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Construct a ABCD, when: $AB = 5.8 cm, diagonal AC = 8.2 cm$ and diagonal $BD = 6.2 cm$ . What is the shaper of this figure?

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Parallelogram

Kite

Hexagon

None

answer is A.

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Detailed Solution

Given, the value

AB = 5.8 cm

Diagonal AC = 8.2 cm

Diagonal

BD = 6.2 cm

We construct a figure stepwise as,
First of all make a line AB = 5.8 cm.
1st In OAB,

OA = \frac{1}{2} AC = \frac{1}{2} \times 8.2 cm = 4.1 cm .

{Since, A parallelogram diagonal bisect each other]

OB = \frac{1}{2} BD = \frac{1}{2} \times 6.2 cm = 3.1 cm .

So draw an arc from A as center with measure 4.1 cm and from B as center with measure 3.1. Give name O to the intersection point of arcs and join A and B to O.
2nd AO produce up to C

OC = OA = 4.1 cm

And BO produces up to D

DO = OB = 3.1 cm

So, extent AO to C such OC=4.1 and OB to D such that DO= 3.1.
3rd we have to joined AD, DC and CB.
C:UserssoftberryDownloadsP0041 JPEGQ.28.jpg

The above figure is the required figure of ABCD.
This figure is parallelogram opposite sides are equal and parallel.
Correct option is 1.

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