## Quadrilaterals -01

We already studied about the triangles in previous chapters. And we know that  joining 3 noncollinear points  in pairs we will get the shape triangle. Same way if we joints such a 4 points in an order we will get shape called quadrilaterals. Quadrilaterals have four side , four angle and four vertices. We can see the figures in page number 135.

Ina quadrilateral ABCD AB, BC, CD and DA are four sides. A,B,C,D are the 4 vertices. And ∠A,  ∠B,  ∠C,  ∠D, are the four angles formed at the vertices.

We can also say that the rectangle, square, parallelogram etc are the special type of quadrilaterals.

###### angle sum property of a quadrilateral

The sum of the angles of a quadrilaterals is 360°. This can be verify by drawing a diagonal AC in the quadrilateral ABC. This will be divide the quadrilateral in two triangles. fig 8.4 at p-136.

In triangle ADC ∠DAC +  ∠ACD +   ∠D = 180°

Similarly in triangle ABC ∠CAB +  ∠ACB +   ∠B = 180° Adding 1 and 2 we will get

∠DAC +  ∠ACD +   ∠D + ∠CAB +  ∠ACB +   ∠B = 180 +180 =360° Also we know

∠DAC + ∠CAB = ∠A,   ∠ACD +  ∠ACB = ∠C

so, ∠A +  ∠D + ∠B + ∠C = 360°.

###### trapezium

If one pair of the opposite side of a quadrilateral is parallel, then it is called trapezium.  ex. fig 8.5-1.

###### Parallelograms

If both pair of opposite sides of a quadrilateral is parallel then it will called parallelograms. ex. Fig. 8.5 – 2,3,4,5.

###### rectangle

If any one of the angle is 90° in a parallelogram, then it will be called rectangle. ex Fig 8.5 – 3.

###### rhombus

If all sides of a parallelogram are same, then it will called as rhombus ex Fif 8.5 – 4.

###### square

If any one of the angle is 90° and all sides are same in a parallelogram, then it will call as square. ex. Fig 8.5 – 5.

###### Kite

If two pairs of adjacent sides of a quadrilateral is same, then it will call as kite. ex Fig.8.5 – 6.

###### Properties of parallelogram

Consider a parallelogram ABCD and diagonal AC. Cut the parallelogram through it diagonal AC. as shown in fig 8.7 in p-139, and we can see that it will divide the parallelogram in two triangle. And  place that two triangle one over other and can notice that both triangle are congruent.

Theorem 8.1.

a diagonal of a parallelogram divide it into two congruent triangle.

Proof : Consider the parallelogram ABCD and AC is the diagonal of it. And the diagonal divide the parallelogram in two triangle ABC AND CDA. Now consider the BC parallel AD and AC transverse

so, ∠BCA=∠DAC (alternative interior angle)

Also AB parallel DC and AC transverse.

so, ∠BAC=∠DCA (alternate interior angle). And AC =CA (commenside)

There for triangle ABC congruent to triangle CDA (ASA rule).

Theorem 8.2.

In a parallelogram opposite side are equal.

this can be proved from above proof itself. Because the opposite side of a parallelogram are the corresponding sides / parts of a congruent triangle.

So, AB=DC and AD=BC.

Theorem 8.3.

If each pair of opposite side of a quadrilateral is equal, then it is a parallelogram.

This theorem is the converse of the theorem 8.2. Here also we can consider the above figure. From the figure assume that AB = CD and AD =BC. Now we can say the ΔABC and ΔCDA   are congruent (SSS).  and we can also observe that

∠BAC= ∠DCA

∠BCA= ∠DAC  (CPCT)

Now ∠BAC + ∠DAC = ∠DCA + ∠BCA

Here the angle BAC + DAC and DCA + BCA  are the opposite corners of the quadrilateral . And we know that if the opposite angle are equal then the quadrilateral will be a parallelogram. and this one we can prove in the next theorem.

Theorem 8.4.

In a parallelogram , opposite angle are equal.

This can be prove by the same explanation of theorem 8.3. Consider AC is the diagonal of parallelogram ABCD. And by using the laws of parallel lines and transverse we can prove that

∠BAC= ∠DCA

∠BCA= ∠DAC

And hence opposite angle of parallelogram are equal.

Theorem 8.5.

This is the converse of theorem 8.4.

If a quadrilateral ,each pair of opposite angle are equal , then it is a parallelogram.

Theorem 8.6.

The diagonals of a parallelogram bisect each other.

This can be check by drawing diagonals of parallelogram and measure the intersecting point. and we can found that the diagonals will intersect at mid of the diagonal each other. This can be also prove by the theorems which we already studied. (parallel and transversal line and congruence of triangle.

Theorem 8.7.

This is the converse of the theorem 8.7

If the diagonals of a quadrilateral bisect each other , then it is a parallelogram.

Proof :

From the fig.8.11 we can see that OA=OC and OB =OD so, Δ AOB congruent to ΔCOD (SAS, vertically opposite angle at O). There for ∠ABO and ∠CDO are same (CPCT). Then we can say that AB parallel to CD (parallel and transversal law). Same way  we can proof AD parallel to CB. And hence it is a parallelogram.

See page 141 for examples

###### Theorem 8.8

A quadrilateral is a parallelogram if a pair of opposite side is equal and parallel.

consider the figure 8.17 (p145) in which AB = CD and AB parallel to CD. and AC is the diagonal of the quadrilateral. ABC and CDA are congruent by SAS rule ( opposite interior angle). So BC  and AD  are equal and hence it is parallel. (equal line between parallel line will parallel). So this quadrilateral is parallel. see p-145 and 146 for examples and exercises.

###### The mid-point theorem

Draw a triangle ABC as shown in fig 8.24, mark the midpoint of the sides AB and AC with E and F respectively. Joint E and F and measure EF, BC, angle E and angle B. We can also repeat this experiment with different triangle also. And we can observe that EF = 1/2 BC, and angle AEF= angle ABC. And hence EF parallel to BC. (corresponding angles same).

###### Theorem 8.9

The line segment joining the mid-point of two side of a triangle is parallel to the third side.

Proof :

In fig 8.25 E and F are midpoints of AB and AC respectively. And CD Parallel to BA . Δ AEF concourent to Δ CDF (ASA rule.) So EF=DF and BE=AE=DC. Also we know that EF = 1/2 BC.(midpoint theorem). there for each opposite pair are equal and hence BCDE is parallelogram and EF is parallel to BC

Theorem 8.10

This is converse of theorem 8.9, The line drawn through the mid-point of one side of triangle, parallel to another side bysects the third side.     See      p-149 and 150 for example and exercises.

## How to find out the surface area of sphere

###### Surface area of sphere = 4πr²

How to find out the surface area of sphere. Before starting to explain the equation, we will see how a sphere is forming. Please consider below figures.

The FIG-1a is a circle cut it from a sheet of paper. Please fix a string on the top side of the circular paper as shown in the fig-1b. And if we rotate the string we can see our circular paper as shown in the fig-1b. Now rotate the string much more speed. And we can see our circular paper as a sphere as shown in fig-1c.

Consider the center of the circle is C in the fig1-a. So the center of the circle C will become the center of sphere in fig-1c. Please remember, the circle is a closed figure whose every point lies at constant distance (called radius). From a fixed point (called center). So we can now say, A sphere is a three dimensional figure (solid figure) .Made up with all points in the space which lie in equal distance (called radius of the sphere). from a fixed point (called center of the sphere).

Surface area of the sphere

Now we can see how to find out the surface area of sphere. Consider a rubber ball (nothing but a sphere). And fixe a nail on it surface as shown in the fig-2a. Now take a long string and start to wind it on the ball from the nail without any space and complete the all surface of the ball with string. cut the remaining string.Now we can unwind the string and keep it in some place.

###### Assigning the Area of the sphere in to circle

Cut four  circular paper with the same radius of the ball. Now start to fill the four circles, one by one  with the string which is used to  wind on the ball. After filling each circle cut the string and use remaining string for the  next circle as shown in fig-2b. After filling four circle, we can see that the entire string is used to cover all the surface area of the four circle. Therefore we can say that the surface area of the sphere is equal to the surface area of four circle with the same radius of the sphere.

As we know the area of the circle is πr²

so the surface area of a sphere is 4πr²

Where r is the radius of the sphere.

###### OBSERVATION

Now all ready we studied how to find out the surface area of sphere.  As we see a sphere have only and only one surface. And that is a curved surface area. The surface area of the sphere can be calculated with equation 4πr² . If cut the sphere through it diameter we can get two supprate figure as shown in the fig-3.

This figure is nothing but a half sphere. It can also be called as hemisphere. But in this case we can see that the hemisphere have two surfaces. One of it surface is curved surface.and the other one is flat circle. The curved surface area of the hemisphere is the half  surface area of the sphere. (as it is half of the sphere). There for the curved surface area of a hemisphere is 2πr²

And the flat circle surface area of the hemisphere is πr²

So the total surface area of the hemisphere  = 2πr²+πr² = 3πr²

If need more clarification please contact through comments or through the forum.

Thanks and regards

## How to find out the surface area of the right circular cone

Surface area of the cone = πr(l+r)

Where r is the radius of the circular top of the cone BD as shown in the fig  FIG-1. And l is the lateral height of the cone AD as shown in FIG-1.

###### What is a cone

A cone is an object as shown in the FIG-1. A cone should have a point “A” as shown in the above fig. And this point is called the vertex of the cone. The cone should also have the height. From the fig. the height is “AB”. In the top there should be a circular shape as shown in the fig. The diameter of the circle is “CD”. The diameter and the height should be in perpendicular. Then only it can call as a right circular cone. the top shape also should be in perfect circle in order to call it “circular cone”. Now we can also observe from the fig., the curved surface of the cone is formed with a slant height. The lateral height in this case is AC and AD.

###### Curved surface area of a cone

As shown in the above fig. we can make the curved surface area of a cone. The curved surface area of the con in the fig-2a can be made of with circular paper in the fig-2b. Her the missing part of the fig-2a is because of the missing part if the fig-2b.

We can cut the circular paper in number of pieces shown in the fig-2b in such a way that we get a smaltrangle with base b1. If the number of the triangle is in hundreds, then the base b1,b2,b3,…… of the triangle can be considered straight line. And the height of the triangle is the lateral height of the cone. that is nothing but the radius OB  of the circle. And if we arrange all that small triangles we can get the curved surface area of the cone.

Now we know the area of each small triangle =   ½*b1*l

Area of all small triangle = ½*b1*l + ½*b1*l + ½*b1*l + ……….

½l(b1+b2+b3+…..)

Now if we add all the b1+b2+b3 ….. we will get the circumference of the circle that is nothing but 2πr.

so the curved surface area =½l*2πr

=πrl

Now we have to consider the area of the top circle of the con . Which is πr².

total surface area of the cone = πrl+πr²  = πr (l+r)

so the total surface area of the right angle circular cone is  πr(l+r)

Where r is the radius of the cone and l is the lateral height of the cone.

###### IMPORTANT notes

From the above fig-3, we can see the triangle ABC  in the cone. Here AB is the height of the cone, BD is the radius of the cone and AD is the slant height. So if we have only the radius and height of the cone we can find out the slant height. Because slant height is required to find out the surface are. By using pythagoras theorem we can find out the slant height of the cone by the below equation

AD² = AB²+BD²

AD = √(AB²+BD²)

Where AD  is the slant height of the cone, AB  is the height of the cone and BD is the radius of the cone.

please contact me in the comment or through the Forum

## How can find the surface area of the cylinder

How can find surface area of cylinder

Surface area of the cylinder =2πr(H+r)

Dear friends,

In this post I will explain how we can find out the surface area of the cylinder.

Consider cylinder having height H and the diameter of the base circle D as shown in the figure FIG1-A. Now consider a rectangular paper having the height “H” as same as the height of the cylinder height “H”. Also assume that the length of the paper  “L” is just enough to cover the complete the curved surface area of the cylinder as shown in the FIG1-B.

So now we know the area of the paper is H*L.

We can also know that the length of the paper is equal to the circumference of the base circle of the cylinder.

So circumference of the circle is = 2πr

so area of the paper = H*2πr.

There for the

Total curved surface area of the cylinder =H*2πr.

Now we have to consider the area of the base and top circle, inorder to get the total surface area of the cylinder.

we know that the area of the circle = πr².

area of 2 circle =2πr².

So the total surface area of the cylinder = H*2πr +2πr²   = 2πr(H+r)

There for

Surface area of the cylinder =2πr(H+r)

Where H is the height of the cylinder and “r’ is the radius of the cylinder.

If some one need more explanation on this topic, please contact through the comment section. Or please ask your question on our forum.