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

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Page 50

Hence we learn , to find the Area of any Quadrilates ral whatever ; the Rúle for

which is , to multiply the

culars ( Fig . 1. ) or , the Sum of the two ' Perpendiculars ( BE and FD ) by half the

...

Hence we learn , to find the Area of any Quadrilates ral whatever ; the Rúle for

which is , to multiply the

**Diagonal**( AC ) by half the Sum of the two Perpendisculars ( Fig . 1. ) or , the Sum of the two ' Perpendiculars ( BE and FD ) by half the

...

Page 253

Suppose the Ratio of A to B and of C to D be as the Side of a Square to the

manifest , that the side of one Square is to its

Suppose the Ratio of A to B and of C to D be as the Side of a Square to the

**Diagonal**; or , as the Diameter of a Circle is to its Circumference , & c . Now , it ismanifest , that the side of one Square is to its

**Diagonal**, as the side of any other ... Page 303

The Perimeters or Circuits of similar Figures are , to each other , as their

corresponding Sides or

consequently , as any one side , or

The Perimeters or Circuits of similar Figures are , to each other , as their

corresponding Sides or

**Diagonals**. ... fo is**Diagonal**to**Diagonal**, AC to FH , & c ;consequently , as any one side , or

**Diagonal**, is to its corresponding Side or**Diagonal**... Page 330

The Diameter or Perpendicular , of a regular Pentagon , is divided in extreme and

mean Proportions by ' a

ABCDE , let the Diameter C F be drawn , and the

in ...

The Diameter or Perpendicular , of a regular Pentagon , is divided in extreme and

mean Proportions by ' a

**Diagonal**cuting it at Right Angles . In the PentagonABCDE , let the Diameter C F be drawn , and the

**Diagonal**G HD BD , cuting C Fin ...

Page 18

The

hoth , is obtained at cnce , by multiplying the

the two Perpendiculars ( Prob.22 . ) The Area of any other two contiguous ...

The

**Diagonal**, BG , is then a common Bafe to the two Triangles and the Area , ofhoth , is obtained at cnce , by multiplying the

**Diagonal**, BC , by half the Sum ofthe two Perpendiculars ( Prob.22 . ) The Area of any other two contiguous ...

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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.