MENSURATION.

SECTION IX.

1. Find the number of turfs 4 ft. by 8 in. required for a gurden plot 50 ft. by 75 ft., allowing for four circular beds, diameter 6 ft.

Area of garden plot = (75 × 50) sq. ft. = 3750 sq. ft.

circular beds = 4(62 × .7854)

=

=4(36 x .7854) sq. ft. (4 x 28.2744)sq. ft. 113.0976 sq. ft.

=

.. Area of garden plot without circular beds

= (3750 - 113.0976) sq. ft. = 3936.9024 sq. ft.

2

Area of one turf = (4 × 3) sq. ft. =3 sq. ft.

No. of turfs reqd. = 3636.9024÷ 3

8

10910.7072÷8=1363.8384

I 20000

=(400 x 300) sq. yards = 120000 sq. yards

3. A schoolroom is 60 ft. long, 20 ft. broad, 10 ft. high to

the wall plate, 16 ft. high to the
ridge of the roof. How many
children would it contain, allow-
ing 80 cubic feet of space to
each child? By how much does
this exceed the workable number
at 8 sq. ft. per child?
Breadth =20 ft.

Cubical space of schoolroom

= (12000+ 3600) or 15600 cub. ft.

No. of children it would contain

Show that the following definitions are incomplete :-' Of quadrilateral figures, a square has all its sides equal.' 'An acute-angled triangle is that which has two acute angles.'

'Parallel straight lines are such as do not meet however far they may be produced.'

(1.) Of quadrilateral figures, a rhombus also has all its sides equal; the complete definition is—a square has all its sides equal, and all its angles right angles.

(2.) A triangle can have two acute angles, and the third angle may be a right angle or an obtuse angle; therefore the definition is incomplete. An acute-angled triangle is that which has three acute angles.

(3.) Straight lines in different planes may not meet however far they may be produced; therefore the complete definition is parallel straight lines are such as are in the same plane, and which, being produced ever so far both ways, do not meet.

SECTION II.

I. If the equal sides of an isosceles triangle be produced, the angles on the other side of the base shall be equal.

Show that this property can be proved by a method similar to that employed in the 4th proposition.

Proof. Apply the triangle MKH to the triangle LHK,

so that the point M be on L, and the base MH on the base LK, then the point H shall coincide with the point K, because MH is equal to LK; then MH coinciding with LK, the line MK shall

L

G

K

M

coincide with LH, because the angle HMK is equal to the angle HLK; wherefore also the point K coincides with the point H, because MK is equal to LH, and the remaining side HK is common to both triangles, therefore the triangle MKH coincides with the triangle LHK, and the angle MKH coincides with the angle LHK (Ax. 8).

Q.E.D.

3. Straight lines, which are parallel to the same straight line, are parallel to each other.

If two adjacent sides of a parallelogram be parallel to two adjacent sides of another parallelogram, the other sides will also be parallel.

Let the two adjacent sides LM and MN of the parallelogram LMNO

be parallel to

the two adja- L

cent sides PQ

Р

S

PQRS, then ON shall be parallel to SR and LO to PS.

Proof.-ON is parallel to LM (def.), and PQ is parallel to LM (hyp.), then ON and PQ are parallel to each other (I. 30).

Again, SR is parallel to PQ (def.), and ON has just been proved parallel to PQ, therefore ON is parallel to SR (I. 30). Similarly it can be proved that LO is parallel to PS.

Q.E.D.

SECTION III.

1. Triangles upon the same base and between the same parallels are equal.

Construct a triangle equal to a given triangle and having a base three times as great.

Trisect the perpendicular LP in Q and R.

Produce the base MN both ways to S and T, making MS and MT each equal to MN (I. 3).

Join RS and RT; then the triangle RST is equal to the given triangle LMN, and the base ST is three times as great as the base MN.

Proof. Join QM, QN, RM, and RN (post. 1).

Because the triangles RMN and RSM are upon equal bases and between the same parallels, therefore they are equal (I. 38).

Similarly the triangle RMN is equal to the triangle RNT.

Therefore the triangle RST is treble the triangle RMN.

Again, the triangles NLQ, NQR, and NRP are upon equal bases and between the same parallels, therefore they are equal to one another (I. 38); therefore the triangle NLP is treble the triangle NRP.

Similarly the triangle MLP is treble the triangle MRP; therefore the whole triangle LMN is treble the whole triangle RMN.

But the triangle RST is treble the triangle RMN, therefore the triangle RST is equal to the triangle LMN, and the base ST is three times as great as the base MN. Q.E.F.

2. To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

To one of the sides of an equilateral triangle apply an equal parallelogram having one of its angles equal to that of the given triangle.