RIGHT-ANGLED TRIANGLE.

35

129. Note 1.-Hence another method of making a triangle equal in all respects to a given triangle

Make any side and the adjacent angles equal, each to each, to like parts in the given triangle; for—

Two triangles are equal in all respects when they have a side and the adjacent angles in one equal, each to each, to like parts in the other.

130. Note 2.-It is an important truth in geometry that

The interior angles of every triangle are together equal to two right angles;—

Whence, any two of them must be less than two right angles.

If the two adjacent angles at the ends of a line were exactly equal to two right angles, their sides would be parallel, and would not meet, however D far produced either way; as, AB, CD.

If they were together greater

than two right angles, their

B

sides would diverge when produced; as, CD, EF.

Problem 20.

131. To construct a right-angled triangle.

E

1. When the base and perpendicular are given; as,

AC and CB (fig. 33).

This is a case of Problem 16.

132. 2. When the base (or perpendicular) and an acute angle are given.

(1.) When the given side is between the two given angles (right and acute); as, AC, and the angles BAC, and C being given.

This is a case of Problem 19.

(2.) When the given side is opposite to the given

FIG. 33

B

C

acute angle; as, AC and the angle required at B.

At A and C draw AD and CB perpendicular to AC. At A in the line AD, make the angle DAB equal to the given acute angle, producing AB till it meets the perpendicular at C in B.

The angle at B will be equal to the given acute angle.

If the acute angle required

at B is given in degrees, subtract it from 90°; the difference will give the angle BAC; we may then proceed by Problems 11 and 19.

133. 3. When the hypotenuse and a side are given ; as, AB and AC.

At one end of the given side, as C, raise a perpendicular. From the other end A, as centre, with a radius equal to the hypotenuse, draw an arc, cutting the perpendicular in B. Join AB.

134. 4. When the hypotenuse and an acute angle are given; as, AB and B.

At one end of the hypotenuse, as B, make an angle equal to the given angle; from the other end, A, let fall a perpendicular, AC, on the other side of the acute angle.

Or, having made the acute angle (at B), bisect the hypotenuse; from its middle point, with its half as radius, describe a semicircle; join the point (C) where the arc cuts the side of the acute angle to the other end (A) of the hypotenuse (72).

135. Note 1.-In a right-angled triangle, there is always one angle known-the right angle. As all the angles of the triangle are together equal to two right

angles, the two acute angles must be together equal to one right angle; and one of them is equal to the difference between the other and one right angle.

Note 2.-In every right-angled triangle, the half of the hypotenuse is equal to the line from its middle point to the right angle.

Problem 21.

136. To describe a rectangle, the diagonal and a side being given.

By case 3 of the preceding problem, make a rightangled triangle, having the hypotenuse equal to the diagonal, and a side equal to the given side. Then, by Problem 15, make an equal triangle on the other side of the hypotenuse, taking care that, in the figure formed, the equal sides are opposite, not adjacent.

The quadrilateral composed of the two triangles will be the required rectangle.

Or, having made the right-angled triangle as above, through the ends of the hypotenuse, draw lines parallel to the sides of the triangle.

137. Note.-The diagonal divides every parallelogram into two triangles equal in all respects, so that if one of these is known, the whole is easily formed.

Problem 22.

138. To make a rectangle, rhombus, rhomboid, or any quadrilateral, when the diagonal and sides are known. It is manifest from the preceding note, that, in the case of the rhombus, rhomboid, or rectangle, this is simply making on opposite sides of the same line (the diagonal) two triangles with equal given sides, as in Problem 15. In the case of the rectangle and rhomboid, care must be taken that the equal sides are opposite, not adjacent. In the case of the trapezium, the triangles are different; but still use Problem 15.

Problem 23.

139. To construct a triangle, given the base, the perpendicular on the base from the opposite angle, and the place where the perpendicular meets the base.

The construction is obvious. Having drawn the base and the perpendicular of the given length and at the given point in the base, join the end of the perpendicular to the ends of the base.

Problem 24.

140. To form a trapezium, the diagonal being given; also the perpendiculars on it from the opposite angles, and the places of these on the diagonal.

The construction is obvious from the preceding problem.

3. PROBLEMS FOR THE CONSTRUCTION OF RECTILINEAL FIGURES EQUAL IN AREA TO OTHERS, THOUGH UNEQUAL IN OTHER RESPECTS.

(Sometimes called equivalent figures.)

Problem 25.

141. To make a rectangle equal to a given parallelogram.

Let ABCD be the given parallelogram.

Through A and D draw AE and DF at right angles to AD, and meeting the opposite side, or that side produced, in E and F.

AEFD will be a rectangle, and it will be equal in area to ABCD.

Another rectangle, equal to ABCD, might be constructed by drawing perpendiculars to AB at A and B to meet CD or that line produced.

EQUIVALENT FIGURES.

39

142. Note. This illustrates the important geometrical truth, that—

Parallelograms on the same base and between the same parallels are equal to one another; that is, equal in

area.

FIC. 34

B

F

M

AEFD and ABCD are on the same base AD and between the same parallels AD and EC. The parts of the whole figure that do not form part of both, are manifestly equal-the triangles AEB, DFC.

Or, parallelograms are equal whose bases and altitudes are equal.

Problem 26.

143. To make a rectangle equal to a given triangle. Let ABD in the preceding figure be the given triangle. Bisect AD in M. Through M and D draw MB and DF at right angles to AD. Through B draw BC parallel to AD.

MBFD will be a rectangle, and it will be equal to the triangle ABD.

144. Note.-A triangle is half of any parallelogram on the same base and between the same parallels (or, of the same base and altitude).

Whence, the triangle ABD is half of the rectangle AEFD, and must be equal to the rectangle MBFD, which is obviously the half of the rectangle AEFD.