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

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Page 84

A Circle being given , to cut off a

a given Line , to construct a

to cut off , from the given Circle FHI , a

...

A Circle being given , to cut off a

**Segment**, similar to a given**Segment**; and , ona given Line , to construct a

**Segment**, fimilar to a given one . First . It is requiredto cut off , from the given Circle FHI , a

**Segment**fimilar to X. Find the Center , c , of...

Page 85

FH is the given Line , on which to construct a

Angle HFG equal to BAD ; which , AB , the Chord of the given

with the Diameter AD ; found as above . Draw HG perpendicular to FH , cuting FG

in ...

FH is the given Line , on which to construct a

**Segment**as required . Make theAngle HFG equal to BAD ; which , AB , the Chord of the given

**Segment**, makeswith the Diameter AD ; found as above . Draw HG perpendicular to FH , cuting FG

in ...

Page 197

An Angle in a

greater Segnient is acute . In the SegmentADCB seeing that theAngleABC is right

, ACB , in that

...

An Angle in a

**Segment**, less then a Semicircle , is obluse ; and an Angle in agreater Segnient is acute . In the SegmentADCB seeing that theAngleABC is right

, ACB , in that

**Segment**, is acute . - C3 . 10. 1 . And , the Angle AGB + ACB = two...

Page 292

PROBLEM I. The two Sides , or Legs , of a right angled Triangle being given ,

how to find the Hypothenuse , the perpendicular , and the two

Hypothenuse , erithmetically . In the right angled Triangle , ABC , is given the two

...

PROBLEM I. The two Sides , or Legs , of a right angled Triangle being given ,

how to find the Hypothenuse , the perpendicular , and the two

**Segments**of theHypothenuse , erithmetically . In the right angled Triangle , ABC , is given the two

...

Page 331

Hence , if a Right Line be divided in extreme and mean Proportion , and from

cither extreme of the greater

also be divided in the fame Ratio ; and the lefser

...

Hence , if a Right Line be divided in extreme and mean Proportion , and from

cither extreme of the greater

**Segment**the measure of the less is set off , it willalso be divided in the fame Ratio ; and the lefser

**Segment**of the first , will be the...

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A Royal Road to Geometry: Or, an Easy and Familiar Introduction to the ... Thomas Malton No preview available - 2016 |

### 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.