Test: Area Of Parallelograms

Two parallelograms are on equal bases and between the same parallels.
i.e., their corresponding heights must be the same.
Thus, areas of both parallelograms are equal. Hence, the ratio of their areas is 1:1.

Let the altitude of the triangle ABC is h

Then altitude of the triangle BED = h/2

Now, ar(Δ FED) = (1/2)*FD*(h/2) = (FD  * h)/2

and ar(Δ AFC) = (1/2)*FC*h

 = (1/2)*(FD + DC)*h  

= (1/2)*(FD + BD)*h

= (1/2)*(FD + BF +  FD)*h

= (1/2)*(2FD + BF)*h

= (1/2)*(2FD + 2FD)*h

= 2FD*H

⇒  ar(Δ AFC)/8 = 2FD*H/8

⇒  ar(Δ AFC)/8 = 2FD*H/4

⇒  ar(Δ AFC)/8 = ar(Δ FED)

Given that ABCD is a IIgm 

AL = 20 cm

CD = 12 cm

CM = 8 cm

Area of IIgm ABCD will be = BASE * ALTITUDE

  = AL * CD

= 20 * 12= 240 cm2

Again taking base AD and Altitude MC

than Ar IIgm ABCD = B * h

  240 = 8 * AD 

 or  AD = 240/8 = 30 cm

now ,

 CD = AB = 12 cm (opp. sides of a IIgm)

AD = BC = 30 cm(opp. sides of a IIgm)

now ,

 Perimeter of IIgm ABCD =

  2a + 2b

  or 2(a+b)

 2 (30+12)

2(42)

or, 42*2 = 84 cm 

If a triangle and a rhombus are on the same base and between the same parallels then the ratio of the areas of the triangle and rhombus is 1 : 2

The base of triangle and rhombus are equal. The rhombus have equal sides, and  one side is the base of triangle.

The are of triangle must be half of the are of rhombus.

Area = B x H
100 = DC x AF
100 = DC x 6.25
DC = 100/6.25
DC = 16cm
AB = DC( in a parallelogram opposite sides are equal )

So AB = 16cm

IIgm ABCD and rectangle ABEF are between the same parallels AB and CF.
AB = EF (For rectangle) and AB = CD (For parallelogram)
? CD = EF ? AB + CD = AB + EF ... (1)
Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.
? AF < AD
similarly we write , BE < BC ? AF + BE < AD + BC ... (2)
From equations (1) and (2), we get
AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD

Now,  Perimeter of parallelogram ABCD = AB + BC + CD + DA = 14 + 12 + 14 + 12 = 52

Thus, Perimeter of rectangle ABEF < 52 cm

ABCD is a parallelogram
In a Triangle ABC and ADC
AB=CD(opposite lines of a parallelogram)
BC=AD(opposite lines of a parallelogram)
AC=AC(common)
Triangle ABC and ADC ( SSS Cong. rule)
ar(ABC) =ar(ADC)=1/2 ar (ABCD)
Hence proved.

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

Here, the area of the parallelogram = base x height = 10 x 8 = 80 cm²

Therefore, the area of the triangle = 1/2 x area of parallelogram

= 1/2 x 80 = 40 cm²

This is a condition of parallelogram

Area = EF×DC =4  cm × 8 cm = 32 sq. cm