Question

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. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops?​

Answer 1

4800 m²

Explanation:

Properties of a Rhombus

• quadrilateral (4 sided shape)
• 2 pairs of parallel sides
• All 4 sides are equal in length
• Opposite angles are equal
• Diagonals bisect each other at 90°
• Adjacent angles sum to 180°

Therefore, a rhombus can be divided into 4 congruent right triangles by drawing the diagonals. The hypotenuse of each right triangle is the side length.

Given:

• Perimeter = 400 m

⇒ side length = 400 ÷ 4 = 100 m

Therefore, the hypotenuse of each right triangle is 100 m.

As the diagonals bisect each other (divide into 2 equal parts) at right angles, a leg length of each right triangle will be half a diagonal.

Given:

• Diagonal = 160 m

⇒ 160 ÷ 2 = 80 m

Therefore, a leg length of the right triangles is 80 m.

To find the other leg length, use Pythagoras' Theorem:

⇒ a² + b² = c²

⇒ a² + 80² = 100²

⇒ a² = 3600

⇒ a = 60 m

The leg lengths are the base and height of the right triangles.

Area of a triangle = 1/2 × base × height

⇒ Area = 1/2 × 60 × 80

⇒ Area = 2400 m²

As the land is divided into 2 equal parts, each area will be twice the area of one right triangle.

Therefore, the area that Sanya's daughter and son each get for their crops is 4800 m²

Answer 2

`{\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 = (400)/(4)`

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

`\therefore` Each side of the field = 100m.

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 (a + b + c)/(2) }`

`\longrightarrow \tt S = (100 + 100 + 160)/(2)`

`\longrightarrow \tt S = \cancel{ (360)/(2)}`

`\longrightarrow \tt S = 180m`

Using herons formula,

`\star \tt Area \: of \: \triangle = \boxed{\bf{{ √(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.

Putting the values,

`\longrightarrow \tt Area_( ( \triangle \: ABD)) = \tt √(180(180 - 100)(180 - 100)(180 - 160) )`

`\longrightarrow \tt Area_( ( \triangle \: ABD)) = \tt √(180(80)(80)(20) )`

`\longrightarrow \tt Area_( ( \triangle \: ABD)) = \tt √(180 * 80 * 80 * 20 )`

`\longrightarrow \tt Area_( ( \triangle \: ABD)) = \tt √(9 * 20 * 20 * 80 * 80)`

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

`\longrightarrow \tt Area_( ( \triangle \: ABD)) = 3 * 20 * 80`

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

Thus, area of
`\triangle` ABD = 4800 m²

As both the triangles have same sides

So,

Area of
`\triangle` BCD = 4800 m²

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

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

★ Figure in attachment.

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