State Board English td-IX Geometry Demo Videos

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Hello student, lets know the concept before we  do a sum. The concept we have learning  is triangle  between two parallel lines . Lets understand very simple, lets consider two lines here   line l and line m . and I m saying  here the two lines  L and M are  what parallel here . so we have two parallel lines L and M. no we are going to . we have consider a triangle ABC and  place it between two parallel lines . when I said  placed it between two parallel lines means the vertices should be lying on the parallel lines. One vertex should be one and other two vertex should be on other . this Is what we understand lying on same two parallel line. Now further gets the some traningle ABC here BC is the base ok. What I m going to observed here I m getting one more triangle .consider one more triangle with the same base BC and placed it between same two parallel lines L and M. And for the this triangle the base has to BC between the same parallel lines now the such a triangle here. So we have the triangle PBC . so let us find the area of two triangles ok observed now we talk about the area of triangle , what is the area of  triangle half*base*height. Now the area of triangle ABC . for triangle ABC if you see  where the base BC we need to have a height from A. so we draw the perpendicular form A that is AM .so now if I can ask you Area of  triangle ABC you will say Area of triangle ABC is equal to Half *BC(base)*AM(height) and mark it statement no. 1. So lets we have area of triangle PBC now, So BC is base obviously  so height is draw form opposite vertex  i.e P. from P draw a perpendicular on that N Wright line N. so now we Can say area of triangle PBC  is nothing but  Half*BC(base)*PN(height) and that is your statement no. 2. Now look at statement no. 1 and statement no.2 on the right hand side half remain as it is into BC . and BC also in second Statement . then what is different AM And PN that is height . observe AM and PN in the figure and tell what can you said the length of AM and PN. Do you know they are equal because the are perpendicular distance between two  same parallel  lines . they are also equal Wright. that is AM=PN  that means the right hand statement one and two are equal means left hand side also equal . hence we can now say that area of triangle ABC = area of triangle PBC we got it from one and Two. From this we got a conclusion we got area of these two triangles equal  why  because the align same two parallel lines and they have the same base now we can generalize statement write that  triangles lying between same two parallel lines and having same base have Equal area that is what we have learned . let us moved head  .and lets do sum base on it And its from HOT and it is sum no. 39 it says the figure given to us A quadrilateral ABCD is a parallelogram is given any line through A cuts Segment BC  at the point P. any segment BC produced at the point Q . so that is figure already given to us we need to proved that Area of triangle BCP  = Area of triangle DPQ. Lets focused on Left  hand side  Area of triangle BCP. Triangle BCP is lying between parallel lines AB and DC . do you agree AB and DC are parallel . because this is opposite sides of parallelogram . because the lying between two parallel lines PC is base of particular triangle is consider. Because  between the two triangles we are taking base  as a parallel lines  of parallelogram . so we consider PC is base . let us create a triangle with PC as a base and lying between same parallel lines . if you create ? how can you created  ? how can created such a triangle now . yes we can create by the construction, to Drawing Segment AC  now we have a triangle APC And triangle BPC both have same base PC and lying between the same parallel lines do you agree with these . yes now we can say that area of these two triangle is equal . lets write it down now so now we know that Seg AB||side DC because it it opposite sides of parallelogram . so now we can say that triangle APC and triangle BPC lie between Same two parallel line AB and DC. And they have a common base PC so the area of triangle APC = area of triangle BPC . that is what we got . it is easy. Wright , make it as statement no. 1 . and that is property which we learn  triangle having same base and parallel lines they have equal area . so now BCP something we want about this  observed now. Now you look at it is equal to area of triangle APC . but when you look at triangle APC now that triangle APC is a part of triangle ADC , agree? Observed now , we can say that area of triangle ADC is nothing but Area of triangle ADP + area of triangle APC  . let us sum of these two triangles so we can write as area addition property . let see we know that area of triangle APC we got which already shown as statement no 1that is equal to area of triangle BCP . let us replace Area of triangle APC  by area of triangle BPC. Now the next statement would be area of triangle ADC =area of triangle ADP + area of triangle BPC  and this is statement no 2 the reason write their from 1 i.e what we got. So now thought about the triangle area of triangle BCP. Now we need to look at right hand side on the right  hand side we have area of triangle DPQ now we do think about that and that get into the sum , how do you get that? Observed that triangle DPQ is the part of  triangle ADQ  so now we can say that area of triangle ADQ = area of triangle ADP+ area of triangle DPQ. The reason for this again it is area addition property  lets mark it statement no. 3. Now student lets us focus of statement no. 2 and   3 carefully in to prove the area of triangle BCP and area of triangle DPQ booth the statement present in statement no 2 And 3. So look at this statement carefully  area of triangle ADP are in both the statement and what we want is area of triangle BCP is equal to area of triangle DPQ if you observed this two statement if you want this two be equal and already the part  on right hand side which is equal , if you get left hand side equal you will be get it left side area of triangle means a that area of triangle  BCP = area of triangle DPQ we can obtain that we get   them equal  now focused on area of triangle ADC and area of triangle ADQ observed this two triangles when you look at this triangle i.e triangle ADC which we have here  and ADQ they both lie between the same to parallel lines AD and BQ because it is opposite sides of parallelogram and also they have same based AD . again their area  are equal . area of triangle ADC = area of triangle ADQ  and that is we got  and that is nothing but statement no. 4 . now when we compare , when we see all the statement no 2,3,4 now we can say that area of triangle BCP  is equal to area of triangle DPQ  and this is what we wanted the reason for this we righted  from 2, 3 and 4 that’s the sum , very important sum but very simple one , which coming in exam for 4 marks

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