Q. 45.0( 2 Votes )

# Draw a square of area 5 square centimetres in three different ways.

Answer :

**Way 1:**

Steps of construction:

1. Draw a line BC = 5 cm, taking B and C as centres, draw

∠ABC = ∠BCD = 90°, such that AB = CD = 1 cm

2. Join AD to get rectangle ABCD, here area(ABCD) = length × breadth = 5 × 1 = 5 cm.

3. From the rectangle drawn, extend AD by 3 cm to D'.

4. From the mid–point of AD'(Let it be O such that OA = OD'=4 cm) Draw a circle taking OA = OD' as radius.

3. Now, draw a chord PQ ⊥ AD', passing through D.

6. Taking DP as side, draw a square DPRS, which is required.

Concept used:

The area of rectangle form of parts into which a diameter of a circle is cut by a perpendicular chord is equal to the area made by the square of half the chord.

And in the diagram, we made, AD' is diameter and PQ is perpendicular chord, therefore

Area(DPRS) = ar(ABCD) = 5 cm^{2}

**Way 2**:

Steps of construction:

1) Make a line segment BC = 2 cm, and C draw AC ⊥ BC such that AC = 1 cm

2) Join AB, here

AB = √(2^{2} + 1^{2}) = √5 cm

3) At points A and B, draw perpendiculars AP ⊥ AB and BQ ⊥ AB, such that

AP = BQ = AB.

4) Join PQ, ABQP is required square and area(ABQP) = (√5)^{2} = 5 cm.

**Way 3**

1) Make a line segment BC = 3 cm, and C draw AC ⊥ BC such that AC = 1 cm

2) Join AB, here

AB = √(3^{2} + 1^{2}) = √10 cm

3) Draw the perpendicular bisector of AB, and name it as XY, XY intersects AB at O.

4) Taking O as centre, and OA = OB as radius, draw a circle which intersects XY at P and Q.

5) Join AP, PB, BQ and QA to get the required square.

Verification:

Here, AB = √10 cm is the diagonal of square, and we know, diagonal of a square of side a is equal to √2 a

⇒ √2 a =√10

⇒ a = √5 cm

Rate this question :

In the picture, chords AB and CD of the circle are extended to meet at P.

i) Prove that the angles of □ APC and □ PBD, formed by joining AC and BD, are the same.

ii) Prove that PA × PB = PC × PD.

iii) Prove that if PB = PD, then ABCD is an isosceles trapezium.

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