Question

Construct a square ,each of whose diagonal measure 6.8 cm

Answer 1

Answer:

construct a square, each of whose diagonals measures 6.8 cm.

Step-by-step explanation:

I hope it helps you

Answer 2

Answer:

Given : Diagonal$=6 \mathrm{~cm}$.

Steps for construction

Step 1: Create a basic diagram and indicate the necessary measurements.

Step 2: Draw a line segment $\mathrm{AC}=6 \mathrm{~cm}$.

Step 3 : Construct a perpendicular bisector $\mathrm{XY}$ of $\overline{\mathrm{AC}}$.

Step $4: \mathrm{XY}$ intersects $\overline{\mathrm{AC}}$at $\mathrm{O}$. We get $\mathrm{OC}=\mathrm{AO}=3 \mathrm{~cm}$.

Step 5: With $O$ as centre draw two arcs of radius $3 \mathrm{~cm}$ cutting the line $X \vec{Y}$ at points$\mathrm{B}$ and $\mathrm{D}$.

Step 6 : Join $\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}$ and$\overline{\mathrm{DA}}$.

$\mathrm{ABCD}$ is the required square

Step-by-step explanation:

Step:1  Create a square with a 6 cm diagonal. The side is measured. locate it as well. Step 1: Sketch a rough diagram and indicate the necessary measurements. Create an XY perpendicular bisector of AC in step 3. Create a square with a 6 cm diagonal. The side is measured. locate it as well. Step 1: Sketch a rough diagram and indicate the necessary measurements. Create an XY perpendicular bisector of AC in step 3. Step 4: At O, XY and AC cross. We obtain OC=AO=3 cm. Create a square with a 6 cm diagonal. The side is measured. locate it as well. Step 1: Sketch a rough diagram and indicate the necessary measurements. Create an XY perpendicular bisector of AC in step 3. Step 4: At O, XY and AC cross. We obtain OC=AO=3 cm.

Step:2 Steps of Construction:

(i) Draw a line segment $A C=5.8 \mathrm{~cm}$.

(ii) Draw its perpendicular bisector intersecting$A C$ at $O$.

(iii) From $\mathrm{O}$, cut off $\mathrm{OD}=\mathrm{OB}=2.9 \mathrm{~cm}$ $=\left(\frac{1}{2} \mathrm{BD}\right)$.

(iv) Join$A B, B C, C D$ and $D A$

$A B C D$ is the required square.

Step:3(1) Draw a line segment $A B=5.8 \mathrm{~cm}$.

(2) At A, draw $\angle X A B=90^{\circ}$ and at $\mathrm{B}$, draw $\angle Y B A=90^{\circ}$ (since each angle in a square is of $90^{\circ}$ )

(3) Taking $A$as centre, draw an arc of radius$5.8 \mathrm{~cm}$that intersects$X A$ at $D$.

(4) Taking B as centre, draw an arc of radius$5.8 \mathrm{~cm}$ that intersects$Y B$ at $C$

(5) Join $D$ and $C$.

The required square $A B C D$ is thus drawn.

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