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Alex787 [66]
3 years ago
7

HCF and LCM of 24 and 30

Mathematics
1 answer:
IgorC [24]3 years ago
5 0
Green beans, cheese and hamburger
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Step-by-step explanation:

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Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and pro
Neporo4naja [7]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Area each of them [son and daughter] will get.

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Let, ABCD be the rhombus shaped field and each side of the field be x

[ All sides of the rhombus are equal, therefore we will let the each side of the field be x ]

Now,

• Perimeter = 400m

\longrightarrow  \tt AB+BC+CD+AD=400m

\longrightarrow  \tt x + x + x + x=400

\longrightarrow  \tt 4x=400

\longrightarrow  \tt  \: x =  \dfrac{400}{4}

\longrightarrow  \tt x= \red{100m}

\therefore Each side of the field = <u>100m</u><u>.</u>

Now, we have to find the area each [son and daughter] will get.

So, For \triangle ABD,

Here,

• a = 100 [AB]

• b = 100 [AD]

• c = 160 [BD]

\therefore \tt Simi \:  perimeter \:  [S] =  \boxed{ \sf \dfrac{a + b + c}{2} }

\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}

\longrightarrow \tt S =  \cancel{ \dfrac{360}{2}}

\longrightarrow \tt S = 180m

Using <u>herons formula</u><u>,</u>

\star \tt Area  \: of  \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

<u>Putt</u><u>ing</u><u> the</u><u> values</u><u>,</u>

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(80)(80)(20) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180 \times 80 \times 80 \times 20 }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{ {3}^{2} \times  {20}^{2}  \times  {80}^{2}  }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  3 \times 20 \times 80

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} = \red{   4800  \: {m}^{2} }

Thus, area of \triangle ABD = <u>4800 m²</u>

As both the triangles have same sides

So,

Area of \triangle BCD = 4800 m²

<u>Therefore, area each of them [son and daughter] will get = 4800 m²</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

{\underline{\rule{290pt}{2pt}}}

7 0
1 year ago
Read 2 more answers
Marco purchased a large box of comic books for $300. He gave 15 of the comic books to his brother and then sold the rest on an i
Yuri [45]

Answer: There are 75 books.

Price of each book = $4.

Step-by-step explanation:

Let x = Number of books in the box.

Then as per given,

Cost of x books = $300

Cost of one book = \$(\dfrac{300}x)

Books left after giving 15 of them = x-15

Selling price of (x-15) books=  $330

Selling price of one book = \$(\dfrac{330}{x-15})

Profit on each book= $1.50

Profit = selling price - cost price

\Rightarrow 1.50=\dfrac{330}{x-15}-\dfrac{300}{x}\\\\\Rightarrow\ 1.50=\dfrac{330(x)-300(x-15)}{x(x-15)}\\\\\Rightarrow\ 1.50=\dfrac{330x-300x+4500}{x^2-15x}\\\\\Rightarrow\ 1.50(x^2-15x)=30x+4500\\\\\Rightarrow\ 1.50x^2-22.5x=30x+4500\\\\\Rightarrow\ 1.50x^2-52.5x-4500=0\\\\\Rightarrow\ 1.50x^2-52.5x-4500=0\\\\\Rightarrow\ x^2-25x-3000=0\ \ [\text{divide by 1.5}]

\Rightarrow (x+40)(x-75)=0\\\\\Rightarrow\ x=-40,75

Number of books cannot be negative.

So, there are 75 books.

Price of each book = \dfrac{300}{75}=\$4

So price of each book = $4.

5 0
3 years ago
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