...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...

...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...

...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...

...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-...

...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...

...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...

...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,...

...generally, are to each other- as the products of their bases and altitudes. PROPOSITION VI. THEOEEM. The area of a triangle is equal to half the product of its base and altitude. For, draw CE parallel to BA, and AE parallel to BC, completing the parallelogram...

...generally, are to each other as the products of their bases and altitudes. PROPOSITION VI. THEOREM. The area of a triangle is equal to half the product of its base and altitude. Let BAC be a triangle, and AD perpendicular to the base: then will its area be equal...

...each other, as AXC BxC: that is, as A : BGEOMETRYAreas of Triangles and Trapczoids1) THEOREM IX% The area of a triangle is equal to half the product of its base by its altitudeLet ABC be any triangle and CD its altitude : then will its area be equal to hall'...