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.

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.1, Ex 10.2, Ex 10.3, Ex 10.4, Ex 10.5.
Number of Questions Solved	16
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry

Chapter 10 Practical Geometry Exercise 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. Take any point P on the line AB.
2. Take any point C outside AB and join CP.
3. With P as centre, draw an arc cutting AB and PC at X and Y respectively.
4. With centre C and the same radius as in step 3, draw an arc on the opposite side of PC to cut PC at Q.
5. With centre Q and radius equal to XY, draw an arc cutting the arc drawn in step 4 at R.
6. Join CP and produce it in both directions to obtain the required line.

Question 2.
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 and take any point P on it.
2. With P as centre and any radius, draw an arc to intersect line l at A and B.
3. With A as centre and radius greater than PA, draw in arc.
4. With centre B and the same radius, as in step 2, draw another arc to intersect the arc drawn in step 2 at C.
5. Join PC and produce it to Q. Then PQ ⊥ l
6. With P as centre and radius equal to 4 cm, draw an arc to intersect PQ at X such than PX = 4 cm.
7. At X, make ∠RXP = ∠BPX.
8. Join XR to obtain the required line m.

Validity : Since ∠BXP = ∠BPX and these are alternate angles, therefore, .XR|| l, i.e., m || l and contain X such that PX = 4 cm and ∠XPB = 90°.

Question 3.
Let l be a line and P be a point 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 any point P outside it.
2. Take any point Q on line l.
3. Join PQ.
4. With as centre, draw an arc cutting l and PQ at C and D respectively.
5. With centre P and the same radius as in step 4, draw an arc on the opposite side of PQ to cut PQ at E.
6. With centre E and radius equal to CD,’ draw an arc cutting the arc of step 5 at F.
7. Join PF and produce it in both directions to obtain the required line m.
8. Take any point R on m.
9. Through P, draw a line PS || PQ by following steps already explained. The shape of the figure endorsed by these lines is a parallelogram RPQS.

Chapter 10 Practical Geometry Exercise 10.2

Question 1.
Construct ∆XYZ in which XY = 45 cm, YZ = 5 cm and ZX = 6 cm.
Solution:
Steps of Construction :

1. Draw a line segment YZ = 5 cm.
2. With Y centre and draw an arc radius cm.
3. With Z as centre and draw another arc intersecting the arc radius = 6 cm, at X.
4. Join XY and XZ to obtain the required triangle.

Question 2.
Construct an equilateral triangle of side 5.5 cm.
Solution:
Steps of Construction :

1. Draw a line segment BC = 5.5 cm.
2. With centre B and radius = 5.5 cm, draw an arc.
3. With centre C and radius = 5.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
4. Join AB and AC to obtain the required triangle.

Question 3.
Draw ∆PQR with PQ = 4, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:
Steps of Construction :

1. Draw a line segment QR = 3.5 cm.
2. With centre Q and radius = 4 cm, draw an arc.
3. With R as centre and radius = 4 cm, draw another arc intersecting the arc drawn in step 2 at P.
4. Join PQ and PR to obtain the required triangle. ∆PQR is an isosceles triangle.

Question 4.
Construct AABC 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 = 6 c.m
   
2. With centre B and radius = 2.5 cm, draw an arc.
3. With centre C and radius =6.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
4. Join AB and AC to obtain the required triangle.
   On measuring, we find that ∠B = 90°.

Chapter 10 Practical Geometry Exercise 10.3

Question 1.
Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m ∠EDF = 90°
Solution:
Steps of Constipation :

1. Draw a line segmerit DE = 5cm.
2. Draw ∠EDX = 90°.
3. With centre D and radius = 3 cm, draw an are to intersect DX at F.
4. Join EF to obtain the required triangle DBF.

Question 2.
Construct an isosceles triangle in which the lengths of each of its equal Sides is 6.5 cm find the angle between them is 110°
Solution:
Steps of Construction :

1. Draw a line segment BC = 6.5 cm.
2. Draw ∠CBX = 110°.
3. With B as centre and radius = 6.5 cm, draw an arc intersecting BX at A.
4. Join AC to obtain the required ∆ABC.

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 =7.5 cm.
2. Draw ∠BCX = 60°.
3. With C as centre and radius = 5 cm, draw an arc intersecting CX at A.
4. Join AB to obtain the required ∆ABC.

Chapter 10 Practical Geometry Exercise 10.4

Question 1.
Construct ∆ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.
Solution:
Steps of Construction :

1. Draw a line segment AB = 5.8 cm.
2. Draw ∠BAX = 60°.
3. Draw ∠ABY, with Y on the same side of AB
such that ∠ABY = 30°.
Let AX and BY interest at C.
Then, ∆ABC is the required triangle.

Question 2.
Construct ∆PQR if PQ = 5cm, m ∠PQR = 105° and m ∠QRP = 40°.
Solution:
Here, we are given the side PQ, ∠Q and ∠R. But to draw the triangle, we require ∠P

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:
Since m∠E +m∠F = 110° + 80° = 190°, so the ∆DEF cannot be drawn as the’sum of all the angles of a triangle is 180°.

Chapter 10 Practical Geometry Exercise 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 a line segment QR = 8 cm.
2. Draw ∠XQR = 90°.
3. With R as centre and radius =10 cm, draw an arc to intersect ray QX at P.
4. Join RP to obtain the required ∆PQR.

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 a line segment QR = 4 cm.
2. Draw ∠XQR = 90°.
3. With R as centre and radius equal to hypotenuse 6 cm, draw an arc to intersect ray QX at P.
4. Join RP to obtain the required ∆PQR.

Question 3.
Construct an isosceles right-angled triangle ABC, where m ∠ACB = 90° and AC = 6 cm.
Solution:
Steps of Construction :

1. Draw a line segment CB = 6 cm (∵ CB = AC = 6 cm)
2. Draw ∠BCX = 90°.
3. With C as centre and radius = 6 cm, draw an arc to intersect ray CX at A.
4. Join BA to obtain the required triangle ABC.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry, drop a comment below and we will get back to you at the earliest.