Choose two of the angles on ∆ABC, and locate the line segment between them. Draw a new line segment, DE , parallel to the line segment you located on ∆ABC. You can draw of any length and place it anywhere on the coordinate plane, but not on top of ∆ABC. From points D and E, create an angle of the same size as the angles you chose on ∆ABC. Then draw a ray from D and a ray from E through the angles such that the rays intersect. You should now have two angles that are congruent to the angles you chose on ∆ABC. Label the point of intersection of the two rays F, and draw ∆DEF by creating a polygon through points D, E, and F.

The two triangles are attached below.

Step-by-step explanation:

Let the ∆ABC be an isosceles right angled triangle.

The two angles will be 45° each.

The triangle ABC is attached below.

Now construct the triangle DEF according to the given instructions.

The two triangles are attached below.

mgxmgzmgzzzlgxhhhx. dlulhx

area of rectangle a is 2 times that of rectangle b that means 2x+x =   36cm^2

3x = 36cm ^2

x = 36/3

x= 12cm^2

and 2x = 24cm^2

now side of square = are divided by sides = 36/4 = 6cm

that means length of rectangle a and b is 6cm too

now for rectangle a,

l*b = area

= 6*y=24

y=24/6 = 4

therefore dimensions of rectangle a are- length = 6cm, breadth = 4cm

and for rectangle b, length = 6cm and breadth = 6-4 = 2cm.

hope this .