MORE ABOUT TRIANGLE -2: AREA OF A TRIANGLE,POLYGONS & CIRCLE

 

       Different units of length may be taken for measuring lengths depending upon the length of the thing whose length is to be measured. For measuring small lengths we take unit of length as millimeter or centimeter while measuring greater lengths we take unit of length as meter. For measuring even larger distances we take unit of length as kilometer.

Just as liquids have their volume, surfaces have their area. Since liquids do not have fixed length, breadth or depth and they take shape of the container in which thery are kept. Therefore, their length, breadth and depth cannot be measured as in the case of solids or surfaces. For measuring volume, we make a container and caliberate it. We can also think of a container having unit length, unit breadth and unit height or depth. This container will represent unit volume. If measuring unit is millimeter, then unit of volume will be 1 millimeter * 1 millimeter *1 millimeter =1*1*1 millimeter*millimeter*millimeter which is 1 millimeter³ (since millimeter * millimeter*millimeter is cubic millimeter written as millimeter cubed). Think os a milkman who supplies milk in liters from a big container. He measures milk with a standard container which is a standard measure.

In case of shapes made on a surfaces, shape can be divided in pieces (small areas) of unit length and unit breadth. Thereafter, we can count number of such small areas of unit length and unit breadth. Total area of surface will be product of number of small areas and area of one such small area. If we take 1 centimeter as unit length and 1 centimeter as unit breadth of small area then unit area will cover space having 1 centimeter length and 1 centimeter breadth and its area will be 1 centimeter * 1 centimeter=1*1 centimeter*centimeter which is 1 centimeter².
      Look at the rectangle, given hereunder, whose length is 3 centimeter and breadth is 2 centimeter.

We have divided it in pieces each having length of 1 centimeter and breadth also 1 centimeter.

Hence area of one piece is 1 centimeter * 1 centimeter = 1 centimeter².

Total number of pieces = 6

Therefore, area of whole rectangle is = area of one piece * number of pieces.

Therefore, total area of rectangle =6 * 1 centimeter² =6 centimeter²

IMPORTANT: Length and breadth distances should be taken in the same units in which we want area to be computed. I mean to say that if area is to be computed in cm² , length and breadth both should be taken in centimeters and if we want to compute area in square meters then length as well as breadth are to be taken in meters only.

       Square is specific form of rectangle whose length and breadth are equal. Area of a square can also be computed in the manner given above. Area of square = length * breadth cm² = (length of side)² * cm². Because in case of a square length and breadth are equal.

 

        Area of atriangle can be computed by converting a triangle into a rectangle or part of single or several rectangles. Look at the following figures. PQR is a triangle in Figure 1 which have been converted into parts of two rectangles APCQ & PBRC. QR is the base of the triangle and point A is vertex. We have drawn a line parallel to base passing through vertex A. Again We have drawn perpendicular on base from vertex A and two other perpendiculars QA & RB from Q & R endpoints of base line QR on parallel line passing through A . Thus we see that two rectangles APCQ & PBRC have been constructed. We see that triangle PQR has been divided into two triangles triangle PQC & triangle PCR.

        Here we see that one part of triangle PQR is part of rectangle APCQ and rest part of the triangle is part of rectangle PBRC and both rectangles constitute bigger rectangle ABRQ. This proves that triangle PQR is part of rectangle ABRQ..

      Let us measure perpendicular distance PC in between base line and line passing through A which is parallel line to base . Let this distance be h cm. Since this is the perpendicular distance in between two parallel lines. Therefore it will be equal at every point of two parallel lines AB & QR. Hence, AQ =BR = PC = h cm.

        Now if we can prove that one part PQC of triangle PQR is congruent to triangle APQ, then triangle PQC will be half of the rectangle APCQ. Similarly, if we can prove that other part PCR of triangle PQR is congruent to triangle PBR, then triangle PCR will be half of the rectangle PBRC. But rectangle APCQ + rectangle PBRC= rectangle ABRQ. Which will give

        1/2APCQ + 1/2 PBRC= 1/2 ABRQ

      But then one part of triangle PQC = 1/2 APCQ ------------------------------- (i)

      And other part of our triangle PCR =1/2 PBRC -------------------------------- (ii)

      Adding these two equations (i) & (ii) (For adding any equations, we add left hand side terems to left hand side terms and right hand side terms to right hand side terms), we get: -

      triangle PQC + triangle PCR = 1/2 rectangle APCQ + 1/2 rectangle PBRC --------- (1)

      But

       triangle PQC + triangle PCR = triangle PQR

      and       in respect of rectangles,we have seen that-

       1/2 + 1/2 PBRC= 1/2 ABRQ

      Putting these values in equation (1), we get:

       triangle PQR = 1/2 ABRQ

       Now for area of rectangle ABRQ, we see that QR is lts length and PC is its breadth. Area of a rectangle = length * breadth. Here length of the triangle is the length of the base of the triangle and breadth of rectangle is the height of the triangle.

       Lines AQ and PC are perpendicular distances in between base and line parallel to base and passing through vertex A. Hence they are equal in length. Also both of them are perpendicular to base & parallel line to base, hence they will make right angles (90 ⁰) from both lines. Therefore, < PAQ = < PCQ = 90⁰. Line PQ is common to both triangles and is hypotenues for both triangles.

       Trapezium is a quadrilateral (polygon having four sides) whose two opposite sides are parallel. We can divide a trapezium in two triangles, one each on a parallel side as shown in the figure below.

        Triangulation is the process by which we can divide any polygon into triangles. If a polygon does not have any interior angle greater than 180⁰ then we can divide the trapezium in triangles by joining all its vetices to a point inside the polygin. For this purpose we take a point inside the polygon and connect this point to all vertices of polygon.In this case nNumber of triangles is equal to number of sides of polygon. We can also join vertices to one vertex of the polygon but then number of triangles will be n -2 where n is the number of sides of polygon.

        In cases of polygons which have one or more interior angles greater than 180⁰ (concave polygons), area can be found by joining vetex of such angle ( angle greater than 180⁰ ) to neibouring second vertex on one side. This will reduce the polygon in triangles and regular polygon. In this case initially, a regular (convex) polygon and triangles will be formed. Now convex polygon may further be divided in triangles.

We know the method for computing area of triangles. We can compute area for all triangles separately. Sum of these areas will give us total area of trapezium.

AREA OF A CIRCLE: Circle has a center, radius and circumference. If length of radius is r, then length of circumference is 2 π r. Also we know that 2 π is the angle in radian formed by circumference at the center. While defining a curved line we have stated that a curved line is made of several small straight lines. One may think about surface of earth. If we travel from East to West or from West to East parallel to equator, we find that path is curved like circumference of a circle. But every small distance along this path is a straight line. So we can always divide a circular path in very small straight lines.

       Thus, every small portion of arc will be a straight line. Therefore, we can divide whole circumference in small straight lines. Now take a small angle θ made at center O of the circle made by two radii OP₁ and OP₂. Length of these lines is equal to r (since r is the radius and OP₁ & OP₂, both are radii of the circle. Let s be the length of small arc made by both radii at the center O. Since the arc is very small (almost near to zero), it will be a straight line joining points P₁ & P₂. Being small line this will be perpendicular to OP₁. Therefore, triangle OP₁P₂ will be a right triangle. Hence area of triangle OP₁P₂=1/2 P₁ P₂* OP₁ which is = 1/2 r * arc s

       Thesefore area of triangle OP₁P₂ =½r*s [half of (radius * arc)].

       Now arc s and radii OP₁& OP₂ make angle θ at the center. Since angle =arc / radius

       Therefore θ= s/r

       We can find out total number of such small arcs of s length made on whole circumference 2 π r by dividing 2 π r by s. Also we can find out total number of θ angles made by such each arc at the center by dividing total angle 2 π by θ. Number of arcs and angles will be equal.

       Number of triangles formed at the center will be same as total number of small arcs. Number of total small angles is 2 π r /s

       Therefore total number of small triangles =2 π r /s

                            Now area on one triangle= ½ r*s

       Therefore area of all triangles = area of one triangle * total number of triangles, which will give

       Total area of all triangles =½ rs * 2 π r/s

       Which gives Total area of all triangles=π*r²

       But total area of all triangles = total area of circle

       Hence, total area of circle = π r² Where r is the radius of circle and π is universal constant having fixed value.