
...the rectangle = AD X DF (7) ; therefore the area of the parallelogram = ADX DF. THEOREM IV. 11, The area, of a triangle is equal to half the product of its lose and altitude. Let BD be the altitude of the triangle ABC ; then the area — ^ACx BD. Draw CE... | |
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The area of a triangle is equal to half the product of its base by its altitude. 
