Finding $\overline {AD} \times \overline {CD}$ in $\triangle APB$

  • MHB
  • Thread starter Albert1
  • Start date
In summary, the formula for finding the length of $\overline {AD} \times \overline {CD}$ in $\triangle APB$ is $\frac{1}{2} \times \overline {AD} \times \overline {CD} \times sin\angle APB$. To find the length of $\overline {AD}$ and $\overline {CD}$ in $\triangle APB$, you can use the Pythagorean theorem or trigonometric ratios. The angle $\angle APB$ is essential in finding the length of $\overline {AD} \times \overline {CD}$ in $\triangle APB$ because it is used in the formula and determines which trigonometric ratio to
  • #1
Albert1
1,221
0
$\triangle APB, \overline {PA}=\overline {PB}, \angle APB=2\angle ACB, $
point $D$ is the intersection of $\overline {AC}$, and $\overline {BP}$
if $\overline {BP}=3 , \overline {PD}=2$
please find the value of $\overline {AD}\times \overline {CD}$
 
Mathematics news on Phys.org
  • #2
Hi Albert. Where is point $C$?
 
  • #3
Since the problem implies (truthfully) that it does not matter where C or A is located up to the constraints (on circle), I picked AC to be vertical.

View attachment 4989

This depicts one of the Pythagorean Means, specifically the Geometric Mean.
That is to say; Length AD is the GM of DB and 6-DB. AD = SQRT(1*5).

AD=DC so AD*DC = 5
(proof of the implication not being required, makes this a simple problem)
 

Attachments

  • Capture.PNG
    Capture.PNG
    6.7 KB · Views: 74
Last edited by a moderator:
  • #4
RLBrown said:
Since the problem implies (truthfully) that it does not matter where C or A is located up to the constraints (on circle), I picked AC to be vertical.
This depicts one of the Pythagorean Means, specifically the Geometric Mean.
That is to say; Length AD is the GM of DB and 6-DB. AD = SQRT(1*5).

AD=DC so AD*DC = 5
(proof of the implication not being required, makes this a simple problem)
very good solution !
 

What is the formula for finding the length of $\overline {AD} \times \overline {CD}$ in $\triangle APB$?

The formula for finding the length of $\overline {AD} \times \overline {CD}$ in $\triangle APB$ is $\frac{1}{2} \times \overline {AD} \times \overline {CD} \times sin\angle APB$.

How do you find the length of $\overline {AD}$ and $\overline {CD}$ in $\triangle APB$?

To find the length of $\overline {AD}$ and $\overline {CD}$ in $\triangle APB$, you can use the Pythagorean theorem or trigonometric ratios (such as sine, cosine, or tangent) depending on the information given in the triangle.

What is the role of the angle $\angle APB$ in finding the length of $\overline {AD} \times \overline {CD}$ in $\triangle APB$?

The angle $\angle APB$ is essential in finding the length of $\overline {AD} \times \overline {CD}$ in $\triangle APB$ because it is used in the formula for finding the length, as well as in determining which trigonometric ratio to use.

How do you label the sides and angles in $\triangle APB$ when finding the length of $\overline {AD} \times \overline {CD}$?

In $\triangle APB$, $\overline {AD}$ and $\overline {CD}$ are the sides that form the right angle, while $\angle APB$ is the angle opposite the hypotenuse, which is typically labeled as $\overline {AB}$.

Can $\overline {AD} \times \overline {CD}$ be negative in $\triangle APB$?

No, $\overline {AD} \times \overline {CD}$ cannot be negative in $\triangle APB$ because it represents the area of a triangle, and area cannot be negative.

Similar threads

Replies
1
Views
1K
  • General Math
Replies
5
Views
2K
Replies
2
Views
1K
Replies
2
Views
1K
  • General Math
Replies
3
Views
1K
  • General Math
Replies
4
Views
1K
  • General Math
Replies
3
Views
1K
Replies
2
Views
1K
  • General Math
Replies
5
Views
2K
Back
Top