AP Board 7th Class Maths Solutions Chapter 10 Construction of Triangles Review Exercise

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Review Exercise

Question 1.
Draw the angles 70° and 110° by using a protractor,
(i) ∠AOB = 70°
Answer:
∠AOB = 70°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\)
2. At ‘O’ erect a ray OB such that ∠AOB = 70°

(ii) ∠AOB = 110°
Answer:
∠AOB = 110°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\).
2. At ‘O’ erect a ray \(\overrightarrow{\mathrm{OB}}\) such that ∠AOB = 110°

Question 2.
Construct the angles 60° and 120° by using ruler and compass.
(i) ∠AOB = 60°
Answer:
∠AOB = 60°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\).
2. Draw an arc with any radius (convenient) taking ‘O’ as a centre and cutting \(\overrightarrow{\mathrm{OA}}\) at X.
3. Draw an arc with the same radius – taking X as a centre and meeting the previous arc at Y.
4. Join OY and produce it to B.
5. ∠AOB is formed.
6. ∠AOB = 60°

(ii) ∠ADI = 120°
Answer:
∠ADI = 120°

1. Draw a ray \(\overrightarrow{\mathrm{DA}}\).
2. Draw an arc with any radius taking ‘D’ as centre intersecting \(\overrightarrow{\mathrm{DA}}\) at X.
3. Now draw two successive arcs from X intersecting the previous arc at Y and Z.
4. Join D and Z and produce DZ to I.
5. Now the angle formed ∠ADI is equal to 120°.

Question 3.
Draw a line segment PQ = 4.5 cm and construct its perpendicular bisector by using ruler and compass.
Answer:

∠PMY = ∠QMY = 90°
So, PQ ⊥ XY

1. Draw a line segment PQ of length 4.5 cm.
2. Draw two arcs with centre P and radius > \(\frac{1}{2}\)PQ on both sides of PQ.
3. Repeat step 2 with centre ‘Q’ intersecting the previous arc at X and Y.
4. Join X and Y and produce it on either sides to form \(\overrightarrow{\mathrm{XY}}\).
5. XY is the perpendicular bisector of PQ.

Question 4.
Construct ∠DEF = 60° and its angular bisector by using ruler and compass.
Answer:

∠DEF = 60°
∠DEK = ∠FEK = \(\frac{\angle \mathrm{DEF}}{2}=\frac{60^{\circ}}{2}\) = 30°

1. Construct ∠DEF = 60°
2. Draw an arc with a convenient radius taking E as centre intersecting ED at X and EF at Y.
3. Draw two arcs with the same radius taking X and Y as centres intersecting at Z.
4. Join E, Z and produce it to K.
5. ∠DEK = ∠KEF = 30°.

Question 5.
Construct an angle of 90° without using a protractor.
Answer:
∠SRI = 90°

1. Draw a ray \(\overrightarrow{R S}\).
2. Draw an arc of convenient radius taking R as a centre and intersecting \(\overrightarrow{R S}\) at X.
3. With the same radius draw two successive arcs from X intersecting the previous arc at Y & Z.
4. Draw bisector to \(\widehat{\mathrm{YZ}}\) intersecting at I.
5. Join R, I.