WBBSE Solutions For Class 7 Maths Geometry Chapter 2 Drawing Angles With Compass Exercise 2 Solved Problems

Class 7 Math Solution WBBSE Geometry Chapter 2 Drawing Angles With Compass Exercise 2 Solved Problems

Question 1. To construct an angle of 90°

Solution:

Method Of Construction

 

 

1. BC is any straight line. With B as a centre and with any radius I draw an arc with a compass so that it cuts the line segment BC at D.
2. With centre D and with the same radius, I draw an arc which cuts the previous arc at E.
3. Again with centre E and with the same radius, I draw another arc, to cut the first arc at F.
4. With E and F as centres and with the same radius two more arcs are drawn to cut each other at A. Points A and B are joined.

Thus ∠ABC = 90°

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Question 2. To construct an angle of 45°

Solution:

Method Of Construction

 

1. At first ∠ABC is drawn where ∠ABC = 90°

2. Then bisect angle ∠ABC with compass, so that ∠ABD = ∠DBC = 45°

Wbbse Class 7 Maths Solutions

Question 3. To construct the angles of 60°, 30° and 15°

Solution:

Method Of Construction

1. BC is any straight line with B as centre and with any radius, an arc is drawn which cut BC at P. P as a centre and with the same radius another arc is drawn to cut the previous arc at Q. I joined B and B produced to A. Thus ∠ABC 60°

 

 

2. I bisect the ∠ABC by BD I get 30° i.e., ∠DBC = 30°

 

 

3.  Again bisecting ∠DBC by BE I get ∠EBC= 15°

 

 

Wbbse Class 7 Maths Solutions

Question 4. To construct the angles of 120° and 75°

Solution:

Method Of Construction

 

 

From the adjacent

∠CQR = 120°

∠PQR = 90° and ∠DQR = 60°

So∠PQD = ∠PQR – ∠DQR

∠PQD = 90° – 60°

∠PQD = 30°

∠PQD = 30°

Bisecting angle ∠PQD by QE we get ∠EQD, where∠EQD = 15°

So ∠EQR = ∠EQD +∠DQR

∠EQR= 15° + 60°

∠EQR= 75°

∠EQR = 75°

Drawing Angles With Compass

Drawing Angles With Compass Exercise 7

Question 1.  At point a on the line segment AB, let’s construct an angle of go with the help of scale, pencil and compass, let’s construct the angle; 120° 75° and 60°= angles at the same point,
Solution :

Given

⇒ At an AB, first, we draw an angle of 60°, 90° & 120°

⇒ Here ∠ BAC = 90°

⇒ ∠ BAD = 120°

⇒ ∠ BAE = 75°

⇒ ∠ BAF = 60°

2. Using scale, pencil and compass, let’s construct the angles, 45 and 22\(\frac{1}{2}\)
Solution:

1. Take a line segment AO. Now with centre O, & with any radius draw an arc which cuts OA at P. Now with centre P, draw an arc which cuts the previous arc at Q.
2. Again with centre Q with the same radius an arc is drawn which cuts the previous arc at R. Now with centre Q and R, draw two arcs which cut each other at B. Join BO & AOB = 90° is formed.
3. Now by bisecting ∠ AOB, two angles ∠ AOC & ∠ COB are formed each equal to 45°.
4. Again by bisecting ∠ AOC, two angles ∠ AOD & ∠ COD two angles are formed, each equal to 22\(\frac{1}{2}\).

Question 3. Using scale, pencil and compass let’s draw the following angles –

1. 30°
2. 60°
3. 75°
4. 105°
5. 120°
6. 135°
7. 150°
8. 15°.

Solution:

Then, At O, draw an angle ∠ EOQ =.90°; ∠ GOQ = 60°& ∠ COQ = 120° (E).

⇒ Now by bisecting ∠ GOQ (= 60°) we get ∠ HOQ = 30s (A).

⇒  Again by bisecting ∠ HOQ (= 30°) we get ∠ KOQ = 15° (H)

⇒ Now by bisecting ∠ EOG, we get ∠ QOF = 75° (C)%

⇒ Again by bisecting ∠ COE we get ∠ DOQ =105° (D)

⇒ Now by bisecting ∠ POC (= 60°), we get Z AOQ = 150° (G)

⇒ Lastly by bisecting ∠AOC (= 30°), we get ∠ BOQ = 1 35° (F)

Class 7 Math Solution WBBSE

Question 4. Using scale, pencil and compass let’s draw an angle ∠ PQR of 60°. Let the side of RQ is produced to any point S, opposite to R. Then ∠ PQS = 120° Let’s also bisect the ∠ PQS and let’s verify using a protractor if ∠ PQS is bisected.
Solution:

Given

⇒  First ∠ PQR = 60° is drawn at Q, then RQ is produced to S, opposite to R.

⇒  ∠ PQS = 180° – 60° = 120° as, straight angle ∠RQS = 180°, Now

⇒  ∠ PQS is bisected by MQ

⇒  ∠MQP = ∠ MQS = 60°

Question 5. Using scale, pencil and compass let’s draw an angle ∠ ABC of measure 30°. Then let’s produce the side CB to a point D opposite to C. Now let’s draw a bisector BE of angle ∠ ABD. Measuring by protractor it is found ∠ DBE = 75° ∠ EBC = 105°
Solution :

Given

⇒ First ∠ ABC = 30° is drawn at B on the line segment BC.

⇒ Now CB is produced to a point D opposite to C.

⇒ Now BE is drawn as the bisector of ∠ ABD

⇒ Measuring by a protractor it is found ∠ DBE = 75° and ∠ EBC = 105°.