Question

The diagonal of a rectangular field exceeds its length by 2 m and exceeds
twice its breadth by 1 m. Find the length and breadth of the field.​

Answer 1

Answer:

Since a diagonal of a rectangle splits the rectangle into 2 right angles triangles, we can apply Pythagoras Theorem to find the length of the diagonal.

From the diagram, On applying the Pythagorean Theorem, we get:

⇒ Length² + Breadth² = Diagonal²

Let the length be 'l' and breadth be 'b'

⇒ Diagonal = √ ( l² + b² )   ...(1)

Also it is given that:

⇒ Diagonal = l + 2 ...(2)

⇒ Diagonal = 2b + 1   ...(3)

Comparing (2) and (3) we get:

⇒ l + 2 = 2b + 1

⇒ l -2b = -1  

⇒ l = 2b - 1   ...(4)

Since we dont have any known values to form a relation between Diagonals and Dimensions, the final answer is in terms of a relation.

Therefore the required relation is:

→ Length = 2 (Breadth) - 1

Answer 2

Given : The diagonal of a rectangular field exceeds its length by 2 m and exceeds twice its breadth by 1 m.

Exigency To Find : The length and breadth of the field.

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❍ Let's Consider the Length , Breadth and Diagonal of Rectangular field be L , B & D , respectively.

⠀⠀⠀⠀⠀⠀CASE I : The Diagonal of a rectangular field exceeds its length by 2 m .

$\qquad:\implies \sf Diagonal \: =\: Length \: + 2m$

$\qquad:\implies \sf D \: =\: L \: + 2$

$\qquad:\implies \bf D \: =\: L \: + 2 \qquad \qquad \:\bigg\lgroup \sf{ \:eq^n \:\: 1 \: }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀CASE II : The Diagonal of Rectangular field exceeds twice its breadth by 1 m.

$\qquad:\implies \sf Diagonal \: =\: 2 \times Breadth \: + 1m$

$\qquad:\implies \sf D \: =\: 2 \times B \: + 1$

$\qquad:\implies \sf D \: =\: 2B \: + 1$

$\qquad:\implies \bf D \: =\: 2B \: + 1 \qquad \qquad \:\bigg\lgroup \sf{ \:eq^n \:\: 2 \: }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀Now ,

⠀⠀⠀⠀⠀⠀ Finding Length of Rectangular field :

⠀⠀⠀⠀⠀⠀From Equation 2 :

$\qquad \dag\:\:\bigg\lgroup \sf{Equation \:2 \:\::\:D \: =\: 2B \: + 1\:\: }\bigg\rgroup$

$\qquad:\implies \sf D \: =\: 2B \: + 1$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: eq^n \: 1 \: \::}}$

$\qquad \dag\:\:\bigg\lgroup \sf{ \:Equation \: 1 \:: \:D \: =\: L \: + 2\: }\bigg\rgroup$

$\qquad:\implies \sf D \: =\: 2B \: + 1$

$\qquad:\implies \sf \: L \: + 2 \: =\: 2B \: + 1$

$\qquad:\implies \sf \: L \: \: =\: 2B \: + 1 - 2$

$\qquad:\implies \sf \: L \: \: =\: 2B \: - 1$

$\qquad:\implies \bf L \: =\: 2B \: - 1 \qquad \qquad \:\bigg\lgroup \sf{ \:eq^n \:\: 3 \: }\bigg\rgroup$

$\qquad:\implies \bf \: L \: \: =\: 2B \: - 1$

$\qquad :\implies \pmb{\underline{\purple{\:L \: \: =\: 2B \: - 1 }} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀▪︎ Here , L denotes Length of Rectangular field which is 2B - 1 OR , 2 × Breadth - 1 .

⠀⠀⠀⠀⠀⠀Now , Finding Breadth of Rectangular field :

⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀From Equation 3 :

$\qquad \dag\:\:\bigg\lgroup \sf{Equation \:3\::\:L\: =\: 2B \: - 1 \:\: }\bigg\rgroup$

$\qquad:\implies \sf \: L \: \: =\: 2B \: - 1$

$\qquad:\implies \sf \: L \: + 1 \: =\: 2B \:$

$\qquad:\implies \sf\: 2B\: = L \: + 1 \: \:$

$\qquad:\implies \sf\: B\:=\:\dfrac{ L \: + 1}{2} \: \:$

$\qquad:\implies \bf\: B\:=\:\dfrac{ L \: + 1}{2} \: \:$

$\qquad :\implies \pmb{\underline{\purple{\:B\:=\:\dfrac{ L \: + 1}{2} }} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀▪︎ Here , B denotes Breadth of Rectangular field which is $\bf \:\dfrac{ L \: + 1}{2}$ OR , $\bf \:\dfrac{ Length \: + 1}{2}$ .

⠀⠀⠀⠀▪︎ Since , we cannot find exact value of Length & Breadth of Rectangular field from the given values so we have found the value of Length and Breadth in terms of relation.

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Length \:of\:Rectangular \:Field \:is\:\bf{ 2 \times (Breadth) - 1\: }}}}$

⠀⠀⠀⠀⠀&

⠀⠀⠀⠀⠀${\underline{ \mathrm {\:Breadth \:of\:Rectangular \:Field \:is\:\bf{ \:\:\dfrac{ Length \: + 1}{2}\: }}}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\qquad \leadsto \sf Area_{(Rectangle)} = Length \times Breadth$

$\qquad \leadsto \sf Perimeter _{(Rectangle)} = 2 (Length + Breadth)$

$\qquad \leadsto \sf Area_{(Square)} = Side \times Side$

$\qquad \leadsto \sf Perimeter _{(Square)} = 4 \times Side$

$\qquad \leadsto \sf Area_{(Trapezium)} = \dfrac{1}{2} \times Height \times (a + b )$

$\qquad \leadsto \sf Area_{(Parallelogram)} = Base \times Height$

$\qquad \leadsto \sf Area_{(Triangle)} = \dfrac{1}{2} \times Base \times Height$

$\qquad \leadsto \sf Area_{(Rhombus)} = \dfrac{1}{2} \times Diagonal _{1}\times Diagonal_{2}$

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