Chapter-11, Constructions - NCERT Exampler

Explanation: 

The correct response is B. With a ruler and compass, we may create multiples of 15 degrees, or angles, such as 90 degrees, 60 degrees, 40 °, 22.5 degrees, 30 degrees, etc.

Hence, it is impossible to create an angle of 40.

Explanation: 

It follows that

 ABC in a triangle

 BC = 6 cm

∠B = 45°

 The fact that

 If the total of the two sides is less than or equal to the third side, a triangle cannot be formed.

 AC = AB + BC

 One way to spell it is as

 BC, AC, and AB

 6 < AC - AB

 hence, we do

AC - AB > 6

Consequently, when the difference between AB and AC is equal to 6.9 cm, construction is not feasible.

Explanation: 

 (D) 2.8 cm

Provided that BC is 3 cm and C is 60,

We are aware that a triangle can be constructed if the sum of its two sides is greater than its third side, or

AB+BC>AC

⇒BC>AC−AB

⇒3>AC−AB

Construction of A B C is therefore feasible under the stated criteria if AC - AB = 2.8 cm.

Explanation: 

The right response is True.

To create a 52.5-degree angle

Create an angle of 90 degrees first, then one of 120 degrees, and finally plot the angle bisector of 120 degrees and 90 degrees to obtain an angle of 105 degrees (90 degrees plus 15 degrees). Then, divide this angle in half to obtain a 52.5 angle.

Explanation: 

The right response is B False.

Having said that,

42.5 Degree=1/2x85 Degree

As it is only possible to construct an angle that is a multiple of 3, such as 30 Degrees, 60 Degrees 90 Degrees and so on, an angle of 85 Degrees cannot be constructed using a ruler and compass.

Hence, it is impossible to form an angle of 42.5. Thus, the claim is untrue.

Explanation: 

The right response is A False.

If the sum of the triangle's two sides is more than the sum of its third side, we know that a triangle can be formed.

BC + AC here equals AB = 5 cm.

Thus, ABC cannot be built.

Explanation: 

It follows that

ABC in a triangle

BC = 6 cm

∠C = 30°

The fact that

If the sum of the first two sides is more than the third side, a triangle can be formed.

A+B+C exceeds A.

AB + 6 > AC

6 > AC - AB

AC - AB < 6

The assertion is accurate as a result.

Explanation: 

The right response is A False.

Here, C = 90 and B = 105.

Likewise, AB + BC + CA = 10 cm.

We are aware that a triangle's total number of angles is 180.

Here, A + B + C equals 180.

The statement "B + C = 105 + 80 = 185" is untrue.

Hence, ABC cannot be formed under the given circumstances.

Explanation: 

 The right answer is A) the Sum of a triangle's angles Equals 180

⟹∠BAC=180−(60+45)=75∘

Now let's verify the assertion.

I Draw a segmented line at XY=11 cm.

ii) From point X, create an angle of B=60 and mark L on this line.

iii) From point Y, create an angle of C=45 and mark M on this line.

iv) Split LXY and MYX in half, then have the halves meet at point A.

v) Create a perpendicular bisector of the line AX, and have it intersect XY at B.

Making a perpendicular bisector of AY and allowing it to cross XY at point C.

vii) Combine AB and AC, then multiply AB+BC+AC.

if it measures 11 cm.

The statement is accurate, and the necessary triangle is ABC.

Explanation: 

A:  Using a protractor, create BXA = 110°

Construction Steps

With the aid of a protractor, draw a 110° angle and divide it in half. Calculate each angle.

1. Using X as the center, create an arc of any radius that intersects the rays XA and XB, or E and D.

2. Create arcs that meet at F using D and E as the centers, and a radius more than half of DE.

3. Build the ray XF.

4. The necessary bisector of the angle BXA is XF.

5. By calculating every angle

AXC = 1/2, BXA = 110/2, and BXC =

BXC equals AXC at 55°.

Hence, each angle is 55 degrees.

Explanation: 

Construction Steps

Create a line segment AB = 4 cm in step one.

2. Create an arc with a radius of more than half of AB and a centre of 4 cm, intersecting AB at point E.

3. Create an arc with the same radius as the last one, with E as the centre, and intersect it at point G.

4. Create arcs that intersect each other at point H using G and F as the centres.

5. Sign up for AH.

6. In point A, AX is perpendicular to AB.

7. Create BY at point B perpendicular to AB in the same manner.

The angle formed by two parallel lines will either be 0° or 180°.

∠XAB = 90° (since XA is perpendicular to AB)

 (as XA is perpendicular to AB)∠YBA = 90°

∠YBA = 90° (as YB is perpendicular to AB)

 (as YB is perpendicular to AB)

The fact that 90° + 90° = 180° for XAB + YBA.

The two lines are parallel because the total internal angle on the same side of the transversal is 180°.

XA and YS are hence parallel.

Create a 4 cm long piece of the line AB. Draw a line between A and B that is perpendicular to AB. Are these lines parallel

The lines are parallel as a result.

Explanation: 

Construction steps include:

1. Sketch an OA ray.

2. Using a protractor, create the following: BOA = 80

3. Draw an arc to connect rays OA and OB at positions P and Q, using O as the center and any reasonable radius.

4. Take a radius that is greater than 50% of the OP radius.

Create an arc with this radius from points P and Q that will meet at X. The bisector of AOB is obtained by connecting O and X.

If ray OC is defined as the bisector of BOA, then COA = 21 and BOA = 21 80 = 40.

5. Draw an arc to cut the extended arc PQ at R with Q as the center and a radius equal to PQ. Join OR and generate it to create the ray OD,therefore DOA = 2 and BOA = 2 times 80, or 160.

6. Bisect DOB as you did in step 4, and let OE be its bisector. This results in EOA = EOB + BOA + DOB + BOA = 21 (80) + 80 = 40 + 80 = 120.

Explanation: 

Creating the triangle is the first step.

Plot a line at AB=4.8 cm.

Now, to draw an arc, use center A and a radius of 3.6 cm. Draw an arc that crosses our previous arc at C, taking into account the radius of 3.0 cm and center B.

By combining CA and CB, we obtain the necessary triangle ABC.

The angle in front of the smallest side, A, is the smallest angle, making it the angle to be divided into two. Thus, we'll divide A in half.

Draw an arc with point A as the center, cutting AC and AB at F and E, respectively.

Draw arcs that intersect at G with E and F as the centers.

Draw a line through D, which is a point on BC, and which follows line AG.

Third step: measuring

With the use of a protractor, we can now measure each other's angles with ease.

We have A=40°, B=50°, and C=90°.

Explanation: 

 Given: ABC, where BC = 5 cm, B = 60 cm, and AC + AB = 7.5 cm.

How to build the triangle in steps ABC

I Sketch the base BC = 5 cm.

(ii) Enter "XBC=60" at point B. (iii) On the ray BE, cut a line segment BD that is AB + AC = 7.5 cm.

(iv) Adopt DC.

(v) Create a DCY=BDC (vi) intersection with CY at A. Thus, ABC is the necessary triangle.

Explanation: 

We are aware that a square's angles are all right angles or 90 degrees.

The methods below should be used to create a square with 3 cm sides.

I Sketch a 3 cm long line segment AB.

(ii) Create a 90-degree angle at the line segment's points A and B, then plot the parallel lines AX and BY there.

(iii) Cut AD and BC on AX and BY, respectively, to a length of 3 cm.

(iv) Draw a 90-degree angle at either point C or D, and connect the two using a line segment CD that is 3 cm long.

ABCD is the necessary square, with a 3-centimeter side.

You can skip step four. simply combine the points DC and if every

Explanation: 

We are aware that a rectangle's opposite sides are equal and parallel, and that each of its angles is correct (i.e., 90°).

Use the steps below to create a rectangle with adjacent sides that are 5 cm and 3.5 cm long.

I Sketch a 5 cm long line segment BC.

(ii) Create a 90-degree angle at locations B and C of the line segment BC, and then plot the parallel lines BX and CY there.

(iii) On BX and CY, cut AB and CD, each measuring 3.5 cm in length.

(iv) Draw a 90-degree angle at either point A or D, then connect the two places with a section of line AD that is 5 cm long.

Hence, ABCD is the necessary rectangle with adjacent sides measuring 5 cm in length and

Explanation: 

We are aware that all sides are equal in a typical rhombus.

Step 1: Draw a 3.4 cm-long line segment AB.

Step 2: Create a 45-degree angle BAX at point A.

Mark ray AX at AD = 3.4 cm in step three.

Step 4: Draw an arc above B with 3.4 cm as the radius and B as the center.

Step 5: Draw another arc that intersects the one from Step 4 with D as the center and a radius of 3.4 cm.

Step 6: Affix a C to this location.

Connect B to C and D to C in step seven.

Hence, the necessary rhombus is ABCD.

Explanation: 

ABC should form a triangle.

Assuming that the perimeter is 10.4 cm (AB + BC + CA), the two angles are 45 and 120 cm.

Let B be 45 and C be 120.

the ABC's construction process (i) Draw a segment of the line XY that is equal to the perimeter, that is, AB + BC + CA = 10.4 cm

Make LXY=B=45 and MYX=C=120 (ii).

(iii) Split LXY and MYX in half, then let their bisectors cross at A.

Draw the PQ and RS perpendicular bisectors of AX and AY, respectively.

(v) Let PQ and RS cross XY at B and C, respectively. Accompany AB and AC. ABC is the necessary triangle as a result.

Make ∠LXY=∠B=45 and ∠MYX=C=120 (ii).

(iii) Split ∠LXY and ∠MYX in half, allowing their bisectors to converge at point A.

Draw the PQ and RS perpendicular bisectors of AX and AY, respectively.

(v) Let PQ and RS cross XY at B and C, respectively. Accompany AB and AC. ABC is the necessary triangle as a result.

Justification

B is on the perpendicular bisector PQ of AX because of this.

∴ XB = AB

Given that C is located on the RS of AY's perpendicular bisector.

∴ CY = AC

AB + BC + CA = XB + BC + CY = XY as a result.

And once more, "BAX=AXB" [∵ΔAXB, AB=XB"] (i)

Moreover, ∠ABC =∠BAX + ∠AXB (where [∵∠ABC is an exterior angle of ΔAXB)

= ∠AXB + ∠AXB (from Eq. I )

= 2 ∠AXB =∠ LXY (where ∵AX is a bisector of ∠LXB)

Furthermore, ∠CAY=∠AYC [in ∵AYC, AC]

∵∠ACB = ∠CAY + ∠AYC [∵ACB is an exterior angle ofΔ AYC] = CAY = CAY = 2 CAY = MYX [AY is an external angle of Δ AYC]

Explanation: 

Drawing a ray OX and a line segment QR=3 cm is the first step.

2. Create an angle at point Q of 45 degrees, ∠YQR. Join the points R and S after subtracting QS=2 cm from the line segment QY.

Step 3: Trace a perpendicular bisector of RS to meet QY at P.

Therefore, the necessary graph is displayed below:

Justify your position.

The base QR and ∠PQR are drawn.

According to the above structure, P is located on the perpendicular bisector of SR.

[From Construction] PS=PR.......1

QP=QS+PS

=> QS=QP-PS

 =>QS-QP-PR (from 1) follows from the construction.

Consequently, the construction is appropriate.

Explanation: 

Let ABC be the right triangle given.

Once BC = 3.5 cm, B = 90°, and the sum of the other side and hypotenuse, AB + AC = 5.5 cm, are known.

Use the steps below to create ΔABC.

1. Draw the BC = 3.5 cm base.

2. At point B of the base BC, create an angle XBC = 90°.

3. Remove the line segment BD=AB+AC=5.5 cm from BX and link it to CD.

4. Create the perpendicular bisector of CD that joins point A on the line BD.

5. Consequently, ΔABC is the necessary right-angled triangle.

Explanation: 

We are aware that an equilateral triangle has 60 degrees for each angle and equal sides on each side.

An equilateral triangle has a given altitude of 3.2 cm.

steps for creating the ΔABC.

(i) PQ line is drawn.

(ii) Draw a ray from a point D on PQ to PQ.

(iii) Cut the 3.2 cm long line segment AD on the ray DE.

(iv) Construct angles of the form ∠DAB=30 at A where line AB intersects DP.

To make DC = BD, cut the line segment DC on DQ.

(vi) Join AC, making ΔABC the necessary triangle.

.

JustificationHere,

∠A = ∠BAD + ∠CAD = 30 + 30 = 60

Furthermore, ∠ADBC ADB=90In ABD

Angle sum property: ∠BAD+∠ADB+∠DBA=180∘

30∘+90∘+∠DBA=180∘

By construction, [∠BAD=30]

∠DBA=60∘Similarly,

∠DCA=60∘Thus,

Because A=B=C=60, ABC is an equilateral triangle.

Explanation: 

We are aware that rhombuses have equal-sized sides and diagonals that are perpendicular to one another.

Use the methods below to build a rhombus with diagonals of 4 cm and 6 cm.

Draw the diagonal with AC equal to 4 cm.

(ii) Draw arcs that intersect each other on either side of the line segment AC with the center's A and C and a radius greater than 1/2AC.

(iii) Let both arcs cross at P and Q, then link P and Q.

(iv) Let PQ and AC converge at O. PQ is hence AC's perpendicular bisector.

(v) Trim the OP and OQ ends to 3 cm, then mark those locations as B and D.

(vi) Join AB, BC, and

Explanation

D and B are on the perpendicular bisector of AC because of this. Since every point on a line segment's perpendicular bisector is equally distant from its endpoints, DA = DC and BA = BC. I

And now, DOC=90

OD = OB = 3 cm as well.

As a result, BD's perpendicular bisector is AC.

CD = CB

Equations I and (ii) show that AB = BC = CD = DA.

ABCD is a rhombus as a result (ii).