 | John Muller - 1773 - 202 sider
...to r ; then unity is to r as the radius PM (y) is to ry its femi-circumference, by Art. 40 ; and as the area of a circle is equal to the product of the radius by half its circumference, by Art. 41 ; ihall have yXry or ryy, for the circle defcribed, defcribed... | |
 | Adrien Marie Legendre - 1819 - 574 sider
...entire circles are as the squares of the.' therefore sect. ACB : sect. DOE : : AC: DO\ THEOREM. 289. The area of a circle is equal to the product of its circumference by half of the radius. Demonstration. Denoting by surf. CA the surface or area of a circle whosc radius is... | |
 | Adrien Marie Legendre - 1822 - 394 sider
...the squares of their radii; therefore sect. ACB : sect. DOE : ACa : DO2. 96 PROPOSITION XII. THEOREM. The area of a circle is equal to the product of its circumference by half the radius Let us designate the surface of the circle whose radius is CA by surf. CA ; we shall have surf. CA=JCA... | |
 | Etienne Bézout - 1824 - 222 sider
...made. Let us apply this theory of indeterminates to some examples. Let it be proposed to prove that the area of a circle is equal to the product of its circumference by half the radius; that is, that calling this radius R, the ratio of the circumference to radius sr, and consequently... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 280 sider
...°ntire circles ; but the entire circles are as the squares of their radii ; therefore THEOREM. 289. The area of a circle is equal to the product of its circumference by half of the radius. Demonstration. Denoting by surf. CA the surface or area of a circle whose radius is... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 294 sider
..."ntire circles ; but the entire circles are as the squares of their radii ; therefore I THEOREM. / 289. The area of a circle is equal to the product of its circumference by half of the radius. Demonstration. Denoting by surf. CA the surface or area of a circle whose radius is... | |
 | Jared Sparks, Edward Everett, James Russell Lowell, Henry Cabot Lodge - 1828 - 598 sider
...circles, the words their radii, and the same theorem is sufficient for both cases. QED Theorem 289. The area of a circle is equal to the product of its circumference by half the radius. It was demonstrated, 280, that ' the area of a regular polygon is equal to the product of its perimeter... | |
 | Jared Sparks, Edward Everett, James Russell Lowell, Henry Cabot Lodge - 1828 - 598 sider
...circles, the words their radii, and the same theorem is sufficient for both cases. QED Theorem 289. The area of a circle is equal to the product of its circumference by half the radius. It was demonstrated, 280, that ' the area of a regular polygon is equal to the product of its perimeter... | |
 | Adrien Marie Legendre - 1828 - 346 sider
...again are as the squares of their radii ; therefore sect. ACB : sect. DOE : AC2 DOJ THEOREM. ' 289. The area of a circle is equal to the product of its circumference by half the radiusl Let us designate the surface of the circle whose radius is CA by surf. CA ; we shall have surf.... | |
 | Timothy Walker - 1829 - 138 sider
...area of the polygon its perimeter multiplied by half the radius of the inscribed . circle. 105. — The area of a circle is equal to the product of its circumference by half the radius — . This follows directly from the preceding. For the circumference of the circle is the perimeter... | |
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The area of a circle is equal to the product of its circumference by half the radius.* Let ACDE be a circle whose centre is O and radius OA : then will area OA— ^OAxcirc. 
