Introduction
You know that the paper is a model for a plane surface. When you join a number of points without lifting a pencil from the paper (and without retracing any portion of the drawing other than single points), you get a plane curve.
Convex and concave polygons
A simple closed curve made up of only line segments is called a polygon.
Here are some convex polygons and some concave polygons. (Fig 3.1)
Can you find how these types of polygons differ from one another? Polygons that are convex have no portions of their diagonals in their exteriors or any line segment joining any two different points, in the interior of the polygon, lies wholly in the interior of it . Is this true with concave polygons? Study the figures given. Then try to describe in your own words what we mean by a convex polygon and what we mean by a concave polygon. Give two rough sketches of each kind. In our work in this class, we will be dealing with convex polygons only
Regular and irregular polygons
A regular polygon is both ‘equiangular’ and ‘equilateral’. For example, a square has sides of equal length and angles of equal measure. Hence it is a regular polygon. A rectangle is equiangular but not equilateral. Is a rectangle a regular polygon? Is an equilateral triangle a regular polygon? Why?
[Note: Use of or indicates segments of equal length].
In the previous classes, have you come across any quadrilateral that is equilateral but not equiangular? Recall the quadrilateral shapes you saw in earlier classes – Rectangle, Square, Rhombus etc. Is there a triangle that is equilateral but not equiangular?
Sum of the Measures of the Exterior Angles of a Polygon
On many occasions a knowledge of exterior angles may throw light on the nature of interior angles and sides.
DO THIS
Draw a polygon on the floor, using a piece of chalk. (In the figure, a pentagon ABCDE is shown) (Fig 3.2).
We want to know the total measure of angles, i.e, m∠1 + m∠2 + m∠3 + m∠4 + m∠5. Start at A. Walk along \(\overline{AB}\).On reaching B, you need to turn through an angle of m∠1, to walk along. \(\overline{BC}\) When you reach at C, you need to turn through an angle of m∠2 to walk along \(\overline{CD}\). You continue to move in this manner, until you return to side AB. You would have in fact made one complete turn
Therefore, m∠1 + m∠2 + m∠3 + m∠4 + m∠5 = 360°. This is true whatever be the number of sides of the polygon. Therefore, the sum of the measures of the external angles of any polygon is 360°
Example : Find measure x in Fig 3.3.
Solution: x + 90° + 50° + 110° = 360° (Why?)
x + 250° = 360°
x = 110°
Example : Find the number of sides of a regular polygon whose each exterior angle has a measure of 45°.
Solution: Total measure of all exterior angles = 360°
Measure of each exterior angle = 45°
Therefore, the number of exterior angles =\(360 \over45\) = 8
The polygon has 8 sides
Kinds of Quadrilaterals
Based on the nature of the sides or angles of a quadrilateral, it gets special names.
Trapezium
Trapezium is a quadrilateral with a pair of parallel sides
Study the above figures and discuss with your friends why some of them are trapeziums while some are not. (Note: The arrow marks indicate parallel lines).
Kite
Kite is a special type of a quadrilateral. The sides with the same markings in each figure are equal. For example AB = AD and BC = CD.
Study these figures and try to describe what a kite is.
Observe that
(i) A kite has 4 sides (It is a quadrilateral).
(ii) There are exactly two distinct consecutive pairs of sides of equal length. Check whether a square is a kite.
Parallelogram
A parallelogram is a quadrilateral. As the name suggests, it has something to do with parallel lines.
Study these figures and try to describe in your own words what we mean by a parallelogram. Share your observations with your friends. Check whether a rectangle is also a parallelogram.
A parallelogram is a quadrilateral whose opposite sides are parallel
Elements of a parallelogram
There are four sides and four angles in a parallelogram. Some of these are equal. There are some terms associated with these elements that you need to remember.
Given a parallelogram ABCD (Fig).
\(\overline{AB}\) and \(\overline{DC}\), are opposite sides. \(\overline{AD}\) and \(\overline{BC}\) form another pair of opposite sides.
∠A and ∠C are a pair of opposite angles; another pair of opposite angles would be ∠B and ∠D.
\(\overline{AB}\) and \(\overline{AB}\) are adjacent sides. This means, one of the sides starts where the other ends. Are \(\overline{BC}\) and \(\overline{CD}\) adjacent sides too? Try to find two more pairs of adjacent sides.
∠A and ∠B are adjacent angles. They are at the ends of the same side. ∠B and ∠C are also adjacent. Identify other pairs of adjacent angles of the parallelogram.
Property:
The opposite sides of a parallelogram are of equal length.
You can further strengthen this idea through a logical argument also.
Consider a parallelogram ABCD (Fig ).
Draw any one diagonal, say \(\overline{AC}\).
Looking at the angles,
∠1 = ∠2 and ∠3 = ∠4 (Why?)
Since in triangles ABC and ADC,∠1 = ∠2, ∠3 = ∠4 and \(\overline{AC}\) is common,
so, by ASA congruency condition,
∆ ABC ≅ ∆ CDA (How is ASA used here?)
This gives AB = DC and BC = AD.
Example : Find the perimeter of the parallelogram PQRS (Fig 3.16).
Solution: In a parallelogram, the opposite sides have same length.
Therefore, PQ = SR = 12 cm and QR = PS = 7 cm
So, Perimeter = PQ + QR + RS + SP
= 12 cm + 7 cm + 12 cm + 7 cm = 38 cm
Angles of a parallelogram
We studied a property of parallelograms concerning the (opposite) sides. What can we say about the angles?
Property: The opposite angles of a parallelogram are of equal measure
You can further justify this idea through logical arguments.
If \(\overline{AC}\) and \(\overline{BD}\) are the diagonals of the parallelogram, (Fig ) you find that
∠1 =∠2 and ∠3 = ∠4 (Why?)
Studying ∆ABC and ∆ADC (Fig 3.19) separately, will help you to see that by ASA congruency condition,
∆ ABC ≅ ∆ CDA(How?)
This shows that ∠B and ∠D have same measure. In the same way you can get m∠A = m ∠C.
Alternatively, ∠1 = ∠2 and ∠3 = ∠4, we have, m∠A = ∠1+∠4 = ∠2+∠C m∠C
Example : In Fig 3.20, BEST is a parallelogram. Find the values x, y and z.
Solution: S is opposite to B.
So, x = 100° (opposite angles property)
y = 100° (measure of angle corresponding to ∠x)
z = 80° (since ∠y, ∠z is a linear pair)
We now turn our attention to adjacent angles of a parallelogram. In parallelogram ABCD, (Fig 3.21).
∠A and ∠D are supplementary
since \(\overline{DC}||\overline{AB}\) and with transversal \(\overline{DA}\) , these two angles are interior opposite.
∠A and ∠B are also supplementary. Can you say ‘why’?
\(\overline{AD}||\overline{BC}\)and \(\overline{BA}\) is a transversal, making ∠A and ∠B interior opposite.
Identify two more pairs of supplementary angles from the figure
Property: The adjacent angles in a parallelogram are supplementary.
Example : In a parallelogram RING, (Fig ) if m∠R = 70°, find all the other angles.
Solution: Given m∠R = 70°
Then m∠N = 70°
because ∠R and ∠N are opposite angles of a parallelogram.
Since ∠R and ∠I are supplementary,
m∠I = 180° – 70° = 110°
Also, m∠G = 110° since ∠G is opposite to ∠I
Thus, m∠R = m∠N = 70° and m∠I = m∠G = 110°
Diagonals of a parallelogram
The diagonals of a parallelogram, in general, are not of equal length. (Did you check this in your earlier activity?) However, the diagonals of a parallelogram have an interesting property.
Property: The diagonals of a parallelogram bisect each other (at the point of their intersection, of course!)
To argue and justify this property is not very difficult. From Fig , applying ASA criterion, it is easy to see that
∆ AOB ≅ ∆ COD (How is ASA used here?)
This gives AO = CO and BO = DO
Example 6: In Fig., HELP is a parallelogram. (Lengths are in cms). Given that OE = 4 and HL is 5 more than PE? Find OH.
Solution : If OE = 4 then OP also is 4 (Why?)
So PE = 8, (Why?)
Therefore HL = 8 + 5 = 13
Hence OH = 1 13 2 × = 6.5 (cms)
Some Special Parallelograms
Rhombus
We obtain a Rhombus (which, you will see, is a parallelogram) as a special case of kite (which is not a a parallelogram).
Note that the sides of rhombus are all of same length; this is not the case with the kite. A rhombus is a quadrilateral with sides of equal length.
Since the opposite sides of a rhombus have the same length, it is also a parallelogram. So, a rhombus has all the properties of a parallelogram and also that of a kite. Try to list them out. You can then verify your list with the check list summarised in the book elsewhere.
The most useful property of a rhombus is that of its diagonals.
Property: The diagonals of a rhombus are perpendicular bisectors of one another.
Here is an outline justifying this property using logical steps. ABCD is a rhombus (Fig ).
Therefore it is a parallelogram too. Since diagonals bisect each other, OA = OC and OB = OD.
We have to show that m∠AOD = m∠COD = 90°
It can be seen that by SSS congruency criterion
∆ AOD ≅ ∆ COD
therefore, m ∠AOD = m ∠COD
Since ∠AOD and ∠COD are a linear pair,
m ∠AOD = m ∠COD = 90°
Example : RICE is a rhombus (Fig ). Find x, y, z. Justify your findings.
Solution:
x = OE y = OR z = side of the rhombus
= OI (diagonals bisect) = OC (diagonals bisect) = 13 (all sides are equal )
= 5 = 12
A rectangle
A rectangle is a parallelogram with equal angles (Fig).
What is the full meaning of this definition? Discuss with your friends.
If the rectangle is to be equiangular, what could be the measure of each angle? Let the measure of each angle be x°.
Then 4x° = 360° (Why)?
Therefore, x° = 90°
Thus each angle of a rectangle is a right angle.
So, a rectangle is a parallelogram in which every angle is a right angle.
Being a parallelogram, the rectangle has opposite sides of equal length and its diagonals bisect each other
In a parallelogram, the diagonals can be of different lengths. (Check this); but surprisingly the rectangle (being a special case) has diagonals of equal length.
Property: The diagonals of a rectangle are of equal length.
This is easy to justify. If ABCD is a rectangle (Fig 3.38), then looking at triangles ABC and ABD separately [(Fig 3.33) and (Fig 3.34) respectively], we have
∆ ABC ≅ ∆ABD
This is because AB = AB (Common)
BC = AD (Why?)
m ∠A = m ∠B = 90° (Why?)
The congruency follows by SAS criterion.
Thus AC = BD and in a rectangle the diagonals, besides being equal in length bisect each other (Why?)
Example : RENT is a rectangle (Fig ). Its diagonals meet at O. Find x, if OR = 2x + 4 and OT = 3x + 1.
Solution: \(\overline{OT}\) is half of the diagonal \(\overline{TE}\) ,
\(\overline{OR}\) is half of the diagonal \(\overline{RN}\) .
Diagonals are equal here. (Why?)
So, their halves are also equal.
Therefore 3x + 1 = 2x + 4 or x = 3
A square
A square is a rectangle with equal sides.
BELT is a square, BE = EL = LT = TB
∠B, ∠E, ∠L, ∠T are right angles.
BL = ET and \(\overline{BL}||\overline{ET}\)
OB = OL and OE = OT
This means a square has all the properties of a rectangle with an additional requirement that all the sides have equal length. The square, like the rectangle, has diagonals of equal length. In a rectangle, there is no requirement for the diagonals to be perpendicular to one another, (Check this).
In a square the diagonals. (i) bisect one another (square being a parallelogram) (ii) are of equal length (square being a rectangle) and (iii) are perpendicular to one another. Hence, we get the following property.
Property: The diagonals of a square are perpendicular bisectors of each other
We can justify this also by arguing logically:
ABCD is a square whose diagonals meet at O (Fig 3.37).
OA = OC (Since the square is a parallelogram)
By SSS congruency condition, we now see that
∆ AOD ≅ ∆ COD (How?)
Therefore, m∠AOD = m∠COD
These angles being a linear pair, each is right angle