*Question-1 Construct an angle of \(90^{\circ}\) at the initial point of a given ray and justify the construction. *

Solution:

Given a ray OA.

Required: To construct an angle of \(90^{\circ}\) at 0 and justify the construction.

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 \(\angle\)EOA = \(60^{\circ}\).

Draw the ray OF passing through D. Then \(\angle\)FOE =\(60^{\circ}\).

- 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 \(\angle\)FOE, i.e., \(\angle\) FOG = \(\angle\)EOG = ; \(\angle\)FOE = (\(60^{\circ}\)) = \(30^{\circ}\).

Thus, \(\angle\)GOA = \(\angle\)GOE + \(\angle\)EOA = \(30^{\circ}\) +\(60^{\circ}\) = \(90^{\circ}\).

Justification:

(i) Join BC.

Then, OC = OB = BC (By construction)

∴ \(\Delta\)COB is an equilateral triangle.

∴ \(\angle\)COB =\(60^{\circ}\).

∴ \(\angle\) EOA = \(60^{\circ}\).

(ii) Join CD.

Then, OD = OC = CD (By construction) \(\Delta\)DOC is an equilateral triangle.

∴ \(\angle\)DOC = \(60^{\circ}\).

∴ \(\angle\) FOE = \(60^{\circ}\). .

(iii) Join CG and DG.

In \(\Delta\)ODG and \(\Delta\)OCG,

OD = OC I Radii of the same arc

DG = CG I Arcs of equal radii

OG = OG l Common ∴ \(\Delta\)ODG\(\cong\) \(\Delta\) OCG ISSS Rule

∴ \(\angle\)DOG =\(\angle\)COG ICPCT

∴ \(\angle\)FOG = \(\angle\)EOG =\(\frac{1}{2}\) \(\angle\)FOE = \(\frac{1}{2}\) (\(60^{\circ}\)) = \(30^{\circ}\)

Thus, \(\angle\)GOA = \(\angle\)GOE + \(\angle\)EOA = \(30^{\circ}\) + \(60^{\circ}\) = \(90^{\circ}\).

**Question-2 **

**Construct an angle of \(45^{\circ}\) at the initial point of a given ray and justify the construction. **

Solution: Given: A ray OA. Required: To construct an angle of \(45^{\circ}\) at 0 and justify the construction. 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 \(\angle\)EOA = \(60^{\circ}\) .
- Draw the ray OF passing through D. Then \(\angle\) FOE = \(60^{\circ}\) .
- 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., \(\angle\)FOG = \(\angle\)EOG = \(\frac{1}{2}\) \(\angle\)FOE = \(\frac{1}{2}\) (\(60^{\circ}\) ) = \(30^{\circ}\) .

Thus, \(\angle\)GOA = \(\angle\)GOE + \(\angle\) EOA = \(30^{\circ}\) + \(60^{\circ}\) = \(90^{\circ}\).

- 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 \(\frac{1}{2}\)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, \(\angle\)GOJ = \(\angle\)AOJ = \(\frac{1}{2}\) \(\angle\)GOA = \(\frac{1}{2}\) (\(90^{\circ}\)) = \(45^{\circ}\).

Justification:

(i)Join BC.

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

∴ \(\angle\)COB is an equilateral triangle.

∴ \(\angle\)COB = \(60^{\circ}\).

∴ \(\angle\)EOA = \(60^{\circ}\).

(ii)Join CD.

Then, OD = OC = CD (By construction)

D DOC is an equilateral triangle.

∴ \(\angle\)DOC = \(60^{\circ}\).

∴ \(\angle\) FOE = \(60^{\circ}\).

(iii)Join CG and DG.

In \(\Delta\)ODG and \(\Delta\)OCG,

OD = OC I Radii of the same arc

DG = CG I Arcs of equal radii

OG = OG I Common

∴ \(\Delta\) ODG = \(\Delta\)OCG I SSS

Rule ∴ \(\angle\) DOG = \(\angle\)COG I CPCT

∴ \(\angle\)FOG = \(\angle\) EOG = \(\frac{1}{2}\) \(\angle\)FOE = \(\frac{1}{2}\) (\(60^{\circ}\)) = \(30^{\circ}\)

Thus, \(\angle\)GOA = \(\angle\)GOE + \(\angle\)EOA = \(30^{\circ}\) + \(60^{\circ}\) = \(90^{\circ}\).

- Join HJ and IJ.

In \(\Delta\)OIJ and \(\Delta\)OHJ,

01 = OH I Radii of the same arc

IJ = HJ I Arcs of equal radii

OJ = OJ | Common ∴ \(\Delta\)OIJ = \(\Delta\) OHJ

Rule ∴ \(\angle\)I0J = \(\angle\) HOJ (\(90^{\circ}\)) = \(45^{\circ}\) I CPCT

∴ \(\angle\)AOJ = ] \(\angle\)GOJ = \(\frac{1}{2}\) \(\angle\)GOA = \(\frac{1}{2}\)

Question-3

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

**\(30^{\circ}\)****\(22\frac{1}{2}\)****\(15^{\circ}\)**

Solution:

- \(30^{\circ}\)

Given: A ray OA

Required:To construct an angle of \(30^{\circ}\) at O.

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 \(\angle\)EOA = \(60^{\circ}\).
- Taking B and C as centres and with the radius more than \(\frac{1}{2}\)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., \(\angle\)EOD = \(\angle\)AOD =\(\frac{1}{2}\) \(\angle\)EOA = \(\frac{1}{2}\)(60°) = \(30^{\circ}\).

(ii) \(22\frac{1}{2}^{\circ}\)

Given: A ray OA.

Required: To construct an angle of \(22\frac{1}{2}^{\circ}\) 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 .

- 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 \(\angle\)EOA = \(60^{\circ}\).
- Draw the ray OF passing through D. Then \(\angle\) FOE = \(60^{\circ}\).
- Next, taking C and D as centres and with radius more than \(\frac{1}{2}\)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., \(\angle\)FOG = \(\angle\)EOG = \(\frac{1}{2}\) \(\angle\)FOE = \(\frac{1}{2}\) (\(60^{\circ}\)) = \(30^{\circ}\).

Thus, \(\angle\)ZGOA = \(\angle\)GOE + \(\angle\)EOA = \(30^{\circ}\) + \(60^{\circ}\) = \(90^{\circ}\).

- 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 \(\frac{1}{2}\)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., \(\angle\)GOJ = \(\angle\)AOJ = \(\frac{1}{2}\) \(\angle\)GOA = \(\frac{1}{2}\)( \(90^{\circ}\)) = \(45^{\circ}\).
- 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 \(\frac{1}{2}\)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., \(\angle\)JOM = \(\angle\)AOM = \(\frac{1}{2}\) \(\angle\)AOJ = \(\frac{1}{2}\)( \(45^{\circ}\) ) = \(22\frac{1}{2}^{\circ}\)

(iii) \(15^{\circ}\)

Given: A ray OA.

Required: To construct an angle of \(15^{\circ}\) 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.

- Draw the ray OE passing through C. Then \(\angle\)EOA =\(60^{\circ}\).
- Now, taking B and C as centres and with the radius more than \(\frac{1}{2}\)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., \(\angle\)EOD = \(\angle\)AOD = \(\frac{1}{2}\) \(\angle\)EOA = \(\frac{1}{2}\) (\(60^{\circ}\)) = \(30^{\circ}\).

- Now, taking B and F as centres and with the radius more than \(\frac{1}{2}\)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., \(\angle\)DOG = \(\angle\)AOG = \(\frac{1}{2}\) \(\angle\)AOD = \(\frac{1}{2}\) (\(30^{\circ}\)) = \(15^{\circ}\).

*Question-4 *

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

*\(75^{\circ}\)**\(105^{\circ}\)**\(135^{\circ}\)*

Solution:

- \(75^{\circ}\)

Given: A ray OA .

Required: To construct an angle of \(75^{\circ}\)

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 \(\angle\)EOA =\(60^{\circ}\).
- Draw the ray OF passing through D. Then \(\angle\)FOE – \(75^{\circ}\).
- Next, taking C and D as centres and with the radius more than \(\frac{1}{2}\)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., \(\angle\)FOG = \(\angle\)EOG =\(\frac{1}{2}\) \(\angle\)FOE = \(\frac{1}{2}\) (\(60^{\circ}\)) = \(30^{\circ}\).
- Next, taking C and H as centres and with the radius more than \(\frac{1}{2}\)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., \(\angle\)GOI= \(\angle\)E0I = \(\frac{1}{2}\) \(\angle\)GOE =\(\frac{1}{2}\) (\(30^{\circ}\)) = \(15^{\circ}\).

Thus, \(\angle\)IOA = \(\angle\)I0E + \(\angle\)EOA = \(15^{\circ}\) + \(60^{\circ}\) = \(75^{\circ}\).

On measuring the \(\angle\)I0A by protractor, we find that \(\angle\)I0A = \(75^{\circ}\).

Thus the construction is verified.

(ii) \(105^{\circ}\)

Given: A ray OA. Required: To construct an angle of \(105^{\circ}\)

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\(\angle\)EOA=\(60^{\circ}\).
- Draw the ray OF passing through D.

Then \(\angle\)FOE =\(60^{\circ}\).

- Next, taking C and D as centres and with the radius more than \(\frac{1}{2}\)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., \(\angle\)FOG = \(\angle\)EOG = ; \(\frac{1}{2}\) \(\angle\)FOE =\(\frac{1}{2}\) (\(60^{\circ}\)) = \(30^{\circ}\)

Thus, \(\angle\)GOA = \(\angle\)GOE + \(\angle\)EOA = \(30^{\circ}\) + \(60^{\circ}\) = \(90^{\circ}\).

- Next, taking H and D as centres and with the radius more than \(\frac{1}{2}\)HD, draw arcs to intersect each other, say at I.
- Draw the ray 0I. This ray 0I is the bisector of the angle \(\angle\)FOG, i.e., \(\angle\)FOI =\(\angle\)GOI = \(\frac{1}{2}\) \(\angle\)FOG = \(\frac{1}{2}\) (\(30^{\circ}\)) = \(15^{\circ}\).

Thus, \(\angle\)I0A = \(\angle\)I0G + \(\angle\)GOA = \(15^{\circ}\) + \(90^{\circ}\) = \(105^{\circ}\).

On measuring the \(\angle\)I0A by protractor, we find that \(\angle\)FOA = \(105^{\circ}\).

Thus the construction is verified.

(iii) \(135^{\circ}\)

Given: A ray OA .

Required: To construct an angle of \(135^{\circ}\)

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 \(\angle\)EOA = \(60^{\circ}\).

- Draw the ray OF passing through D.

Then \(\angle\)FOE =\(60^{\circ}\).

- Next, taking C and D as centres and with the radius more than \(\frac{1}{2}\) 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., \(\angle\)FOG = \(\angle\)EOG = \(\frac{1}{2}\) \(\angle\) FOE = \(\frac{1}{2}\) (\(60^{\circ}\)) = \(30^{\circ}\).

∴ \(\angle\)B’OH = \(90^{\circ}\).

- Next, taking B’ and H as centres and with the radius more than . \(\frac{1}{2}\) 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., \(\angle\)B’0I = \(\angle\)G0I=\(\frac{1}{2}\) \(\angle\)B’OG = \(\frac{1}{2}\) (\(90^{\circ}\)) = \(45^{\circ}\).

Thus, \(\angle\)IOA = \(\angle\)I0G + \(\angle\)GOA = \(45^{\circ}\) + \(90^{\circ}\) = \(135^{\circ}\).

On measuring the \(\angle\)I0A by protractor, we find that \(\angle\)IOA = \(135^{\circ}\).

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

∴ \(\Delta\) ABC is an equilateral triangle.

∴ 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

\(F=\left ( \frac{9}{5} \right )C+32\)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,

\(F=\left ( \frac{9}{5} \right )C+32\)\(\Rightarrow\) \(x=\left ( \frac{9}{5} \right )x+32\)

\(\Rightarrow\) \(\left ( \frac{9}{5} \right )x-x=-32\)

\(\Rightarrow\) \(x=-\frac{32X5}{4}\)

\(\Rightarrow\) \(\left ( \frac{4}{5} \right )x=-32\)

\(\Rightarrow\) \(x=\frac{-32X5}{4}\)

\(\Rightarrow\) x=-40

∴ Numerical value of required temperature=-40

Nice!

plese entertain us with some easier justifcation

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