Question

11. Area of a rhombus is 90 cm2. If the length of one diagonal is 10 cm then what is length of second diagonal​

Answer 1

Given

Area of rhombus is 90cm²

Length of one diagonal is 10cm

To Find

we have to find the length of second diagonal

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Since,Area is given so,we will apply the formula for area of rhombus and assume some Variable in place of Diagonal 2 we have to find .

Area of Rhombus= D1 × D2 ÷ 2

↣Area of Rhombus= 10× Diagonal 2 ÷ 2

let length of other diagonal be 'x'

↣90= 10 × x ÷2

↣90= 10x/2

↣10x= 90×2

↣10x=180

↣x=180/10=18

↣x= 18

hence,length of other diagonal is 18cm

Check:

Area of Rhombus= 18×10÷2

Area of Rhombus= 180÷2=90cm²

Extra information=>

Area is Space or region which is covered by a solid object/figure it is said to be area of that object.

Area depends upon the dimensions of the figure.If dimensions is larger then larger will be the area of that object/figure and vice -versa.

Properties of Rhombus:

• All sides of rhombus are equal and opposite sides are parallel.
• Diagonals of Rhombus bisects each other at 90°.
• There are four congruent right angled triangles formed between the two diagonals

Answer 2

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$\LARGE\textsf{\underline\textcolor{aqua}{↭ GiVeN :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{Area of the Rhombus = 90 cm²}$

$\qquad\tt{:}\longrightarrow\large\textsf{Length of one diagonal = 10 cm}$

$\large\textsf{ }$

$\LARGE\textsf{\underline\textcolor{aqua}{↭ To FiNd :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{Length of the second diagonal = ?}$

$\large\textsf{ }$

$\LARGE\textsf{\underline\textcolor{aqua}{↭ FoRmUlA :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{purple}{ {\large\textsf{Area}}_{\normalsize\textsf{( \; Rhombus \; )}} = \cfrac{\large\textsf{1}}{\large\textsf{2}} × \large\textsf{ D1 × D2} }}$

$\large\textsf{ }$

$\LARGE\textsf{\underline\textcolor{aqua}{↭ SoLuTiOn :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{ {\large\textsf{Area}}_{\normalsize\textsf{( \; Rhombus \; )}} = \cfrac{\large\textsf{1}}{\large\textsf{2}} × \large\textsf{ D1 × D2} }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{90 = \cfrac{\large\textsf{1}}{\large\textsf{2}} × \large\textsf{ 10 × D2} }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ \cfrac{\large\textsf{90 × 2}}{\large\textsf{10}} = \large\textsf{ D2 } }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ \cancel{\cfrac{\large\textsf{180}}{\large\textsf{10}}} = \large\textsf{ D2 } }\\\\\\\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{18 = D2}}$

$\large\textsf{ }$

$\large\textsf\textcolor{orange}{∴ The length of the second diagonal = 18 cm}$

$\large\textsf{ }$

$\LARGE\textsf{\underline\textcolor{aqua}{↭ MoRe FoRmUlAs :-}}$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{L.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2h ( l + b )} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2 ( lb + bh + hl )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{l×b×h} }$

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$\large\textsf{ {\large\textsf{L.S.A.}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{4×l²} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{6 × l²} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{l²} }$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{C.S.A.}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{2 × πrh} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{2πr × ( r + h )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{πr²h} }$

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$\large\textsf{ {\large\textsf{C.S.A.}}_{\large\textsf{( \; Cone \; )}} = \large\textsf{πrl} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cone \; )}} = \large\textsf{πr × ( r + l )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cone \; )}} } \large\textsf{ = \cfrac{\large\textsf{1}}{\large\textsf{3}} }\large\textsf{× πr²h}$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Sphere \; )}} = \large\textsf{4πr²} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Sphere \; )}} } \large\textsf{ = \cfrac{\large\textsf{4}}{\large\textsf{3}} }\large\textsf{× πr³}$

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