Trapped Triangle

Assuming ABCD to be a parallelogram:

-Let U be the midpoint of AB , V the midpoint of AD and O the midpoint of both UX and VY .

-As Y is the midpoint of BC , ABY is $\frac{1}{2}$ of ABYV , which is $\frac{1}{2}$ of ABCD ., thus ABY is $\frac{1}{4}$ of ABCD .

-Similarly as X is the midpoint of AD , ADX is $\frac{1}{2}$ of ADXU which is $\frac{1}{2}$ of ABCD ., thus ADX is $\frac{1}{4}$ of ABCD ..

- CXOY is $\frac{1}{4}$ of ABCD . and CXY is $\frac{1}{2}$ of this area so CXY is $\frac{1}{8}$ of ABCD .

Finally AXY = (1- ( $\frac{1}{4}$ + $\frac{1}{4}$ + $\frac{1}{8}$ )) x ABCD , so S = $\frac{3}{8}$

Thus 8 $\times$ S =3

Let a be the height of ABCD; b be the length CD=AB; ar of ABCD = ab; ar of triangle ADX=ab/4; ar of triangle XCY= ab/8; ar of triangle ABY= ab/4; ar of triangle AXY= Sab=ab-(ab/4)-(ab/8)-(ab/4)=3ab/8; S=3/8;S 8=(3/8) 8=3

If area of //gm is Q, Area of ADX is Q/4. Area of ABY = Q/4. Area of XCY = Q/4. Remaining area = Q-Q/4-Q/4-Q/8=3*Q/8

Area of the given triangle is equal to the area of the parallelogram minus sum of areas of the remaining three triangles. And, two of the triangles share a common base (with the parallelogram) but half the altitude, so their areas are both 1/4 of parallelogram, and the third triangle has both the base and the altitude half of those of parallelogram so its area is 1/8 of that of parallelogram. Therefore, required area = (1-1/4-1/4-1/8)xparallelogram.

Assume the longest side is 4 units and smaller side is 2 units. So area of parallelogram is 8 Sq.units.

X is mid point of larger side, so BX =2, XC=2

Y is mid point of smaller side, so CY=1and ZB=1.

Area of 🔺 ADX =1/2(3

Explained simply:

Axyz = ab bc - ab bc/4 - ab bc/4 - ab bc/8

Axyz = ab bc - (2 ab bc/8 + 2 ab bc/8 - ab bc/8)

Axyz = 3ab*bc/8

Axyz/Aabcd = 3/8

Times 8.... --> 32x 2)=2 Sq.units Similarly Area of 🔺 XCY = 1. Sq unit And area of 🔺 ABZ =2 Sq units

Area of square - area of these 3 triangles give area of 🔺 AXZ . i .e = 8-2-2-1=3 Sq units .

As per given relation, AREA AXY= S (AREA ABCD) 3= S X 8 S= 3/8

So 8s=3

OR

Axyz = ab x bc - ab x bc/4 - ab x bc/4 - ab x bc/8

Axyz = ab x bc - (2 x ab x bc/8 + 2 x ab x bc/8 - ab x bc/8)

Axyz = 3ab x bc/8

Axyz/Aabcd = 3/8

Times 8.... --> 3

Assuming $ABCD$ is a parallelogram, the value of $S$ can be proven as follows.
Join $B$ and $D$ to form segment $BD$ which is parallel to $XY$ . The triangles $\Delta XYC$ and $\Delta DBC$ are therefore similar, and since the side lengths of $\Delta XYC$ are half those of $\Delta DBC$ , its area must be $\frac{1}{4}$ of $\Delta DBC$ , or $\frac{1}{8}$ of $ABCD$ .

Join $A$ and $C$ now to form segment $AC$ . Since $BY$ is half the length of $BC$ , the area of $\Delta ABY$ is half that of $\Delta ABC$ , or $\frac{1}{4}$ of $ABCD$ . Similarly, the area of $\Delta ABX$ is $\frac{1}{4}$ of $ABCD$ . Therefore, the answer is

$S = 1 - \frac{1}{4} - \frac{1}{4} - \frac{1}{8} = \frac{3}{8}$ $8S = \boxed{3}$