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 :
**

- Take any point P on the line AB.
- Take any point C outside AB and join CP.
- With P as centre, draw an arc cutting AB and PC at X and Y respectively.
- 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.
- With centre Q and radius equal to XY, draw an arc cutting the arc drawn in step 4 at R.
- 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 :
**

- Draw a line l and take any point P on it.
- With P as centre and any radius, draw an arc to intersect line l at A and B.
- With A as centre and radius greater than PA, draw in arc.
- With centre B and the same radius, as in step 2, draw another arc to intersect the arc drawn in step 2 at C.
- Join PC and produce it to Q. Then PQ ⊥ l
- With P as centre and radius equal to 4 cm, draw an arc to intersect PQ at X such than PX = 4 cm.
- At X, make ∠RXP = ∠BPX.
- 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 :
**

- Draw a line l and take any point P outside it.
- Take any point Q on line l.
- Join PQ.
- With as centre, draw an arc cutting l and PQ at C and D respectively.
- 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.
- With centre E and radius equal to CD,’ draw an arc cutting the arc of step 5 at F.
- Join PF and produce it in both directions to obtain the required line m.
- Take any point R on m.
- 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 :
**

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

**Question 2.**

Construct an equilateral triangle of side 5.5 cm.

**Solution:**

**Steps of Construction :
**

- Draw a line segment BC = 5.5 cm.
- With centre B and radius = 5.5 cm, draw an arc.
- With centre C and radius = 5.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
- 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 :
**

- Draw a line segment QR = 3.5 cm.
- With centre Q and radius = 4 cm, draw an arc.
- With R as centre and radius = 4 cm, draw another arc intersecting the arc drawn in step 2 at P.
- 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 :**

- Draw a line segment BC = 6 c.m

- With centre B and radius = 2.5 cm, draw an arc.
- With centre C and radius =6.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
- 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 :
**

- Draw a line segmerit DE = 5cm.
- Draw ∠EDX = 90°.
- With centre D and radius = 3 cm, draw an are to intersect DX at F.
- 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 :
**

- Draw a line segment BC = 6.5 cm.
- Draw ∠CBX = 110°.
- With B as centre and radius = 6.5 cm, draw an arc intersecting BX at A.
- 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 :
**

- Draw a line segment BC =7.5 cm.
- Draw ∠BCX = 60°.
- With C as centre and radius = 5 cm, draw an arc intersecting CX at A.
- 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 :
**

- Draw a line segment QR = 8 cm.
- Draw ∠XQR = 90°.
- With R as centre and radius =10 cm, draw an arc to intersect ray QX at P.
- 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 :
**

- Draw a line segment QR = 4 cm.
- Draw ∠XQR = 90°.
- With R as centre and radius equal to hypotenuse 6 cm, draw an arc to intersect ray QX at P.
- 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 :
**

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

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