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

### From inside the book

Page 97

will form a Rectangle ; and if the Diagonals of the Rectangle , FH and IG , be

drawn , they will pass through the Center of the Ellipsis ; the parts , KM and LN ,

which are terminated by the Curve , are

is ...

will form a Rectangle ; and if the Diagonals of the Rectangle , FH and IG , be

drawn , they will pass through the Center of the Ellipsis ; the parts , KM and LN ,

which are terminated by the Curve , are

**Diameters**of the Ellipsis ; each of which ,is ...

Page 107

If from the Point of Coutact , K , of any Tangent , KN , to an Ellipsis , an Ordinate ,

KI , be drawn to any

Proportional between the Segos ment El , made by the Ordinate , and the

Dirtance ...

If from the Point of Coutact , K , of any Tangent , KN , to an Ellipsis , an Ordinate ,

KI , be drawn to any

**Diameter**, cuting it in I ; half that**Diameter**, AE , is a meanProportional between the Segos ment El , made by the Ordinate , and the

Dirtance ...

Page 202

But four times AFO = ALO ; i . e . of the

Segments , AE , EB , CE , & ED , are equal to the Square of the

Having joined the extremes AD , DB , AC , and CB . Then , the Squares of AD ...

But four times AFO = ALO ; i . e . of the

**Diameter**. - 4.2 Th . the Squares of the fourSegments , AE , EB , CE , & ED , are equal to the Square of the

**Diameter**. 2nd .Having joined the extremes AD , DB , AC , and CB . Then , the Squares of AD ...

Page 405

Consequently , a Circle whose Radius is equal to the

the Surface of a Sphere , described by the revolution of the ... But , Circles are to

each other , or amongst themselves , as Squares of their

Consequently , a Circle whose Radius is equal to the

**Diameter**, AB , is equal tothe Surface of a Sphere , described by the revolution of the ... But , Circles are to

each other , or amongst themselves , as Squares of their

**Diameters**, Cor . 1. 14. Page 26

1 Having obtained its

) which being multiplied , by the

whose height and

a ...

1 Having obtained its

**Diameter**; find the 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 . The Sphere is equal to two thirds of sucha ...

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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.