NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry, are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1

Question 1.

Draw a line, say, AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

Solution:

Steps of Construction

1. Draw a line AB.
2. Take a point C outside it.
3. Take any point D on AB.
4. Join C to D.
   
   
5. with D as centre and a convenient radius, draw an arc cutting AB at F and CD at E.
6. Now with C as centre and the same radius as in step 5, draw an arc GH cutting CD at I.
7. Place the pointed tip of the compasses at F and adjust the opening so that the pencil tip is at E.
8. With the same opening as in step 7 and with I as centre, draw an arc cutting the arc GH at J.
9. Now join CJ to draw a line ‘KL’. Then KL is the required line.

Question 2.

Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.

Solution:

Steps of Construction

1. Draw a line l.
2. Take any point A on line l.
   
   
3. Construct an angle of 90° at point A of line l and draw a line AL perpendicular to line l.
4. Mark a point X on AL such that AX = 4 cm.
5. At X construct an angle of 90° and draw a line XC perpendicular to line AL.
6. Then line XC (line m) is the required line through X such that m || l.

Question 3.

Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?

Solution:

Steps of Construction

1. Draw a line l and take a point P not on it.
2. Take any point Q on l.
3. Join Q to P.
   
   
4. Draw a line m parallel to the line l, as shown in figure. Then line m || line l.
5. Join P to any point Q on l.
6. Choose any point Ronm.
7. Join R to Q.
8. Through R, draw a line n parallel to the line PQ.
9. Let the line n meet the line l at S.
10. Then, the shape enclosed by the two sets of parallel lines is a parallelogram.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

Question 1.

Construct A XYZ in which XY = 4.5 cm, YZ = 5 cm and, ZX = 6 cm.

Solution:

Steps of Construction

1. Draw a line segment YZ of length 5 cm.
   
   
2. With Y as centre, draw an arc of radius 4.5 cm.
3. With Z as centre, draw an arc of radius 6 cm,
4. Mark the point of intersection of arcs as X.
5. Join XY and XZ. ∆ XYZ is now ready.

Question 2.

Construct an equilateral triangle of side 5.5 cm.

Solution:

Steps of Construction

1. Draw a line segment BC of length 5.5 cm.
2. With B as centre, draw an arc of radius 5.5 cm.
   
   
3. With C as centre, draw an arc of radius 5.5 cm.
4. Mark the point of intersection of arcs as A.
5. Join AB and AC. Equilateral ∆ ABC is now ready.

Question 3.

Draw ∆ PQR with PQ = 4 cm, QR =3.5 cm and PR = 4 cm. What type of triangle is this?

Solution:

Steps of Construction

1. Draw a line segment QR of length 3.5 cm.
2. With Q as centre, draw an arc of radius 4 cm.
3. With R as centre, draw an arc of radius 4 cm.
   
   
4. Mark the point of intersection of arcs as P.
5. Join PQ and PR.

∆ PQR is now ready,

∵ PQ = PR

∴ ∆ PQR is isosceles.

Question 4.

Construct ∆ ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

Solution:

Steps of Construction

1. Draw a line segment BC of length 6 cm.
   
   
2. With B as centre, draw an arc of radius 2.5 cm.
3. With C as centre, draw an arc of radius 6.5 cm.
4. Mark the point of intersection of arcs as A.
5. Join AB and AC.

∆ ABC is now ready.

On measurement, ∠B = 90°.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 1.

Construct ADEF such that DE = 5 cm, DF 3 cm, and m ∠EDF = 90°.

Solution:

Steps of Construction

1. Draw a line segment DE of length 5 cm.
   
   
2. At D, draw DX making 90° with DE.
3. With D as centre, draw an arc of radius 3 cm. It cuts DX at the point F.
4. Join FE.
   
   ∆ DEF is now obtained.

Question 2.

Construct an isosceles triangle in which the length of each of its equal sides is 6.5 cm and the angle between them is 110°.

Solution:

Steps of Construction

1. Draw a line segment QR of length 6.5 cm.
2. At Q, draw QX making 110° with QR, using a protractor.
   
   
3. With Q as centre, draw an arc of radius 6.5 cm. It cuts QX at P.
4. Join PR
   
   ∆ PQR is now obtained.

Question 3.

Construct ∆ ABC with BC = 7.5 cm, AC = 5 cm and m ∠C = 60°.

Solution:

Steps of Construction

1. Draw a line segment BC of length 7.5 cm.
   
   
2. At C, draw CX making 60° with CB
3. With C as centre, draw an arc of radius 5 cm. It cuts CX at A.
4. Join AB. ∆ ABC is now obtained.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

Question 1.

Construct ∆ ABC, given m ∠A = 60°, m ∠B = 30° and AB = 5.8 cm.

Solution:

Steps of Construction

1. Draw AB of length 5.8 cm.
   
   
2. At A, draw a ray AP making an angle of 60° with AB.
3. At B, draw a ray BQ making an angle of 30° with BA.
4. Mark the point of intersection of two rays as C.

∆ ABC is now completed.

Question 2.

Construct ∆ PQR if PQ = 5 cm, m ∠PQR = 105° and m ∠QRP = 40°.

(Hint: Recall angle-sum property of a triangle).

Solution:

By angle-sum property of a triangle

m ∠RPQ + m ∠PQR + m ∠QRP = 180°

⇒ m ∠RPQ + 105° + 40° = 180°

⇒ m ∠RPQ + 145° = 180°

⇒ m ∠RPQ = 35°

Steps of Construction

1. Draw PQ of length 5 cm.
2. At Q, draw a ray QX making an angle of 105° with QP.
   
   
3. At P draw a ray PY making an angle of 35° with PQ.
4. Mark the point of intersection of two rays as R.

∆ PQR is now completed.

Question 3.

Examine whether you can construct ∆DEF such that EF = 7.2 cm, m ∠E = 110° and m ∠F = 80°. Justify your answer.

Solution:

m ∠E + m ∠F = 110° + 80° = 190° > 180°

This is not possible since the sum of the measures of the three angles of a triangle is 180°. As such, the sum of two angles of a triangle cannot exceed 180°.

Hence, ∆ DEF cannot be constructed.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1.

Construct the right-angled ∆ PQR where m ∠Q = 90°, QR = 8 cm and PR = 10 cm.

Solution:

Steps of Construction

1. Draw QR of length 8 cm.
   
   
2. At Q, draw QX ⊥ QR.
3. With R as centre, draw an arc of radius 10 cm.
4. Mark the meeting point of these two as P.

∆ PQR is now obtained.

Question 2.

Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Solution:

Steps of Construction

1. Draw QR of length 4 cm.
2. At Q, draw QX ⊥ QR.
   
   
3. With R as centre, draw an arc of radius 6 cm.
4. Mark the meeting point of arc and QX as P.

∆ PQR is now obtained.

Question 3.

Construct an isosceles right-angled triangle ABC where m ∠ACB = 90° and AC = 6 cm.

Solution:

Steps of Construction

1. Draw AC of length 6 cm.
2. At C, draw CX ⊥ CA.
   
   
3. With C as centre, draw an arc of radius 6 cm to intersect CX at B.
4. Join AB.

∆ ACB is now obtained.

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