 alright guys so in this question we have been given a quadrilateral APCD whose sides are also given and the sides are 940, 28, 15 centimeter respectively and the angle between the first two sides is 90 degrees so we have already drawn the figure here now the question is you have to find out the area of this quadrilateral now this question can be solved either directly using Brett Snyder's formula which we have discussed in session one but then since angles are not provided over here so we'll have to use a lot of trigonometry but then we have another method of doing it and that is by you know breaking this shape into two triangles okay two triangles and then and then we can find out the area very easily let's see what do I mean so I'm just going to connect AC okay so here I go so I connected AC now what I need to do I have to just simply find out the areas of both the triangles and add so I okay right area quadrilateral ABCD will be equal to area of triangle ABC which happens to be a right triangle hence I will be having some piece and that of ACD now here this will be a little linear task but never mind we'll try to do that but in ACD we have two sides given so we cannot use here on formula until unless we know AC but if you notice carefully ABC is right angle triangle and two sides are given so AC happens to be the hypotenuse and wow so we can find out AC and then proceed to find out the area of ACD so let's first find it out let's find out the value of AC so AC square is equal to AB square plus BC square and this is by our very famous Pythagoras uncle's theorem Pythagoras theorem okay so what is AC squared so AC squared is 40 squared plus 9 squared which happens to be 1681 and if you see 1681 happens to be 41 squared so thank God that we got AC as an integer 41 this is what all of you love to see isn't it if it was our irrational number then it would have given you nightmares but yes thankfully we have got an integer so let's utilize the value of this integer what next so let us find out area of triangle ABC first low hanging fruit so let's you know pluck it and this is nothing but half into base base could be AB and height is BC so it's low hanging fruit half into AB is 9 into BC which is 40 so 20 into 9 so hence it is 180 centimeters squared so we got we got the first triangles area that's so nice so let's go to finding out the area of triangle triangle ACD okay how to find this out here on formula will come for the rescue SS minus a S minus P S minus C now what's S semi-parameter so now you know that AC is 41 so let us write it 41 here so semi-parameter is 41 plus 15 plus 28 upon 2 and this value comes out to be 84 I believe yes 84 upon 2 is equal to 42 so we got the semi-parameter guys now S minus a let's calculate so it is 42 minus 41 let's say this is one beautiful so hence load is reduced S minus B is 42 minus 15 not that great but never mind let's see 42 minus 15 is 27 cool and S minus C is 42 minus 28 which happens to be 14 my dear friends okay now what so let's find out area of triangle ACD is under root S what is SS is 42 into S minus A is 1 27 and 14 okay now thankfully you'll again get factors which are having even power so 42 is 14 into 3 27 is 3 cubed and this is 14 so you can very easily find out the area to be 14 times 9 correct yes sir correct so hence it is 126 centimeter square so now it's a cakewalk now area of triangle it's not quite quadrilateral ABCD ABCD is equal to the sum of the two areas first area value was how much guys first area value we calculated here 180 centimeter square and the second one was 126 centimeter square so 180 plus 126 which happens to be 306 centimeter square right so what is the learning learning is let's jot down it's a good practice to like write learning at the end of the day so learning is area of a quadrilateral can be found out by breaking it down into triangles area of triangles comprising the quadrilateral right and then obviously we know the formula of finding out the area of triangle whether it is through half into base into height or through heron's formula okay so that is all about this question