Question

construct the Rhombus ABCD ?
AC=6 cm
BD=4 cm

( Class 8 ka ha ) answer bhejna please

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Answer 1

Answer:

40.7

Step-by-step explanation:

1. Draw a diagonal AC=6 cm

2. Draw a perpendicular bisector of AC, let the perpendicular bisector and AC intersects at O (∵ Diagonals of rhombus intersects at right angle)

3. Taking O as centre and half of the length of another diagonal (i.e 3.5 cm) as radius, mark arc on the perpendicular bisector at both sides of AC. Name these intersection points as B and D

4. Join AB,AD,BC,CD.

5. ABCD is the required rhombus.

6. Measure the ∠ABD which come out to be 40.7˚.

Answer 2

Given :  AC=6 cm

BD=4 cm

To Find : construct the Rhombus ABCD

Solution:

Diagonal of Rhombus bisect each other.

Step  1: Draw  a line segment AC = 6  cm

Step 2 : Using compass width more than 3 cm , and taking A as center , draw arcs on both sides of AC

Step 3 : Keeping compass width same and  taking A as center , draw arcs on both sides of AC such that it intersect arcs drawn in previous step at X and Y

Step 4 : Join X and Y , intersecting AC at O

Step 5 : Using compass width = 2 cm  (BD/2 = 4/2 = 2)  and taking O as center  cut XY at  B and D

Step 6 : Join AB , BC , CD and AD

Rhombus ABCD is constructed

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