These 3 please!!! Properties of Kites and Trapzoids exercises 1-3

Step-by-step explanation:

Triangles ADB and CDB are congruent, so each is half the total area. The area of ADB is ...

  A = 1/2bh

  A = 1/2(DB)(1/2AC) = 1/4(DB)(AC)

The area of the kite is twice this value:

  A = 1/2(DB)(AC)

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In order to use the formula derived in part 1, we need to know the length of BE. That can be found using the Pythagorean theorem:

  BE² +AE² = AB²

We know that AE = 1/2AC, so the desired dimension is ...

  BE = √(AB² -(1/4)AC²)

Then the area of the kite is ...

  area = 1/2(DB)(AC)

  area = (1/2)(AC)(ED +√(AB² -(1/4)AC²)

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Given three sides of a triangle, the area of it can be found using Heron's formula. If you have not yet seen that in your geometry class, you can always use the Pythagorean theorem and some algebra.

As before, we need to know the lengths of the diagonals. One is given. The other can be found considering the Pythagorean relations:

  AD² = AE² +DE²

  AB² = AE² +BE²

  BE +DE = BD . . . . the sum of the horizontal diagonal lengths

Subtracting the second equation from the first gives ...

  AD² -AB² = DE² -BE²

Using the third to write an expression for DE, we get an equation for BE:

  AD² -AB² = (BD -BE)² -BE²

  AD² -AB² = BD² -2BD·BE +BE² -BE²

  (AB² +BD² -AD²)/(2BD) = BE

Substituting this into the original second equation, we can find AE, hence the area of the kite.

  AE² = AB² -((AB² +BD² -AD²)/(2BD))²

Which means the kite area is ...

  area = BD·AE = BD√(AB² -((AB² +BD² -AD²)/(2BD))²)

  area = (1/2)√(4·AB²·BD² -(AB² +BD² -AD²)²

This area formula demonstrates it is possible to find the area of the kite given the three lengths AB, BD, DA.

Additional comment

We have no doubt that the area formula of Part 3 can be rearranged to match Heron's formula. Here, we're interested in demonstrating area can be found.

m∠vru = 80°

m∠urw = 15°

step-by-step explanation: