Question 1041012
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Q: two line segment in the x y plane form a right triangle with the x axis, having vertices (2,a),(0,0),(10,0). 
What is the area of triangle?
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<pre>
It is clear that the vertex (2,a) is the right angle vertex and the segment [(0,0 - (10,0)] is the hypotenuse.

Then you can write the Pythagorean equation in the form

{{{2^2 + a^2 + (10-2)^2 + a^2}}} = {{{10^2}}}.       (1)

It is your equation to find the unknown value of "a".

Simplify and solve it:

{{{4 + a^2 + 64 + a^2}}} = {{{100}}},

{{{2a^2}}} = {{{100 - 64 - 4}}},

{{{2a^2}}} = {{{32}}}  --->  {{{a^2}}} = {{{16}}}  --->  a = 4.

Now, the hypotenuse has the length 10, while the altitude drawn to the hypotenuse has the length 4.

Hence, the area of the triangle is {{{(1/2)*10*4}}} = 20 square units.

<U>Answer</U>.  The area of the triangle is 20 square units.
</pre>


<U>Comment from student</U>: can you tell me how you made this equation.



<U>My response</U>:

<pre>
Let A be the point (0,0) in the coordinate plane (the origin);
let B be the point (2,a), and let C be the point (10,0).
Then the triangle ABC is right-angled triangle, according to the condition.
with the vertex B as the right angle and the segment AC as the hypotenuse.

Next, draw the perpendicular from the vertex B to the hypotenuse AC. The foot of this perpendicular is the point D = (2,0).

I recommend you to make a sketch of the triangle ABC and draw this perpendicular CD.

Now you have two right angled triangles, ADB and CDB.

From the triangle ADB you have {{{abs(AB)^2}}} = {{{2^2 + a^2}}}.
From the triangle CDB you have {{{abs(BC)^2}}} = {{{(10-2)^2 + a^2}}}.

The equation (1) in my very first post is

{{{abs(AB)^2}}} + {{{abs(BC)^2}}} = {{{abs(AC)^2}}}.

Everything is so obvious . . . 
</pre>

I have a contr-question to you.


Why, when sending your question, you didn't mention the ID number of the problem?