*Question-1 Construct an angle of 90∘ at the initial point of a given ray and justify the construction. *

Solution:

Given a ray OA.

Required: To construct an angle of

Steps of Construction:

- Taking 0 as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
- Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

- Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
- Draw the ray OE passing through C. Then
∠ EOA =60∘ .

Draw the ray OF passing through D. Then

- Next, taking C and D as centres and with the radius more than CD, draw arcs to intersect each other, say at G.
- Draw the ray 0G. This ray OG is the bisector of the angle
∠ FOE, i.e.,∠ FOG =∠ EOG = ;∠ FOE = (60∘ ) =30∘ .

Thus,

Justification:

(i) Join BC.

Then, OC = OB = BC (By construction)

∴

∴

∴

(ii) Join CD.

Then, OD = OC = CD (By construction)

∴

∴

(iii) Join CG and DG.

In

OD = OC I Radii of the same arc

DG = CG I Arcs of equal radii

OG = OG l Common ∴

∴

∴

Thus,

**Question-2 **

**Construct an angle of 45∘ at the initial point of a given ray and justify the construction. **

Solution: Given: A ray OA. Required: To construct an angle of

- Taking 0 as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
- Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
- Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

- Draw the ray OE passing through C. Then
∠ EOA =60∘ . - Draw the ray OF passing through D. Then
∠ FOE =60∘ . - Next, taking C and D as centres and with radius more than 1CD, draw arcs to intersect each other, say at G.
- Draw the ray OG. This ray OG is the bisector of the angle FOE, i.e.,
∠ FOG =∠ EOG =12 ∠ FOE =12 (60∘ ) =30∘ .

Thus,

- Now, taking 0 as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I respectively.
- Next, taking H and I as centres and with the radius more than
12 HI, draw arcs to intersect each other, say at J. - Draw the ray OJ. This ray OJ is the required bisector of the angle GOA. Thus,
∠ GOJ =∠ AOJ =12 ∠ GOA =12 (90∘ ) =45∘ .

Justification:

(i)Join BC.

Then, OC = OB = BC triangle. (By construction)

∴

∴

∴

(ii)Join CD.

Then, OD = OC = CD (By construction)

D DOC is an equilateral triangle.

∴

∴

(iii)Join CG and DG.

In

OD = OC I Radii of the same arc

DG = CG I Arcs of equal radii

OG = OG I Common

∴

Rule ∴

∴

Thus,

- Join HJ and IJ.

In

01 = OH I Radii of the same arc

IJ = HJ I Arcs of equal radii

OJ = OJ | Common ∴

Rule ∴

∴

Question-3

**Construct the angles of the following measurement:**

30∘ 2212 15∘

Solution:

30∘

Given: A ray OA

Required:To construct an angle of

Steps of Construction:

- Taking 0 as centre and some radius , draw an arc of a circle, which intersects OA, say at a point B.
- Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
- Draw the ray OE passing through C. Then
∠ EOA =60∘ . - Taking B and C as centres and with the radius more than
12 BC, draw arcs to intersect each other, say at D. - Draw the ray OD. This ray OD is the bisector of the angle EOA, i.e.,
∠ EOD =∠ AOD =12 ∠ EOA =12 (60°) =30∘ .

(ii)

Given: A ray OA.

Required: To construct an angle of

Steps of Construction:

- Taking 0 as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
- Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C .

- Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
- Draw the ray OE passing through C. Then
∠ EOA =60∘ . - Draw the ray OF passing through D. Then
∠ FOE =60∘ . - Next, taking C and D as centres and with radius more than
12 CD, draw arcs to intersect each other, say at G. - Draw the ray OG. This ray OG is the bisector of the angle FOE, i.e.,
∠ FOG =∠ EOG =12 ∠ FOE =12 (60∘ ) =30∘ .

Thus,

- Now, taking 0 as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I respectively.
- Next, taking H and I as centres and with the radius more than
12 HI, draw arcs to intersect each other, say at J. - Draw the ray OJ. This ray OJ is the bisector of the angle GOA. i.e.,
∠ GOJ =∠ AOJ =12 ∠ GOA =12 (90∘ ) =45∘ . - Now, taking 0 as centre and any radius, draw an arc to intersect the rays OA and OJ, say at K and L respectively.
- Next, taking K and Las centres and with the radius more than
12 KL, draw arcs to intersect each other, say at M. - Draw the ray OM. This ray OM is the bisector of the angle AOJ, i.e.,
∠ JOM =∠ AOM =12 ∠ AOJ =12 (45∘ ) =2212∘

(iii)

Given: A ray OA.

Required: To construct an angle of

Steps of construction:

- Taking 0 as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.

- Draw the ray OE passing through C. Then
∠ EOA =60∘ . - Now, taking B and C as centres and with the radius more than
12 BC, draw arcs to intersect each other, say at D. - Draw the ray OD intersecting the arc drawn in step 1 at F. This ray OD is the bisector of the angle EOA,

i.e.,

- Now, taking B and F as centres and with the radius more than
12 BF, draw arcs to intersect each other, say at G. - Draw the ray OG. This ray OG is the bisector of the angle AOD, i.e.,
∠ DOG =∠ AOG =12 ∠ AOD =12 (30∘ ) =15∘ .

*Question-4 *

*Construct the following angles and verify by measuring them by a protractor: *

75∘ 105∘ 135∘

Solution:

75∘

Given: A ray OA .

Required: To construct an angle of

at 0.

Steps of Construction:

- Taking 0 as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.

- Join the ray OE passing through C. Then
∠ EOA =60∘ . - Draw the ray OF passing through D. Then
∠ FOE –75∘ . - Next, taking C and D as centres and with the radius more than
12 CD, draw arcs to intersect each other, say at G. - Draw the ray OG intersecting the arc of step 1 at H. This ray OG is the bisector of the angle FOE, i.e.,
∠ FOG =∠ EOG =12 ∠ FOE =12 (60∘ ) =30∘ . - Next, taking C and H as centres and with the radius more than
12 CH, draw arcs to intersect each other, say at I. - Draw the ray 01. This ray 01 is the bisector of the angle GOE, i.e.,
∠ GOI=∠ E0I =12 ∠ GOE =12 (30∘ ) =15∘ .

Thus,

On measuring the

Thus the construction is verified.

(ii)

Given: A ray OA. Required: To construct an angle of

at 0.

Steps of Construction:

- Taking 0 as centre and some radius , draw an arc of a circle, which intersects OA, say at a point B.

- Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C .
- Draw the ray OE passing through C. Then
∠ EOA=60∘ . - Draw the ray OF passing through D.

Then

- Next, taking C and D as centres and with the radius more than
12 CD, draw arcs to intersect each other, say at G. - Draw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE, i.e.,
∠ FOG =∠ EOG = ;12 ∠ FOE =12 (60∘ ) =30∘

Thus,

- Next, taking H and D as centres and with the radius more than
12 HD, draw arcs to intersect each other, say at I. - Draw the ray 0I. This ray 0I is the bisector of the angle
∠ FOG, i.e.,∠ FOI =∠ GOI =12 ∠ FOG =12 (30∘ ) =15∘ .

Thus,

On measuring the

Thus the construction is verified.

(iii)

Given: A ray OA .

Required: To construct an angle of

at 0.

Steps of Construction:

- Produce AO to A’ to form ray OA’.

- Taking 0 as centre and some radius, draw an arc of a circle, which intersects OA at a point B and OA at a point B’.
- Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.
- Draw the ray OE passing through C.

Then

- Draw the ray OF passing through D.

Then

- Next, taking C and D as centres and with the radius more than
12 CD, draw arcs to intersect each other, say at G. - Draw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE, i.e.,
∠ FOG =∠ EOG =12 ∠ FOE =12 (60∘ ) =30∘ .

Thus,

∴

- Next, taking B’ and H as centres and with the radius more than .
12 B’H, draw arcs to intersect each other, say at I. - Draw the ray 01. This ray 01 is the bisector of the angle B’OG, i.e.,
∠ B’0I =∠ G0I=12 ∠ B’OG =12 (90∘ ) =45∘ .

Thus,

On measuring the

Thus the construction is verified.

*Question-5 *

*Construct an equilateral triangle, given its side and justify the Construction. *

Solution:

Given: Side (say 6 cm) of an equilateral triangle .

Required: To construct the equilateral triangle and justify the construction.

Steps of Construction:

- Take a ray AX with initial point A.
- Taking A as centre and radius (= 6 cm), draw an arc of a circle, which intersects AX, say at a point B.

- Join AC and BC.

Justification:

AB = BC I By construction

AB = AC I By construction

∴ AB = BC = CA

∴

∴ The construction is justified.

In countries like USA and Canada, temperature is measured in Fahrenheit, Whereas in countries like India. It is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius

Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it?

Solution:-

Let the temperature be x numerically .Then,

∴ Numerical value of required temperature=-40