maths1996l
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Does the line segment AB need to be changed to AD?
BC and AB should be the same length?
Can this be solved -
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Does the line segment AB need to be changed to AD?
BC and AB should be the same length?
Your sketch claims AB = BC
yet your data says different. According to your sketch none of the sides can have a length of 2√2
Last edited:
Otis
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Hi. No, but their label BC needs to be changed to AD. Use these lengths:
AB = BC = 2√2 cm
AD = CD = 9 cm
They didn't provide any angle measures, so you'll first need to find the two diagonal lengths AC and BD. The kite area is half the product of those diagonal lengths.
Here's a hint: Let E be the intersection point of AC and BD. Then we have right triangle ABE with hypotenuse 2√2. The legs AE and BE are equal. Use the Pythagorean Theorem to find AE (and then use it again, to find ED).
Please post your work, if you need more help. Cheers
[imath]\;[/imath]
If [imath]AB = BC[/imath] and [imath]AD = CD[/imath], I think that Otis method by dividing the kite into four right triangles is the quickest way to find the area of the kite. But I have another idea which is more fun and it does not matter what is the length of each leg as long as you know the lengths of all legs. Draw a segment [imath]AC[/imath] and find the length of this segment, then you will have two triangles, each with three known legs. Use this formula to find the area of each triangle. I will prepare it for the smaller triangle:
[imath]\displaystyle \text{Area} \ = \sqrt{\frac{AB+BC+AC}{2}\left(\frac{AB+BC+AC}{2} - AB\right)\left(\frac{AB+BC+AC}{2} - BC\right)\left(\frac{AB+BC+AC}{2} - AC\right)}[/imath]
Do the same for the larger triangle. Find their areas and add them together to find the area of the kite.