Area of square ABCD is 169 and that of square EFGD is 49. Find area of quadrilateral FBCG
I am stuck and just thinking which way can be helpful for me finding this area of quadrilateral FBCG. Instead of hint can please suggest me perfect answer , so using this idea answer, i will solve other questions which are near about similar.
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$\begingroup$By symmetry, the quadrilaterals $FBCG$ and $FBAE$ are congruent, so they have the same area. Their combined area is the area of the region which is in the big square but outside the little square. So their combined area is $169-49$, that is, $120$. It follows that quadrilateral $FBCG$ has area $60$.
Remark: The above is the quickest way to find the answer.
But we can do it in various other ways, that may be useful to you in other problems. For example, the little square has area $49$, so it has sides $7$ each. The big square has area $169$, so has sides $13$.
It follows that the height $CG$ of the trapezoid $FBCG$ is $13-7=6$. The two bases are $7$ and $13$. So by a standard formula for the area of a trapezoid, the area is $(6)\left(\frac{7+13}{2} \right)$.
Or else you can think of the region $FBCG$ as made up of a $6\times 7$ rectangle, with half of a $6\times 6$ square on top. That gives area $(6)(7)+\left(\frac{1}{2}\right)(6)(6)$.
Another computational approach is by using triangles. Because of your picture, we calculate the area of $ABFE$, which is the same as the area of our target region. This is made up, as your picture shows, of two triangles. Each has height $13-7=6$. Their bases are $7$ and $13$, Now huse the standard formula for the area of a triangle.
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