Area of a kite wich is divided into four triangles

Asad Abbas

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I need some help in my maths homework.In the question the kite is divided into four triangles.I also need help in another question but it was to complicated to explain so I was trying to attach an attachment but I couldn't.I was hoping if you could help me.

Q.ABCD is a kite.AC=10cm BD=12cm.AC is a horizontal line and BD is a vertical line.BD is also the axis of symmentry.Find the area.
 
Asad Abbas said:
I need some help in my maths homework.In the question the kite is divided into four triangles.I also need help in another question but it was to complicated to explain so I was trying to attach an attachment but I couldn't.I was hoping if you could help me.

Q.ABCD is a kite.AC=10cm BD=12cm.AC is a horizontal line and BD is a vertical line.BD is also the axis of symmentry.Find the area.

Did you draw a sketch? If you do, you will see that the diagonals of the kite intersect at a point I'll call E, creating four small triangles....triangle AEB, triangle CEB, triangle DEC and triangle DEA.

Because BD is a line of symmetry, the four angles at E are right angles, meaning those four small triangles are all right triangles.

Remember that the area of a right triangle is (1/2)*(product of the legs). Applying this to the small triangles, we see that

area triangle AEB = (1/2)*AE*BE
area triangle CEB = (1/2)*CE*BE
area triangle CED = (1/2)*CE*DE
area triangle DEA = (1/2)*AE*DE

The area of the kite is the sum of the areas of all four of these triangles, so

area kite ABCD = (1/2)*AE*BE + (1/2)*CE*BE + (1/2)*CE*DE + (1/2)*AE*DE

The first two terms on the right side have a common factor of (1/2)*BE, and the last two have a common factor of (1/2)*DE. Let's remove the common factor from each pair of terms:

area kite ABCD = (1/2)*BE(AE + CE) + (1/2)*DE(CE + AE)

AND, AE + CE = AC.....

area kite ABCD = (1/2)*BE*AC + (1/2)*DE*AC

Now, (1/2)*AC is a common factor in both terms of the right side....and we have

area kite ABCD = (1/2)*AC*(BE + DE)

Since BE + DE = BD, we can make a substitution.....

area kite ABCD = (1/2)*AC*BD

I know...you are saying "Wait....that's a WHOLE LOT of work!" Well, yes, it was. But look what we have proved!

AC and BD are the two diagonals of a kite....and the area of that kite turns out to be (1/2)*(sum of the diagonals)!

You may have proved this already...but did not realize you could apply it to any problem where you know the diagonals of a kite, and need the area of the kite.

It should be easy for you to find the area of kite ABCD now, I hope.
 
Thank you for the help but I only understood what you said until the part where you have to add the areas of the 4 triangles.Since I am in grade 7 can you please explain me what is after that in an easier way.The perpendicular height is not equal in the lower triangles and the upper ones.I was wondering whether I would use the total perpendicular height which is BD(12cm).I also wanted to know how can I attach an attachment.
 
is the kite formula works for any convex quadrilateral?

Is the kite formula works for any quadrilateral whose diagonals are perpendicular?
Pls. prove
Thanks
 
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