Question

rhombus is 30 m and its longer diagonal is 48 m. how much area of grass field will each
A rhombus shaped field has green grass for 18 cows to graze. If each side of the
207
cow be getting
thy stitching 10tan​

Answer 1

Correct Question :-

• The Diagonals of a Rhombus are 30m and 48m, then Find it's Area.

Given :-

• Shape of the Object = Rhombus
• First Diagonal of Rhombus = 30m
• Second Diagonal of Rhombus = 48m

To Find :-

• Area of Rhombus.

Solution :-

⇒ Area = ½ × product of its diagonals

⇒ Area = ½ × ( 30 × 48 )

⇒ Area = ½ × 1440

⇒ Area = 720m²

________________

Therefore, Area of Rhombus is 720m²

________________

★ Additional Info :

Formulas Related to Area :-

• Area of Square = Side x Side
• Area of Rectangle = Length × Breadth
• Area of Triangle = ½ × base x height
• Area of parallelogram = base x height
• Area of circle = πr²
• Area of Rhombus = ½ × product of its diagonals
• Area of Trapezium = ½ × height × sum of parallel sides
• Area of Polygon = sum of the area of all regions into which it is divided

________________

Answer 2

$\large\underline{\underline{ \sf \maltese{\:appropriate \: question : }}}$

• A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

__________________________________________

$\large\underline{\underline{ \sf \maltese{\:Given:-}}}$

• each side of the rhombus is 30M
• it's longer diagonal measures 48M
• 18 cows are to be grazed

__________________________________________

$\large\underline{\underline{ \sf \maltese{\:to \: find:}}}$

• how much area of grass field will each cow be getting?

__________________________________________

$\large\underline{\underline{ \sf \maltese{\:understanding \: the \: question : }}}$

➢ Now here we first find the area of the rhombus and then divide it by the number of cows so that, it we get the area of grass each cow gets to graze :)

__________________________________________

$\large\underline{\underline{ \sf \maltese{\:solution : }}}$

➳ each cow gets 48m² of land to graze.

__________________________________________

$\large\underline{\underline{ \sf \maltese{\:full \: solution : }}}$

• Now let's name the rhombus field as PQRS for easy identification.

• As each side measures 30cm. ( PQ = QR = RS = PS )

• Diagonal is 48cm ( QS )

$\orange{ \underline {\boxed {\mathfrak{ area \: of \: the \: rhombus = \triangle \: pqs \: + \triangle \: qrs}}}} \bigstar$

➺ Now by using heron's therom

For triangle PQS,

$\sf \: area \: of \triangle = \sqrt{s(s - p)(s - q)(s - s)}$

❍ Here ,

★ S = semi perimeter

★ P,Q,S are the sides

✬Let's find the value of the semi perimeter now✬

$\longrightarrow \tt \: semiperimeter = \frac{p + q + s}{2} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \tt \: semiperimeter = \frac{30 + 30 + 48}{2} \: \: \: \\ \\ \longrightarrow \tt \: semiperimeter = \frac{108}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \tt \: \orange{semiperimeter = 54m \bigstar} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

✏️ Let's substitute the values now:

$: \implies\sf \: \sqrt{54(54 - 30)(54 - 30)(54 - 48)} \\ \\ \\ : \implies \sf \: \sqrt{54 \times 24 \times 24 \times 6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \implies \sf \sqrt{186624} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \implies \sf \: \pink{ \underline{ \boxed{432 }}\bigstar} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\therefore {\underline{ \sf{the \: area \: of \triangle \: pqs = 432 {m}^{2} }}} \: \star \:$

❒ Now as the sides of triangle QRS are the same too,

• area of triangle PQS = QRS

$: \longrightarrow \sf area \: of \: \triangle \: pqs = qrs \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \longrightarrow \sf area \: of \: rhombus = 432 {m}^{2} + 432 {m}^{2} \\ \\ \\ : \longrightarrow \sf area \: of \: rhombus = 864 {m}^{2} \bigstar\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\therefore {\underline{ \sf{the \: area \: of \: rhombus = 864 {m}^{2} }}}$

$\\ \\ \\ { : \implies}\sf \: area \: each \: cow \: will \: get = \cancel \frac{864}{18} \: \: \: \: \: \: \\ \\ \\ { : \implies}\sf \: area \: each \: cow \: will \: get = 48 {m}^{2} \bigstar$

$\blue{ \underline{ \boxed{ \pink{ \mathfrak{ \therefore \:\: area \: each \: cow \: will \: get = 48 {m}^{2} }}}}}$

__________________________________________

$\large\underline{\underline{ \sf \maltese{\:formulas \: used:}}}$

$\sf area \: of \: a \: triangle = \sqrt{s(s - a)(s - b)(s - c)}$

__________________________________________

Knowledge triangle:

$\leadsto \sf \: area \: of \: a \: triangle = \frac{1}{2} \times bh \: \: \: \: \: \\ \\ \leadsto \sf \: area \: of \: a \: rhombus = \frac{ d_{1}d_{2} }{2} \: \: \: \: \: \: \: \\ \\ \leadsto \sf \: area \: of \: a \: circle = \pi \: {r}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \leadsto \sf \: area \: of \: a \: paralleogram = b \times h$