Key to the Exercises in Wells' Plane and Solid Geometry |
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AB+ BC AB² ABCD AC and BD AC² ACD and BCD AD² altitude apothem arc BD arc cutting area ABC base BC² BD² BE² bisector CD bisect CD² centre chord circumference circumscribed const describe an arc diagonals AC diameter diedral Draw AD bisecting Draw AE Draw CD Draw the line equally distant equilateral figure EFGH frustum GHKL hexagon hypotenuse intersecting isosceles join AC join OA KLMN lateral area Let AB Let ABC Let the line line AB equal measured by arc middle point parallelopiped pentagon plane MN polygon pyramid quadrilateral quadrilateral ABCD radius equal rectangle rhombus right angles segment semicircle spherical polygon square straight line tangent vertex volume Whence
Popular passages
Page 67 - The area of a regular inscribed hexagon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles.
Page 59 - In any triangle the sum of the squares on the sides is equal to twice the square on half the base together with twice the square on the straight line drawn from the vertex to the middle point of the base.
Page 39 - A, with a radius equal to the sum of the radii of the given circles, describe a circle.
Page 21 - BCA are together less than two right angles. 3. Not more than two equal straight lines can be drawn from a given point to a given straight line. 4.
Page 11 - If two triangles have the three sides of one equal respectively to the three sides of the other, the triangles are congruent, (sss) A*. Given the triangles ABC and A'B'C', in which AB equals A'B', BC equals B'C', and AC equals A'C'.
Page 73 - R, S are the middle points of the sides AB, BC, CD, and DA of a square. Compare the area of the square with that of the square formed by the joins AQ, BR, CS, and DP. 24. ABCDEFGH is a regular octagon, and AD and GE are produced to meet in K. Compare the area of the triangle DKE with that of the octagon. 25. The rectangle on the chord of an arc and the chord...
Page 39 - With 0 as a centre, and a radius equal to the difference...
Page 49 - Fig. 127. Construction for centroid of quadrilateral. Fig. 126. Construction for centroid of quadrilateral. First method. Let E be the point of intersection of the diagonals AC and BD, Fig. 126 ; from C set off CE' = AE and join DE' and BE', then the centroid of the quadrilateral will be the same as that of the A BE'D. Therefore bisect BE' and E'D in K, H and join DK, BH; then their point G of intersection gives the required centroid.
Page 43 - The volume of a cone of revolution is equal to its lateral area, multiplied by one-third the perpendicular from the vertex of the right angle to the hypotenuse of the generating triangle.
Page 24 - The two tangents to a circumference from an outside point are equal. Let AB and AC be the tangents from the point A to the circumference whose centre is at O. To prove that AB = AC.