Mathematics: 3. In a figure a small square is cutting out from the largesquare. The area of the remaining part is 91 cm². The sum of Sides equal to 52cm. (a). If the side of large square is x and the side of smallest square is y then find the area of the remaining part?(b). Find the sides of the smallest square?(c). Find the areas of the two squares?

Sam would have m of rope leftAmount of rope he had = (otherwise expressed as )Amount of rope he cut for a project = Further ExplanationTo determine how much rope Sam has left, the amount of rope he had ( ) will be subtracted from the amount of rope he cut for a project ( ). The final value is the how much rope he has left.Notably, the numerical values to be added are in fractions, and since fraction has to be solved by its numerator and denominator, we therefore find the lowest common denominators...

3. In a figure a small square is cutting out from

Ответ:

alyssabalinnt

09.10.2021

Mathematics

Step-by-step explanation:

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

Given that,

ABCD is a square of side x cmEFGH is a square of side y cm

Further given that,

Sum of all sides = 52 cm

It means

Perimeter of square ABCD + Perimeter of square EFGH is 52 cm

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ Perimeter_{[square]} \: = \: 4 \times side \: }}}$

So, using this, we have

$\rm :\longmapsto\:4x + 4y = 52$

$\rm :\longmapsto\:4(x + y) = 52$

$\rm\implies \:\boxed{\tt{ x + y = 13}} - - - - (1)$

Now, Further given that,

If square EFGH is cutting out from square ABCD, the area of remaining part is 91 square cm.

It means

$\rm :\longmapsto\:Area_{[ABCD]} - Area_{[EFGH]} = 91$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ Area_{[square]} = 4 \times side}}}$

So, using this, we get

$\rm :\longmapsto\: {x}^{2} - {y}^{2} = 91$

can be further rewritten as using algebraic Identity,

$\rm :\longmapsto\:(x + y)(x - y) = 91$

$\rm :\longmapsto\:13(x - y) = 91$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (1) \bigg\}}$

$\rm\implies \:\boxed{\tt{ x - y = 7}} - - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2x = 13 + 7$

$\rm :\longmapsto\:2x = 20$

$\rm\implies \:\boxed{ \: \bf \: \: x \: = \: 10 \:cm \: \: }$

On substituting the value of x in equation (1), we have

$\rm :\longmapsto\:10 + y = 13$

$\rm :\longmapsto\:y = 13 - 10$

$\rm\implies \:\boxed{ \: \bf \: \: y \: = \: 3 \:cm \: \: }$

So,

$\rm\implies \:\boxed{Side_{[square \: EFGH]} = y \: = 3 \: cm}$

and

$\rm\implies \:\boxed{Side_{[square \: ABCD]} = x \: = 10 \: cm}$

Also,

$\rm :\longmapsto\:\boxed{Area_{[square \: ABCD]} = {10}^{2} = 100 \: {cm}^{2} }$

$\rm :\longmapsto\:\boxed{Area_{[squareEFGH] }\: = {3}^{2} = 9 \: {cm}^{2} }$

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

3. In a figure a small square is cutting out from the largesquare. The area of the remaining part is

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Ответ:

Pranav2004

16.09.2022

Mathematics

Sam would have $\frac{7}{8}$ m of rope left

Amount of rope he had = $1\frac{1}{2}$ (otherwise expressed as $\frac{3}{2}$ )Amount of rope he cut for a project = $\frac{5}{8}$ Further Explanation

To determine how much rope Sam has left, the amount of rope he had ( $\frac{3}{2}$ ) will be subtracted from the amount of rope he cut for a project ( $\frac{5}{8}$ ). The final value is the how much rope he has left.

Notably, the numerical values to be added are in fractions, and since fraction has to be solved by its numerator and denominator, we therefore find the lowest common denominators (LCD) of both denominators (2 and 8), and then use it to factor out the numerators, the resulting values will be subtracted to get the final value.

Therefore,

The LCD of 2 and 8 = 8

= $\frac{3}{2}$ – $\frac{5}{8}$

= (4 x 3) – (1 x 5) / 8

= (12) – (5) / 8

= 12 – 5 / 8

= $\frac{7}{8}$

Therefore, the total amount of rope Sam has left = $\frac{7}{8}$ m

The above math problem is expressed in fraction. In order to solve the math problem, an order of operation i.e. BODMAS had to be applied. Significantly, Multiplication and Subtraction are part of the arms of BODMAS that were used to derive the solution to the given problem.

LEARN MORE:

mixed numbers Lowest common denominator fraction Sam had 1 1/2 m of rope. He cut off 5/8 m and used it for a project. How much rope does Sam have left

KEYWORDS:

1 1/2 m of ropemixed fractionlowest common denominatorfractionnumbersnumerator5/8 mSam

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