Question

please help anybody who can​

Answer 1

$\red{\large\text{\underline{Footnote}}}$

Join the point of intersections of two squares. Then, the shaded region will be divided into two congruent triangles, both of which are right-angled. This is the area of one triangle.

$\implies\dfrac{1}{2}\times12\times6=36\text{ m}^{2}$

Then, we can calculate the area of two triangles, which is the area of the shaded region.

$\implies2\times36=72\text{ m}^{2}$

So, the area of the shaded region is $72\text{ m}^{2}$.

$\red{\large\text{\underline{End of footnote.}}}$

$\red{\large\text{\underline{Solution}}}$

The area of one square is $12\times12=144\text{ m}^{2}$.

We can find the unshaded area using the following method.

$\red{\large\text{\underline{Footnote}}}$

We can calculate the whole area in the following method.

$\implies\text{(Whole area)}=\text{(Area of two squares)}-\text{(Shaded region)}$

We can calculate the unshaded region in the following method.

$\implies\text{(Unshaded region)}=\text{(Whole area)}-\text{(Shaded region)}$

This gives the result that the area is the following.

$\red{\bigstar}\text{(Unshaded region)}=\text{(Area of two squares)}-2\times\text{(Shaded region)}$

$\red{\large\text{\underline{End of footnote.}}}$

$\red{\bigstar}\text{(Unshaded region)}=\text{(Area of two squares)}-2\times\text{(Shaded region)}$

$\implies\text{(Unshaded region)}=2\times144-2\times72=144\text{ m}^{2}$

$\red{\large\text{\underline{Conclusion}}}$

Hence, the area of the unshaded region is $144\text{ m}^{2}$.

Answer 2

Given,

Two identical squares with side-length 12 m intersect at M point and M is the midpoint of the corresponding side of both the squares.

We have to find :

• The area of the unshaded region.

Solution :

From the figure,

The sides of quadrilateral ABMD is :

AB = AD = 12 m

BM = MD = 6 m

Now at first we have to find the black shaded region.

By Brahmagupta's formula, we know that,

The area of a quadrilateral with sides a, b, c, d is given by

$\rm \: area = \sqrt{(s - a)(s - b)(s - c)(s - d)}$

Where,

a, b, c and d are the four sides.

and

$\rm \: s = \frac{a + b + c + d}{2}$

Now for the quadrilateral ABMD

$\begin{array}{cc} \rm \: s = \frac{AB +MB + MD + AD }{2} \\ \\ \rm = \frac{12 + 6 + 6 + 12}{2} \\ \\ = 18 \: m \end{array}$

${\boxed{\begin{array}{cc}\rm \: area \: \: A_{ {ABMD}} = \sqrt{(18 - 12)(18 - 6)(18 - 6)(18 - 12)} \\ \\ = \sqrt{6 \times 12 \times 12 \times 6} \\ \\ = 72 \: {m}^{2} \end{array}}}$

Again, the area of both square are same

${\boxed{\boxed{\begin{array}{cc}\bf \: area \: of \: square \: \: \: A_{AEFD} =A _{ABGH} = {(12)}^{2} \\ \\ = 144 \: {m}^{2} \end{array}}}}$

Now area of unshaded region :

${\boxed{\boxed{\begin{array}{cc}\rm \: A_{unshaded} =A_{AEFD} + A _{ABGH} - 2 \: A_{ABMD}\end{array}}}}$

here,

$\orange{{\boxed{\begin{array}{cc}\sf \: (2 \times \: A{_{ABMD}} )\: \: cause \: A_{ABMD} \: part \: is \: located \\ \\ \sf \: both \: part \: of \: those \: square\end{array}}}}$

So,

${\boxed{\boxed{\begin{array}{cc}\rm \: A_{unshaded} = 144 + 144 - 2 \times 72 \\ \\ = 144 + 144 - 144 \\ \\ = 144 \: {m }^{2} \end{array}}}}$

So the area of unshaded region is 144 m²