Question

draw a square of area 5 centimeters in three different ways

Answer 1

there will be only one side that is
$\sqrt{5}$

Answer 2

Answer:

There are 3 ways.

Step-by-step explanation:

Way 1

Steps of construction

1) Create a segment of a line BC = 3 cm, and then have C draw AC ⊥ BC so that AC = 1 cm.

2) Join AB 

$AB = \sqrt{3^{2}+1^2}\\AB= \sqrt{10}cm$

3) Draw the perpendicular bisector of AB, label it XY, and place O at the intersection of XY and AB.

4) Draw a circle using P and Q where it intersects XY using O as the centre and OA = OB as the radius.

5) AP, PB, BQ, and QA can be joined to form the necessary square.

Here, the square's diagonal measures AB = 10 cm, and we already know that the diagonal correspond to $\sqrt{2}$ the times square's side.

$\sqrt{2a} =\sqrt{10}$

$a=\sqrt{5}cm$

$area= 5cm^{2}$

Way 2

Steps of construction

1) Create the line segment BC = 2 cm, and then have C draw AC ⊥  BC so that AC = 1 cm.

2)Join AB 

and $AB= \sqrt{ (22 + 12) }$

$AB=\sqrt{5}$

3)Draw the perpendiculars AP ⊥  AB and BQ ⊥ AB at locations A and B so that

$BQ=AB=AP$

4) Join PQ, ABQP is a necessary square, and its area is $=[\sqrt{5}]^{2}$

$=5cm$

$area= 5cm^{2}$

Way 3

Steps of construction

1. With B and C as the centres, draw a line BC = 5 cm.

AB = CD = 1 cm when ∠ABC = ∠BCD = 90°.

(Image 1)

2. Join the two sides of AD to create the rectangle ABCD, where the area(ABCD) = length x breadth = 5 x 1 = 5 cm

3. Draw a rectangle and extend AD to D' by 3 cm.

4. Draw a circle with OA = OD' as the radius starting at the midpoint of AD' (OA = OD'=4 cm).

5. Draw a chord PQ ⊥ AD' now, going through D.

6. Draw the necessary square DPRS using DP as the side.

Idea employed:

A perpendicular chord cuts a circle's diameter into rectangle-shaped pieces, and the area created by the square of half the chord is equal to that area.

Additionally, we created a graphic in which AD' denotes diameter and PQ denotes a perpendicular chord.

Area(DPRS) equals area(ABCD) as $5cm^{2}$

$area= 5cm^{2}$

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