Geometry finding the measure of a kite

Since the diagonals of a kite form a right triangle, use the formula for area of a right triangle to solve for the missing length. A = 1/2bh
 
Rachel

I think part of your problem here is that you started by thinking about what equation to use before you had fully worked out the geometric relationships within the kite. The equation you want will be based on geometry so you need to work that out first. Make sense?

Can you show that

[MATH]\triangle BCD \text { and } \triangle ABD \text { are both isosceles.}[/MATH]
From that can you show that

[MATH]\triangle AOD \cong AOB,\ \triangle COB \cong \triangle COD, \text { and all four are right triangles.}[/MATH]
Now assign letters to the various lengths, many of which are equal due to the congruences that you have established. And you can go crazy with the Pythagorean Theorem because every triangle is a right triangle.

On complex problems, it frequently pays to do some preliminary work before trying to formulate equations.
 
JeffM said:
Rachel

I think part of your problem here is that you started by thinking about what equation to use before you had fully worked out the geometric relationships within the kite. The equation you want will be based on geometry so you need to work that out first. Make sense?

Can you show that

[MATH]\triangle BCD \text { and } \triangle ABD \text { are both isosceles.}[/MATH]
From that can you show that

[MATH]\triangle AOD \cong AOB,\ \triangle COB \cong \triangle COD, \text { and all four are right triangles.}[/MATH]
Now assign letters to the various lengths, many of which are equal due to the congruences that you have established. And you can go crazy with the Pythagorean Theorem because every triangle is a right triangle.

On complex problems, it frequently pays to do some preliminary work before trying to formulate equations.
Click to expand...
A kite-shaped figure is a quadrilateral with two distinct pairs of equal adjacent sides(sharing a common vertex and side.)
The diagonals of a kite are perpendicular and intersect forming 90 degree right angles. A kite is made up of two isosceles triangles joined base to base. An isosceles triangle is a triangle that has at least two congruent sides
OC = 5 in OD = √24 in

CD = 7 in OP = x in
 
rachelmaddie said:
I still need clarification in terms of justification with this problem.
Click to expand...
The problem with answering you is that we do not know what your teacher believes is sufficient justification.

Based on the definitions of a kite, you should be able to prove in whatever detail is required that the length of OD is

[MATH]2\sqrt{6}.[/MATH]
That is the geometric task.

Then it is easy to show that the area of triangle COD =

[MATH]\dfrac{1}{2} * 5 * 2\sqrt{6} = 5\sqrt{6}.[/MATH]
But you can compute that area another way, namely

[MATH]\dfrac{1}{2} * 7 * x.[/MATH]
[MATH]\therefore \dfrac{1}{2} * 7 * x = 5 \sqrt{6} \implies x = \dfrac{10\sqrt{6}}{7}.[/MATH]
What more do you need?
 
JeffM said:
The problem with answering you is that we do not know what your teacher believes is sufficient justification.

Based on the definitions of a kite, you should be able to prove in whatever detail is required that the length of OD is

[MATH]2\sqrt{6}.[/MATH]
That is the geometric task.

Then it is easy to show that the area of triangle COD =

[MATH]\dfrac{1}{2} * 5 * 2\sqrt{6} = 5\sqrt{6}.[/MATH]
But you can compute that area another way, namely

[MATH]\dfrac{1}{2} * 7 * x.[/MATH]
[MATH]\therefore \dfrac{1}{2} * 7 * x = 5 \sqrt{6} \implies x = \dfrac{10\sqrt{6}}{7}.[/MATH]
What more do you need?
Click to expand...
In terms of justification, the formula or process used should be specified and why.
 
The justification essentially is a statement of the work you did (in post #14, if I recall correctly) with words stating why you did each step. Why is the area calculated this way? Why is it also calculated that way? What are you saying when you write your equation?

I see so many students with a habit of just scribbling calculations and equations across a paper with no words, so even they don't know just what they are doing. If you just narrate what and why, rather than just showing bare symbols, you will be showing the justification.

Give it a try. Start from the top and show all your calculations, with words attached. Then we can help you refine that if necessary.
 
A kite-shaped figure is a quadrilateral with two distinct pairs of equal adjacent sides(sharing a common vertex and side). The diagonals of a kite are perpendicular and intersect forming 90 degree right angles
Since DOC is a right triangle
First use Pythagorean theorem to solve for the measure of DO.
DC^2 = DO^2 + OC^2
Substitute
7^2 = DO^2 + 5^2
DO^2 = 7^2 - 5^2
DO^2 = 49 - 25
DO^2 = 24
DO = √(7^2-5^2) = √24

To find the area of DOC use the right triangle formula. A = 1/2bh Set up an equation to solve for x. Multiply the base by the height, and divide by 2 since the area of each triangle is equal to one-half the area of the parallelogram.
(1/2)(OC × OD) = (1/2)(CD×OP)
5×√24 = 7 × x
x = 10√6/7
 
rachelmaddie said:
A kite-shaped figure is a quadrilateral with two distinct pairs of equal adjacent sides(sharing a common vertex and side). The diagonals of a kite are perpendicular and intersect forming 90 degree right angles
Since DOC is a right triangle
First use Pythagorean theorem to solve for the measure of DO.
DC^2 = DO^2 + OC^2
Substitute
7^2 = DO^2 + 5^2
DO^2 = 7^2 - 5^2
DO^2 = 49 - 25
DO^2 = 24
DO = √(7^2-5^2) = √24

To find the area of DOC use the right triangle formula. A = 1/2bh Set up an equation to solve for x. Multiply the base by the height, and divide by 2 since the area of each triangle is equal to one-half the area of the parallelogram.
(1/2)(OC × OD) = (1/2)(CD×OP)
5×√24 = 7 × x
x = 10√6/7
Click to expand...
Good....

However, in algebra - use * as multiplication (to avoid confusion with variable "x".)
 
You must log in or register to reply here.
Top