If the points A (1,-2) B (2,3)C (-3,2) D (-4,-3) are the vretices of parallelogram ABCD, then taking AB as the base find the height of the parallelogram

Area of the triangle ABC

Since ABCD is a parallelogram,

therefore area of parallelogram ABCD = 2 x area of triangle ABC = 2 x 12 = 24 square units

Area of the parallelogram = 

Thus height of the parallelogram = 

Let us find the base AB of the parallelogram.

Thus,

I need an easy ans pls someone

zero is right Asha . go according to what zero had posted!!

          For Diagram:
                       * Draw a parallelogram ABCD.
                       * Draw DE perpendicular to AB.
                       * Join BD.
                    Note: DE is the height.

             Answer:
                     Now,
                            1   * b * h = Area of triangle ABD
                            2
                            1   * AB * DE                                      = Area of triangle ABD
                            2
                            1  { Root (x₂ -x₁)² + (y₂ - y₁ )² } * h
                            2
                   =         1 { 1 ( 3 + 3 ) + 2 (-3 + 2  ) + (-4)(-2-3) }
                              2
                              Root 1² + 5²   * h = { 6-2+20 }
                                  Root 26   * h  = { 24 }
                                                         h =    24          units.           
                                                                Root 26
               Therefore, Height of the parallelogram =     24        
                                                                                    Root 26   
               I'm sorry for not bringing out the diagram.      
               Hope this helps.    


     
                       

   

          

Since ll^gm ABCD and ∆ABC lie on the same base and  the same parallels, thus,
​ar(||^gmABCD)=2ar(∆ABC)
 By using the area formula we have,
​Area of a triangle=½|x₁(y_​2-y₃)+x₂(y_​3-y₁)+x₃(y_1​-y₂)
=½|1(3-2)+2(2+2)+(-3)(-2-3)|
=½|1+8+15|
=½×24=12 square units
Therefore, ar(||^gm ABCD)= 2×12 =24 square units


Now,
Using the distance formula we  have,
AB=√(x₂-x₁)²+(y₂-y₁)²
=√(2-1)²+(3+2)²
=√1+25
=√26 units
We know that,
Area of a parallelogram=height×base
=>ar(ll^gm ABCD)=height ×AB
=>24=height ×√26
=>height=24/√26 
=>height=12√26/13 units.                                (rationalizing the denominator)

 ZERO IS WRONG

root 50*root26
 

The points A(4, –2), B(7, 2), C(0, 9) and D(–3, 5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB

I also need easy ans

AB = y2 - y1/x2 - x1 = 3 - (-2)/2 - 1 = 5/1 DC = y2 - y1/x2 - x1 = 2 - (-3)/a - (-4) = 2 + 3/a + 4 AC has the same gradient as DC (parallelogram opposite sides) DC = 5/ a + 4 = 5/1 Cross multiply 5 = 5(a + 4) 5 = 5a + 20 5a = -15 a = -3 C (-3,2) Height of parallelogram Using DC D(-4,-3) C(-3,2) Distance formula: √(x2 - x1)^2 + y^2 - y1)^2 √ (-3 - (-4))^2 + (2 - (-3) √(1)^2 + (5)^2 √ 26 = 5.1 units (1 decimal place) Hope this helps :-)

thanks this answer help me in assingment
 

24/root 26 is right answer

24/root 26 is the right answer

Height of the parallelogram = 24/root 26

H= 24/root 26

HOW TO DO THIS MERITNATION EXPERT PLZZANSWER
 

Ok haha

Nic aap provider

kbcfgxgfxgxxx
 

If points a(1,-2) b(2,3) c(-3,2) and d(-4,-3) are the vertices of parallelogram abcd then taking ab as base find the height of parallelogram

answer 

I hope this helps you!

answer please...........

answer please .......