Constructions: Definition and Important Questions

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Very Short Answer Questions [1 Marks Question]

Ques 1: Draw an AB = 8 cm line segment. Part 13 should be drawn. Measure the 13th portion of AB's length.

Ans:

1. Draw an AB = 8 cm line segment.
2. Create a perpendicular bisector that intersects AB in M.
3. In N, draw MB's perpendicular bisector and let it intersect AB.

As a result, AN = 13 and AB = 6 cm.

Ques 2: Why can we not build an ABC if A = 60°, AB = 6 cm, and AC + BC = 5 cm, but we can build an ABC if A = 60°, AB = 6 cm, and AC – BC = 5 cm?

Ans: Triangle inequality property states that if the total of two triangle sides is greater than the third side, a triangle can be constructed. AC + BC equals 5 cm, which is less than AB (6 cm) As a result, ABC is not conceivable.

Also, if the difference between two sides of a triangle is less than the third side, a triangle can be constructed using the triangle inequality property.

AC - BC equals 5 cm, which is less than AB (6 cm)

ABC is therefore possible.

Ques 3. Create a 90° angle at the starting point of the provided ray.

Ans: Construction Procedures:

1. Draw an OA ray.
2. Draw an arc with O as the centre and any convenient radius, cutting OA at P.
3. Draw an arc with P as the centre and the same radius as the arc drawn in step 2 at Q.
4. Draw an arc with Q as the centre and the same radius as steps 2 and 3, cutting the arc formed in step 2 at R.
5. Draw two arcs, cutting each other in S, with Q and R as centres and the same radius.
6. Connect to OS and output to B. As a result, AOB is the requisite 90° angle.

Ques 4. Draw a straight angle in question 4. Divide it with a compass. Name the angles you've discovered.

Ans: Steps of Construction:

1. Make any straight angle (for example, AOC).
2. Split AOC in half and connect BO.
3. The needed bisector of the straight angle AOC is AOB.

Ques 5. Draw any reflex angle you like. Divide it with a compass. Name the angles you've just discovered.

Ans: 

Construction Procedures:

1. Choose any reflex angle for AOB.
2. Draw an arc with O as the centre and any convenient radius, cutting OA in P and OB in Q.
3. Draw two arcs of radius a little more than half the radius of P and Q, and let them connect in C. Join the OC. As a result, OC is the necessary bisector. AOC and COB are the angles obtained in this way.

Ques 6. Construct a right triangle with a 12 cm base and an 18 cm hypotenuse and opposite side.

Ans: To construct the required triangle, follow the steps listed below.

Step 1: Draw a 12-cm line segment AB. Draw a ray AX with AB at a 90° angle.

Step II: Cut an 18-cm line segment AD from ray AX (the total of the other two sides is 18).

Step III: Join DB and produce an angle DBY equal to ADB by joining them together.

Step IV: Bring BY and AX together at C. BC, join AC.

The needed triangle is ABC.

ΔABC is the required triangle.

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Short Answer Questions [2 Marks Question]

Ques 1: Make a triangle with sides in the ratio 2: 3: 4 and a circumference of 18 cm.

Ans: 

Steps of Construction :

1. Draw a line segment AB =18 cm.
2. At A, construct an acute angle ∠BAX (< 90°).
3. Mark 9 points on AX, such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆
   = A₆A₇ = A₇A₈ = A₈A₉.
4. Join A₉B.
5. From A₂ and A₅, draw A₂M || A₅N || A₉B, intersecting AB in M and N respectively.
6. With M as centre and radius AM, draw an arc.
7. With N as centre and radius NB, draw another arc intersecting the previous arc at L.
8. Join LM and LN. Thus, ∆LMN is the required triangle.

Construction Procedures:

1. Draw an AB =18 cm line segment.
2. Create an acute angle BAX (90°) at A.
3. On AX, make 9 marks so that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆
   = A₆A₇ = A₇A₈ = A₈A₉..
4. Register with A₉B.
5. Draw A₂M || A₅N ||A₉B from A₂ and A₅, intersecting AB in M and N, respectively.
6. Draw an arc with M as the centre and AM as the radius.
7. Draw another circle with N as the centre and NB as the radius, intersecting the previous arc at L.
8. Make friends with LM and LN. As a result, the needed triangle is LMN.

Ques 2. Make an ABC with BC = 8 cm, B = 45 degrees, and AB – AC = 3.1 cm.

Ans: 

Construction Procedures:

1. Draw any BC = 8 cm line segment.
2. Create an angle CBX = 45° at B.
3. Cut off BD = 3.1 cm from BX.
4. Join the DC.
5. In A, draw DC's perpendicular bisector 'p' and have it cross BX.
6. Join the AC. As a result, the needed triangle is ABC.

Ques 3: Make an ABC with BC = 3.2 cm, B = 45 degrees, and AC – AB = 2.1 cm.

Ans: 

Construction Procedures:

1. Draw a BC = 3.2 cm line segment.
2. Create an angle CBX = 45° at B and produce it at point X'.
3. Join CD and cut off BD = 2.1 cm.
4. In A, draw the perpendicular bisector of CD and intersect it with X'BX.
5. Join the AC. As a result, the needed triangle is ABC.

Ques 4: Draw a QR = 5 cm line segment. Make perpendiculars to it at points Q and R. QX and RY are their respective names. Are they both connected?

Ans: 

Construction Procedures:

1. Draw a QR = 5 cm line segment.
2. Make a 90° angle with Q as the centre and let this line through Q be QX.
3. Create a 90° angle with R as the centre and let this line through R be RY. Yes, the parallel lines QX and RY are perpendicular.

Ques 5: Create an isosceles triangle with two equal sides of 6 cm each and a 5 cm base. Show that the perpendicular bisector of its base passes through the opposite vertex by drawing it.

Ans: 

Steps of Construction:

1. Draw a line segment with the length AB = 5 cm.
2. Draw two arcs of radius 6 cm with A and B as the centres, and let them intersect in C.
3. Combine AC and BC to form ABC.
4. Draw two arcs with radius somewhat less than half of AB with A and B as the centres. Allow them to intersect in P and Q. To get through C, join PQ and produce.

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Long Answer Questions [3 Marks Question]

Ques . Make a triangle ABC with BC = 4.7 cm, AB + AC = 8.2 cm, and C = 60 degrees.

Ans: BC = 4.7 cm, AB + AC = 8.2 cm, and C = 60° are given in ABC.

Construction of ABC is required.

Construction Procedures:

1. Make a BC of 4.7 cm.
2. Draw
3. Cut off CD = 8.2 cm from ray CX.
4. Join the BD.
5. At point A, draw the perpendicular bisector of BD and CD.
6. Combine AB and ABC to get the needed triangle ABC.

Justification:

Because A is on BD's perpendicular bisector, AB = AD.

CD now equals 8.2 cm, AC + AD equals 8.2 cm, and AC + AB equals 8.2 cm.

Ques 2: Make XYZ with a perimeter of 14 cm, one side of 5 cm, and an X angle of 45°.

Ans: YZ + XZ = 14 – 5 = 9 cm and X = 45°, with a perimeter of 14 cm and one side of 5 cm.

Construction Procedures:

1. Draw a 5 cm XY line segment.
2. Using a compass and ruler, construct a YXA = 45°.
3. Cut off ray XB = 9 cm from ray XA.
4. Join the BY.
5. In Z, draw a perpendicular bisector of BY that intersects XB.
6. Register with ZY. As a result, the needed triangle is XYZ.

Ques 3. Build a triangle with a perimeter of 10 cm and base angles of 60° and 45°.

Ans: 

Given: AB + BC + CA = 10 cm, B = 60°, and C = 45° in ABC.

Requirements: To put together ABC.

Construction Procedures:

1. Make a DE of 10 cm.
2. Construct EDP= 5 of 60°= 30° at D, and DEQ = 1 of 45^o = 22° at E.
3. Arrange for DP and EQ to meet at A.
4. Draw a perpendicular bisector from AD to B.
5. Draw a perpendicular bisector from AE to C.
6. Assemble AB and AC. As a result, the needed triangle is ABC.

Ques 4: Make an ABC triangle with BC = 8 cm, B = 45 degrees, and AB AC = 3.5 cm.

Ans: To draw the required triangle, follow the procedures outlined below.

Step I: Draw the line segment BC = 8 cm and draw a 45° angle at point B, XBC.

Step II: On ray BX, cut the line segment BD = 3.5 cm (equivalent to AB AC).

Step III: Join DC and draw PQ, DC's perpendicular bisector.

Step IV: At point A, let it intersect BX. Come join AC. The needed triangle is ABC.

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Very Long Answer Questions [5 Marks Question]

Ques 1: Make an equilateral triangle with a 6 cm long altitude.

Ans: 

Steps of Construction :

1. Draw a line PQ and pick any point S on it as a starting point.
2. On PQ, construct the perpendicular SR.
3. Cut a line segment SA = 6 cm from SR.
4. Construct SAB = 30° and SAC = 30° at the beginning point A of the line segment AS.
5. In B and C, the angles SAB and SAC have arms AB and AC that meet PQ. The needed equilateral triangle with an altitude of 6 cm is thus ABC.

Ques 2: Make a rhombus with diagonals of 8 cm and 6 cm in length. Measure the length of each rhombus side.

Ans: 

Construction Procedures:

1. Draw a line segment with the length PR = 8 cm.
2. Draw the line segment PR's perpendicular bisector XY. Let O be the place where PR and XY connect, and O be the 8 cm midpoint of PR.
3. Cut a line segment OS = 3 cm from OX and a line segment OQ = 3 cm from OY.
4. PQRS is the needed rhombus after joining PS, SR, RQ, and QP.
5. Measure the length of PQ, QR, RS, and SP segments; each is 5 cm long.

Ques 3. Construct an XYZ triangle with Y = 30°, Z = 90°, and XY + YZ + ZX = 11 cm.

Ans: To construct the required triangle, follow the steps listed below.

Step 1: Draw an 11-cm line segment AB.

(Considering that XY + YZ + ZX = 11 cm)

Step II: Create a 30° angle, PAB, at point A and a 90° angle, QBA, at point B.

Step 3: Divide PAB and QBA in half. Allow these bisectors to intersect at point X.

Step IV: Create a perpendicular bisector between AX's ST and BX's UV.

Step V: Let ST and UV intersect at Y and Z, respectively.

XY, XZ, join us.

ΔXYZ is the required triangle.

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