Question

It would be great if someone could help me with the question on the right. Please include working and solution

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

• Bigger square is represented as ABCD

• Smaller square is represented as EFGH

Let further assume that

• EFGH intersects the sides of square ABCD at I, J, K, L, M, N, O, P respectively as shown in attachment.

According to given statement,

$\purple{\rm :\longmapsto\:Area_{(EFGH)} = 85 \: {cm}^{2} \: - - - - (1)}$

Now,

• Side of square ABCD, = 10 cm

So,

$\purple{\rm :\longmapsto\: Area_{(ABCD)} = {(10)}^{2} = 100 \: {cm}^{2} - - - (2) \: }$

Now,

$\green{\rm :\longmapsto\:Area_{(dark \: green \: \triangle)} = 4 \times \dfrac{1}{2} \times 4 \times 3 = 24 \: {cm}^{2}}$

Now,

$\red{\rm :\longmapsto\:Area_{(IJKLMNOP)}}$

$\rm \:  =  \: Area_{(ABCD)} - Area_{(dark \: green \: \triangle's)}$

$\rm \:  =  \: 100 - 24$

$\rm \:  =  \: 76 \: {cm}^{2}$

$\rm\implies \:\boxed{\tt{ Area_{(IJKLMNOP)} = 76 \: {cm}^{2}}} - - - - (3)$

Now,

$\green{\rm :\longmapsto\:Area_{(light \: green \: \triangle's)}}$

$\rm \:  =  \: Area_{(EFGH)} - Area_{(IJKLMNOP)}$

$\rm \:  =  \: 85 - 76$

$\rm \:  =  \: 9 \: {cm}^{2}$

$\rm\implies \:\boxed{\tt{ Area_{(light \: green \: \triangle's)} = 9 \: {cm}^{2} }}$

Thus,

$\rm\implies \:Area_{(dark \: green \: \triangle's)} + Area_{(light \: green \: \triangle's)}$

$\rm \:  =  \: 24 + 9$

$\rm \:  =  \: 33 \: {cm}^{2}$

So,

$\green{\rm\implies \boxed{\sf{ \:Area_{(dark \: green \: \triangle's)} + Area_{(light \: green \: \triangle's)} = 33 \: {cm}^{2} }}}$

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Solution :-

→ Side of larger square = 10 cm

So,

→ Area of larger square = (side)² = (10)² = 100 cm² .

now,

→ Area of dark green ∆ = (1/2) * Base * Perpendicular height { since angle of a square is equal to 90° , so it is a right angled ∆ . }

So,

→ Area of 4 dark green identical ∆'s = 4 * (1/2) * 3 * 4 = 24 cm² .

then,

→ Area of larger square which is not shaded (white portion) = Total area - Area of 4 dark ∆'s = 100 - 24 = 76 cm² .

now, given that,

→ Area of smaller square = 85 cm²

So,

→ Area of 4 light green ∆'s = Area of smaller square - White portion area = 85 - 76 = 9 cm²

therefore,

→ Total shaded area of 8 ∆'s = Area of 4 dark green identical ∆'s + Area of 4 light green ∆'s = 24 + 9 = 33 cm² (Ans.)

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