Construction

Option (A):

37.5° = 1/2 × 75°

75° can be constructed with the help of a ruler and a compass.

Option (B):

40° = 1/2 × 20°

This is not possible.

20° cannot be constructed with the help of a ruler and a compass.

Option (C):

22.5° = 1/2 × 45°

45° can be constructed with the help of a ruler and a compass.

Option (D):

67.5° = 1/2 × 135° = 1/2 × (90° + 45°)

Both 90° and 45° can be constructed with the help of a ruler and a compass.

Ans. (B) It is not possible to construct an angle of 40° with the help of ruler and compass.

We know that the difference of any two sides of a triangle is less than the third side.

So, (AB – AC) must be less than 6 cm.

Options (B), (C) and (D) satisfy the above condition.

When AB – AC = 6.9 cm, the condition doesn’t satisfy.

Ans. The construction of the given ΔABC is not possible when difference of AB and AC is equal to 6.9 cm.

We know that the difference of any two sides of a triangle is less than the third side.

So, (AB – AC) must be less than 3 cm.

Options (A), (B) and (C) do not satisfy the above condition.

When AB – AC = 2.8 cm, the condition satisfies.

Ans. The construction of the given ΔABC is possible when difference of AB and AC is equal to 2.8 cm.

True

Explanation:

52.5° = 1/2 × 105° = 1/2 × (90° + 15°)

We can construct both 90° and 15° with the help of ruler and compass.

∴ Angle of 52.5° can be constructed.

False

42.5° = 1/2 × 85°

We cannot construct 85° with the help of ruler and compass.

∴ Angle of 42.5° cannot be constructed.

False

Explanation:

We know that the sum of any two sides of a triangle must be greater than the third side.

Here, BC + AC = 5cm = AB which does not satisfy the above condition as the sum is equal to the third side.

True

Explanation:

We know that the difference of any two sides of a triangle is less than the third side.

Here, (AC – AB) = 4 cm < BC = 6 cm.

∴ The above condition satisfies.

False

Explanation:

Given, AB + BC + AC = 10 cm

∠B = 105° and ∠C = 90°

We know that the sum of angles of a triangle is 180°.

But ∠B + ∠C = 105° + 90° = 195° > 180° which is not possible.

True

Explanation:

Given, AB + BC + AC = 12 cm

∠B = 60° and ∠C = 45°

We know that the sum of angles of a triangle is 180°.

But ∠B + ∠C = 60° + 45° = 105° < 180° which is possible.

Step 1:

Draw ∠AOB=110° using Protractor.

Step 2:
With Center O and any Radius Draw an arc MN.

Step 3:

With M as center and radius more than half of MN draw an arc and with N as center and equal radius draw an arc and let the arcs intersect at P.

Step 4:

Join OP.

Step 1:

Draw a line and mark AB=4cm

Step 2:

With A as center draw an arc that intersects the line at P and Q.

Similarly with B as center draw an arc that intersects the line at R and S.

Step 3:

With P as center and radius more than half of PQ draw an arc and with Q as center and equal radius draw an arc and let the arcs intersect at X.

Similarly, With R as center and radius more than half of RS draw an arc and with S as center and equal radius draw an arc and let the arcs intersect at Y.

Step 4:

Join AX and BY.

Yes the lines XX’ and YY’ are parallel to each other as they both are perpendicular to the same line AB.

(i)

Step 1:

Draw ∠AOB = 80° and with any radius and A as center draw an arc.

Step 2:

With X as center and radius more than half of XY draw an arc and with Y as center and equal radius draw an arc and let the arcs intersect at C.

Step 3:

Join OC.

∠COB is equal to 40°.

Step 4:

With Y as center and radius more than half of PY draw an arc and with P as center and equal radius draw an arc and let the arcs intersect at D.

Step 5:

Join OD.

∠DOB is 60°.

Step 6:

With O as center and OX as radius extend the Arc XY to Arc XY’.

Step 7:

With Q as center and XQ as radius, draw an arc which cuts the Arc XY’ at R.

Step 8:

Join RO.

∠ROB is 120°.

Step 1:

Draw a Ray BX and with B as center and radius 4.8cm cut an arc on the line and name it as C.

Step 2:

With B as center and 3.6cm as radius draw an arc.

Step 3:

With C as center and 3cm as radius cut the previous arc at A.

Step 4:

Join AB and AC.

Our Triangle is constructed.

Now we know that the angle opposite to the smallest side is the smallest angle in a triangle.

So ∠ABC is the smallest.

Now bisecting the ∠ABC

Step 5:

With B as center draw an arc that cuts AB and BC at P and Q respectively.

Step 6:

With P as center and radius more than PQ draw an arc then with Q as center and equal radius draw another arc that cuts the previous arc at Z.

Step 7:

Join BZ.

Now by cosine rule we get to know that ∠ABC=40°

So, ∠ABY=∠YBC=20°.

Step 1:

Draw ∠YBX = 60° and BC=5cm

Step 2:

With B as center and 7.5cm as radius draw an arc that intersects BY at Z.

Step 3:

Join CZ and then with C as center draw two arc at either side of the line CZ and repeat the same with Z as center.

Name the intersecting points as P and Q.

Step 4:

Name the intersecting point of PR and BZ as A and join AC.

So ABC is the required Triangle and here we can observe that AZ=AC (as AZC is isosceles).

Step 1:
Draw a Ray AX. With A as center and radius 3cm cut an arc and name the intersecting point B.

Step 2:

With B as an center draw an arc intersecting the ray AX at P and Q.

Then with P as center and radius more than half of PQ cut an arc which intersects the previous arc at R and repeat the same for Q as center and name the intersecting point S.

Step 3:

Now with R as center draw an arc and then with S as center.

Name the intersecting point O.

Step 4:

Now B as center and 3cm as radius draw an arc that cuts the ray BO at C.

Step 5:

With C as center and radius 3cm draw an arc and repeat the same with A and name the intersecting point D.

Step 6:

Join AD and CD.

ABCD is required Square.

Step 1:
Draw a Ray AX. With A as center and radius 5cm cut an arc and name the intersecting point B.

Step 2:

With B as an center draw an arc intersecting the ray AX at P and Q.

Then with P as center and radius more than half of PQ cut an arc which intersects the previous arc at R and repeat the same for Q as center and name the intersecting point S.

Step 3:

Now with R as center draw an arc and then with S as center.

Name the intersecting point O.

Step 4:

Now B as center and 3.5cm as radius draw an arc that cuts the ray BO at C.

Step 5:

With C as center and radius 5cm draw an arc and repeat the same with A as center and radius 3.5cm and name the intersecting point D.

Step 6:

Join AD and CD.

ABCD is the required Rectangle.

Step 1:
Draw ∠YBW = 45°.

Thus can be done by bisecting a 90°.

Step 2:

With B as center and a radius of 3.4cm and cut an arc at C.

Step 3:

With A as center and radius 3.4cm draw an arc and then C as a center and radius 3.4cm draw another arc and name the intersecting point as D. Then join AD and CD.

ABCD is required Rhombus.

Step 1:

Draw a line XY=10.4 i.e. the perimeter.

Step 2:

Construct an angle equal to ∠B=45° and another angle equal to ∠C=120°

Step 3:

Bisect these angles and name the intersecting point as A.

Step 4:

Construct perpendicular bisectors of AX and AY and name the PQ and RS respectively.

Step 5:

Name the intersecting point of PQ and XY as B and RS the intersecting point of RS and XY as C.

Join AB and AC.

ABC is the required triangle.

Justification

As B is on line PQ which is the perpendicular bisector of AX,

AB+BC+CA = XB+BC+CY=XY

Then,∠BAX=∠AXB (as in triangle AXB, AB is equal to XB)

As ∠ABC is the external angle of triangle AXB

Then,∠ABC=∠BAX + ∠AXB(exterior angle sum property)

∠ABC=∠AXB+∠AXB

∠ABC=2∠AXB=45° or ∠B.

Similarly,

∠ACB=2∠CAY=120° or ∠C.

Thus the construction is justified.

Step 1:

Draw a line QR=3cm.

Step 2:
Construct an angle of 45° at Q.

Step 3:

With Q as center and 2cm as radius cut an arc on QX at S.

Step 4:
Join SR and bisect it.

Name the intersecting point of QX and the bisector P.

Step 5:

Join PR.

PQR is the required Triangle.

Justification

QR and ∠PQR =45°.

P lies on the perpendicular bisector of SR.

PS = PR

Given,

QS = PQ – PS

= PQ -PR (as PS=PR)

Thus The construction is justified.

Step 1:
Draw ∠YBX = 90°.

Step 2:

With B as center and radius 3.5cm and 5.5cm respectively cut arcs intersecting BX at C and BY at Z.

Step 3:

Join ZC and construct its perpendicular bisector intersecting BY at A.

Step 4:
Join AC.

ABC is the required triangle.

Justification

Here ∠ABC=90° , BC=3.5cm by construction.

As A is the point of intersection of the perpendicular bisector and CZ,

So, AZ=AC which means AB+AZ=5.5cm

And at the same time AB+AC=5.5cm

Hence the triangle Construction is justified.

Step 1:
Draw a line XY.

Step 2:

Mark a point D on XY and construct DZ perpendicular to XY.

Step 3:

With center D and radius 3.2cm cut an arc on DZ at A.

Step 4:

With A as center, draw ∠UAD=30 intersecting XY at B and ∠VAD=30° intersecting XY at C.

ABC is the required triangle.

Justification

By construction we can say,

∠ZDY=90°

∠BAC = ∠BAD + ∠CAD

∠BAC = 30°+30° =60°.

In ΔABD,

∠ABD + ∠BAD + ∠DBA = 180° (By Angle Sum Property)

30° + 90° + ∠DBA = 180°

∠DBA = 60°

Similary, ∠DCA = 60°

Thus, ∠A = ∠B = ∠C = 60°

Thus our construction is justified.

Step 1:

Draw a line AX and with A as center and radius 6cm cut an arc on AX at C.

Step 2:

Construct perpendicular bisector of AC and name it as PQ.

Name the intersecting point O.

Step 3:

With O as the center and radius 2cm draw arcs to cut OP and OQ at D and B respectively.

Step 4:

Join AB,BC,CD and DA.

Justification

AC=6cm (By Construction)

PQ is perpendicular to AC (By Construction)

As AC=6cm is one the diagonals So BD is the other diagonal with length 4cm.

As it is a rhombus we know that the diagonals bisect each other so

So, OD=OB=2cm.

Thus our construction is justified.