# 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 a triangleABC  = 1/2  x1 (y2-y3) + x2 (y3-y1) + x3 (y1-y2)

= 1/2  1(3-2) + 2(2+2) + (-3)(-2-3)  =  1/2 x 1 + 8 + 15  =  1/2 x 24  =  12

since ABCD is a parallelogram

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

and AB = root of (2-1)2 + (3+2)2 = root of 1 + 25 = root 26

Now, we know that

Area of a parallelogram = base x height

24 = root 26 + height

or height = 24 - root26

• -4
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, • 5
I need an easy ans pls someone
• -9
zero is right Asha . go according to what zero had posted!!
• -13
For Diagram:
* Draw a parallelogram ABCD.
* Draw DE perpendicular to AB.
* Join BD.
Note: DE is the height.

Now,
* b * h = Area of triangle ABD
2
* AB * DE                                      = Area of triangle ABD
2
{ 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.

• -8
Since llgm 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(y3-y1)+x3(y1-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(llgm ABCD)=height ×AB
=>24=height ×√26
=>height=24/√26
=>height=12√26/13 units.                                (rationalizing the denominator)
• 86
ZERO IS WRONG
• -12
root 50*root26

• -9
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
• -3
I also need easy ans
• -9
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 :-)
• -2
thanks this answer help me in assingment

• -3
• -4
24/root 26 is the right answer
• -4
Height of the parallelogram = 24/root 26
• -1
H= 24/root 26
• -2
HOW TO DO THIS MERITNATION EXPERT PLZZANSWER

• 3
PLS HELPPPP
• -5
Ok haha
• -2
Nic aap provider
• -2
kbcfgxgfxgxxx

• -2
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 • 28 