Question

Draw a line 7.1cm long and draw it's perpendicular bisector. Using this bisector as a side, draw a square of side 3.55cm. (please explain how)​

Answer 1

Step-by-step explanation:

Let given right triangle be ABC.

Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps

1.Draw the base BC = 3.5 cm

2.Make an angle XBC = 90° at the point B of base BC.

3.Cut the line segment BD equal to AB + AC i.e., 5.5 cm from the ray XB.

4.Join DC and make an ∠DCY equal to ∠BDC.

5.Let Y intersect BX at A.

Thus, ΔABC is the required triangle.

Justification

Base BC and ∠B are drawn as given.

In ΔACD, ∠ACD = ∠ADC [by construction]

AD = AC …(i)

[sides opposite to equal angles are equal] Now, AB = BD – AD = BD – AC [from Eq. (i)]

=> BD = AB + AC

Thus, our construction is justified.

Answer 2

Step-by-step explanation:

Let given right triangle be ABC.

Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps

Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps1.Draw the base BC = 3.5 cm

2.Make an angle XBC = 90° at the point B of base BC.

3.Cut the line segment BD equal to AB + AC i.e., 5.5 cm from the ray XB.

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4.Join DC and make an ∠DCY equal to ∠BDC.

4.Join DC and make an ∠DCY equal to ∠BDC.5.Let Y intersect BX at A.

4.Join DC and make an ∠DCY equal to ∠BDC.5.Let Y intersect BX at A.Thus, ΔABC is the required triangle.

Justification

Base BC and ∠B are drawn as given.

In ΔACD, ∠ACD = ∠ADC [by construction]

AD = AC …(i)

[sides opposite to equal angles are equal] Now, AB = BD – AD = BD – AC [from Eq. (i)]

=> BD = AB + AC

Thus, our construction is justified.