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THE AREA OF ANY RECTILINEAL FIGURE.

1st METHOD. A rectilineal figure may be divided into triangles whose areas can be separately calculated from suitable measurements. The

sum of these areas will be the area of the given figure.

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Example. The measurements quired to find the area of the figure ABCDE are AC, AD, and the offsets BX, DY, EZ.

A

B

2nd METHOD. The area of a rectilineal figure is also found by taking a base-line (AD in the diagram below) and offsets from it. These divide the figure into right-angled triangles and right-angled trapeziums, whose areas may be found after measuring the offsets and the various sections of the base-line. Example. Find the area of the enclosure ABCDEF from the plan and measurements tabulated below.

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= × 10 × 15 =

=× 18×20= 180
=× 16 × 18= 144
=× 6×12=

36

▲ DVC. DV x VC
trapm XFEZ. XZ × (XF + ZE) = 1 × 30 × 33 =
trapm YBCV. YV × (YB + VC) = 1 × 32 × 32 = 512

..., by addition, the fig. ABCDEF=

495

1442 sq. yds.

EXERCISES.

1. Calculate the areas of the figures (i) and (ii) from the plans and dimensions (in cms.) given below.

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2. Draw full size the figures whose plans and dimensions are given below; and calculate the area in each case.

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3. Find the area of the figure ABCDEF from the following measurements and draw a plan in which 1 cm. represents 20 metres.

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EXERCISES ON QUADRILATERALS.

(Theoretical.)

1. ABCD is a rectangle, and PQRS the figure formed by joining in order the middle points of the sides.

Prove (i) that PQRS is a rhombus ;

(ii) that the area of PQRS is half that of ABCD.

Hence shew that the area of a rhombus is half the product of its diagonals.

Is this true of any quadrilateral whose diagonals cut at right angles? Illustrate your answer by a diagram.

2. Prove that a parallelogram is bisected by any straight line which passes through the middle point of one of its diagonals.

Hence shew how a parallelogram ABCD may be bisected by a straight line drawn

(i) through a given point P;

(ii) perpendicular to the side AB;

(iii) parallel to a given line QR.

3. In the trapezium ABCD, AB is parallel to DC; and X is the middle point of BC. Through X draw PQ parallel to AD to meet AB and DC produced at P and Q. Then prove

(i) trapezium ABCD = parm APQD.

(ii) trapezium ABCD=twice the AXD.

(Graphical.)

4. The diagonals of a quadrilateral ABCD cut at right angles, and measure 3.0" and 2.2" respectively. Find the area.

Shew by a figure that the area is the same wherever the diagonals cut, so long as they are at right angles.

5. In the parallelogram ABCD, AB=80 cm., AD=3.2 cm., and the perpendicular distance between AB and DC=3.0 cm. Draw the par allelogram. Calculate the distance between AD and BC; and check your result by measurement.

6

One side of a parallelogram is 2.5", and its diagonals are 34" and 24". Construct the parallelogram; and, after making any necessary measurement, calculate the area.

7. ABCD is a parallelogram on a fixed base AB and of constant Find the locus of the intersection of its diagonals.

area.

EXERCISES LEADING TO THEOREM 29.

In the adjoining diagram, ABC is a triangle right-angled at C; and squares are drawn on the three sides. Let us compare the area of the square on the hypotenuse AB with the sum of the squares on the sides AC, CB which contain the right angle.

A

B

1. Draw the above diagram, making AC-3 cm., and BC=4 cm.; Then the area of the square on AC=32, or 9 sq. cm. and .... ..... the square on BC=42, or

.. the sum of the squares on AC, BC=

16 sq. cm. J

25 sq. cm.

Now measure AB; hence calculate the area of the square on AB, and compare the result with the sum already obtained.

2. Repeat the process of the last exercise, making AC=1.0", and BC=2.4".

3. If a=15, b=8, c=17, shew arithmetically that c2=a2+b2.

Now draw on squared paper a triangle ABC, whose sides a, b, and c are 15, 8, and 17 units of length; and measure the angle ACB.

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In any right-angled triangle the square on the hypotenuse is equal

to the sum of the squares on the other two sides.

A formal proof of this theorem is given on the next page.

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THEOREM 29. [Euclid I. 47.]

In a right-angled triangle the square described on the hypotenuse is equal to the sum of the squares described on the other two sides.

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Let ABC be a right-angled ▲, having the angle ACB a rt. L.

It is required to prove that the square on the hypotenuse AB = the sum of the squares on AC, CB.

On AB describe the sq. ADEB; and on AC, CB describe the sqq. ACGF, CBKH.

Through C draw CL par to AD or BE.
Join CD, FB.

Proof. Because each of the 4 ACB, ACG is a rt. 4,
... BC and CG are in the same st. line.

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