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" Prove that the area of a triangle is equal to half the product of its perimeter by the radius of the inscribed circle. "
Plane and Solid Geometry: Inductive Method - Page 231
by Arthur A. Dodd, B. Thomas Chace - 1898 - 406 pages
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Library of Useful Knowledge: Natural philosophy, Volume 3

1834 - 578 pages
...Si t, being very small, differs insensibly from the straight line joining the points s, t ; and the area of a triangle is equal to half the product of its base, and the perpendicular from its vertex. Again, the angle t E s, being very small, the perpendicular...
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Elements of the Differential and Integral Calculus

Charles Davies - Calculus - 1836 - 296 pages
...straight line passing through the origin of co-ordinates, and we shall have fF(x)dx-—xy, which proves that the area of a triangle is equal to half the product of the base and perpendicular. 286. It is frequently necessary to find the integral or function, between...
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Mechanics of Fluids for Practical Men: Comprising Hydrostatics, Descriptive ...

Alexander Jamieson - Fluid mechanics - 1837 - 516 pages
...the triangle are given, its area can easily be found. Thus, the writers on mensuration have shown, that the area of a triangle is equal to half the product of the base drawn into the perpendicular altitude ; consequently, if a be put to denote the area of the...
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An Elementary Treatise on Plane and Solid Geometry

Benjamin Peirce - Geometry - 1837 - 216 pages
...radii of the given and of the required circles. 277. Theorem. The area of any circumscribed polygon is half the product of its perimeter by the radius of the inscribed circle. Demonstration. From the centre O (fig. 134) of the circle draw OA, OB, OC, &c. to the vertices of the...
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Elements of the Differential and Integral Calculus

Charles Davies - Calculus - 1838 - 298 pages
...line passing through the origin of co-ordinates, and we shall have • fF(x)dx = — xy, which proves that the area of a triangle is equal to half the product of the base and perpendicular. 286. It is frequently necessary to find the integral or function, between...
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An Elementary Treatise on Plane and Solid Geometry

Benjamin Peirce - Geometry - 1847 - 204 pages
...radii of the given and of the required circles. 277. Theorem. The area of any circumscribed polygon is half the product of its perimeter by the radius of the inscribed circle. Proof. From the centre O (fig. 134) of the circle draw Area of a Circle. OA, OB, OC, &c. to the vertices...
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Elements of Geometry and Conic Sections

Elias Loomis - Conic sections - 1849 - 252 pages
...Schol.); therefore the area of the parallelogram ABCD is equal to AB X AF. PROPOSITION VI. THEOREM. The area of a triangle is equal to half the product of its base by its altitude. For, complete the parallelogram ABCE. The triangle ABC is half of the parallelo-...
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Elementary Geometry: With Applications in Mensuration

Charles Davies - Geometry - 1850 - 218 pages
...altitude, then they will be to each other, as GEOMETRY. Areas of Triangles and Trapezoids. THEOREM IX. The area of a triangle is equal to half the product of its base by its altitude. Let AE C be any triangle and CD its altitude : then will its area be equal to...
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Elementary Geometry: With Applications in Mensuration

Charles Davies - Geometry - 1850 - 238 pages
...other, as A x C : BxC: that is, as A : B. GEOMETRY. Areta of Triangles and Trapezoids. THEOREM IX. The area of a triangle is equal to half the product of its base by its altitude. Let ABC be any triangle and CD its altitude : then will its area be equal to...
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Elements of Analytical Geometry and of the Differential and Integral Calculus

Elias Loomis - Calculus - 1851 - 296 pages
...and the area equals fxy. If n=1, the figure becomes a triangle, and the area equals %xy; that is, the area of a triangle is equal to half the product of its base and perpendicular. Ex. 3. It is required to find the area of a circle. The equation of the circle,...
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