Warm up problem 16: area of quad easy-medium

Lemma: if two opposite mid points are connected on a parallelogram, then, regardless of the other two points' position on the other two sides, the area of the quadrilateral created will be 1/2.

Proof:

Because the new triangle will be made of two triangles with the same base and 1/2 the height of the reference quadrilateral, we can express the area of the original quad as $bh$ and the new one as $2\left(\frac{\frac{h}{2}*b}{2}\right)\Rightarrow\frac{bh}{2}$

As their is no details about a point in diagram thus the area will not change if we change the location of tha point . So I have moved this poin to D . Thus figure is divided into 3 triangles . And further I used the property that AREA OF TWO TRIANGLE IS SAME IF THEY HAVE COMMON BASE AND ARE BETWEEN TWO PARALLEL LINES. thus I have moved the point F to point C and thus the required area is half of the area of the parallelogram i.e. 42 . :)

1. Join E and F. DC = EF = AB = 12. Let points E and F meet DC at X. Given DC = base = 12. Also given, height of the parallelogram = perpendicular height = 7.

2. Since E and F are the mid-points of AD and BC, the perpendicular distance to AB and DC are the same. This is same as the perpendicular distance from X to EF. That means, perpendicular distance from X to EF and perpendicular distance from F to AB are one and the same. Also EF = AB.

3. Thus, area of triangle EFX = area of triangle ABF.

4. Hence, area of the shaded region = half the area of the whole parallelogram = (1/2)xbasexheight = (1/2)x12x7 = 42.

OR

Let's take the shaded regions itself.

1. Area of shaded region = Area of triangle EDX + area of triangle FXC + area of triangle ABF.

2. The altitudes for triangles EDX and FXC are one and the same. Thus, the sum of their areas = 1/2 * DX * h + 1/2 * XC * h = 1/2* h * (DX+XC) = 1/2 * h * DC = 1/2 * 3.5* 12 = 21.

3. Area of triangle ABF = 1/2 * 12 * 3.5 = 21.

4. Finally, the area of the shaded region = 21+21 = 42.

In the picture, $\overline { BC } = 7 \text { cm }$ , and $F$ is a mid point of $\overline { BC }$ , and we know that $\triangle { ABF } = \triangle { EGF }$ , where the point $G$ is a point on $\overline { CD }$ . Then, the shaded area is equal to $\square { EDCF }$ , so we get $\square { EDCF } = 12 \times 3.5 = 6 \times 7 = \boxed { 42 }$ .

Note: the reason that $\triangle { ABF } = \triangle { EGF }$ is if we draw a line $\overline { EF }$ , then $\square { AEFB } = \square { EDCF }$ because $F$ and $E$ are a mid point. Then, we draw a line $\overline { AF }$ and $\overline { EC }$ , then we get $\triangle { ABF } = \triangle { AEF } = \triangle { EFC } = \triangle { EDC }$ .

Hence, the area of shaded area is $\boxed { 42 \text { cm } ^ { 2 } }$ .

Triangle AFB is congruent with/to Triangle AFE, which in turn is congruent to EMF [ M being the vertex point on DC ]. Congruent triangles have equal areas. So required area = ar(EDM) + ar(EMf) + ar(FMC) = ar(EDCF) = 1/2 * ar(ABCD) = 1/2 * DC * 7 = 42 [ area of a parallelogram = base * altitude ]

Join E and F Then the area of the white portion can be took out by 3.5x12 The area of the quadrilateral ABCD is 12x7 Area of shaded portion is 12x7 - 3.5x12 = 42