Math, asked by miyeong55, 11 months ago

여보세요 chingus!!!!
pls solve this
The lenght of a side of square field is 4m.what will be the altitude of rhombus if the area of rhombus is equal to the area of square field and one of its diagonal is 2m( rhombus)

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Answered by Anonymous

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Answered by Anonymous

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$\huge \pink {\tt {여보세요 \: Miyeong ,}}$

The length of the side of a square field PQRS = 4 m

The Area of the square field PQRS = $\large {side}^{2} = {4}^{2} = 16 {m}^{2}$

The Diagonal1 of Rhombus ABCD = 2 m

Area of Rhombus ABCD = $\large \frac{1}{2} \times d1 \times d2 = 16{m}^{2}$

$\large = \frac{2 \times d2}{2} = 16 {m}^{2}$

$\large = d2 = 16m$

Diagonal of rhombus are perpendicular bisector of each other.

By Using Pythagoras Theorem in Triangle AOB, we get

$\large {AO}^{2} + {BO}^{2} = {AB}^{2} = base \: of \: the \: rhombus \: ABCD$

$\large = {1}^{2} + {8}^{2} = {AB}^{2}$

$\large = \sqrt{1 + 64} = AB =base \: of \: the \: Rhombus$

$\large \sqrt{65} = AB$

Rhombus is also a parallelogram because all it's sides are equal.

Area of a Rhombus = $\large base \times height$ = Area of a parallelogram

Area of rhombus =

$\large \sqrt{65} \times height = 16 {m}^{2}$

Height of the Rhombus ABCD =

$\large \frac{16}{ \sqrt{65} } = 1.98456 \: metre$

Hope that helps you Chingu

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