## A Royal Road to Geometry: Or, an Easy and Familiar Introduction to the Mathematics. ... By Thomas Malton. ... |

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Page 406

с EX G If a

parallel to the Bases of the Cylinder ; each Segment of the

be equal to the portion of the cylindrical , intercepted between the same parallel ...

с EX G If a

**Sphere**be inscribed in a Cylinder , and being both cut by Planes ,parallel to the Bases of the Cylinder ; each Segment of the

**spherical**Surface willbe equal to the portion of the cylindrical , intercepted between the same parallel ...

Page 407

The Segments of a

have the fame Ratio , between ... H A

Cone , whose Base is equal to the Surface of the

The Segments of a

**spherical**Surface , intercepted between parallel Circles ,have the fame Ratio , between ... H A

**Sphere**is equal , in its solid Contents , to aCone , whose Base is equal to the Surface of the

**Sphere**, and its Altitude to the ... Page 408

I say , the Cone is equal to the

Cone is equal to the Surface of the

Figures , as the circumscribing Solid has Faces ; each equal , respectively , to ...

I say , the Cone is equal to the

**Sphere**. B F Dem . For , because the Base of theCone is equal to the Surface of the

**Sphere**, it may be divided into as many planeFigures , as the circumscribing Solid has Faces ; each equal , respectively , to ...

Page 26

To find the solid Contents of a

Area of a Circle of that Diameter ( Art . 8. ) which being multiplied , by the

Diameter , gives the Contents of a Cylinder , whose height and Diameter are

equal .

To find the solid Contents of a

**Sphere**. 1 Having obtained its Diameter ; find theArea of a Circle of that Diameter ( Art . 8. ) which being multiplied , by the

Diameter , gives the Contents of a Cylinder , whose height and Diameter are

equal .

Page 27

47,1 + 15 Multiply 47,14 by 15 , the Diameter of the

Area of the Surface of the

equal to a Cone , the Area of whose Base is 707,1 ; and its Altitude equal to its

Radius ...

47,1 + 15 Multiply 47,14 by 15 , the Diameter of the

**Sphere**, the Product is theArea of the Surface of the

**Sphere**. Now , since the Contents of the**Sphere**isequal to a Cone , the Area of whose Base is 707,1 ; and its Altitude equal to its

Radius ...

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### Common terms and phrases

ABCD added alſo Altitudes analogous Area Baſe becauſe biſected Book called Center Chord Circle Circumference common Cone conf conſequently Conſtruction contained cuting Cylinder Demonſtration deſcribe Diagonal Diameter difference divided draw drawn equal Euclid evident extreme fame Feet Figure firſt formed four fourth given given Line greater half Hence Inches inſcribed join laſt leſs manner mean meaſure multiplied muſt oppoſite parallel Parallelogram Parallelopiped Pentagon perpendicular Plane Point Poligon Priſm Prob PROBLEM produced Proportion Propoſition proved Pyramid Quantities Radius Ratio Rect Rectangle reſpectively Right Angles Right Line ſame ſame Ratio ſay ſeeing Segment Sides ſimilar Solid ſome Sphere Square ſuch Surface taken Terms THEOREM third thoſe touch Triangle uſe wherefore whole whoſe

### Popular passages

Page 124 - When you have proved that the three angles of every triangle are equal to two right angles...

Page 221 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Page 285 - EG, let fall from a point in the circumference upon the diameter, is a mean proportional between the two segments of the diameter DS, EF (p.

Page 284 - IN a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.

Page 186 - From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are equal, they are right angles.

Page 248 - To express that the ratio of A to B is equal to the ratio of C to D, we write the quantities thus : A : B : : C : D; and read, A is to B as C to D.

Page 161 - In any triangle, if a line be drawn from the vertex at right angles to the base; the difference of the squares of the sides is equal to the difference of the squares of the segments of the base.

Page 160 - In any isosceles triangle, the square of one of the equal sides is equal to the square of any straight line drawn from the vertex to the base plus the product of the segments of the base.

Page 250 - Ratios that are the same to the same ratio, are the same to one another. Let A be to B as C is to D ; and as C to D, so let E be to F.

Page 124 - Angles, taken together, is equal to Twice as many Right Angles, wanting four, as the Figure has Sides.