**Download NCERT Solutions for Class 7 Maths Chapter 10 – Practical Geometry**

### NCERT Solutions for Class 7 Maths Chapter 10 Exercise 10.1

**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:**

**Step 1.**Draw a line AB and take a point C outside it.**Step 2.**Take any point D on AB and join CD.**Step 3.**With D as centre and a convenient radius, draw an arc cutting AB at E and CD at F.**Step 4.**With C as centre and the same radius, draw an arc GH cutting CD at I**Step 5.**Place the pointed tip of the compasses at E and adjust the opening so that the pencil tip is at F.**Step 6.**With the same opening and I as centre, draw an arc cutting GH at J.**Step 7.**Draw a line passing through C and J.

The line so formed is parallel to AB.

**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**

- Draw a line l and take a point A on it.
- At point A, draw a perpendicular p to l.
- Take a point X on the perpendicular such that AX = 4 cm
- At point X, draw a perpendicular m.

The perpendicular to the line p is the line m which is parallel to the line l.

**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**

- Draw a line (l) and take a point P, not on l.
- Take any point Q on l and join PQ.
- With Q as centre and a convenient radius, draw an arc intersecting the line l at A and PQ at B.
- With the same radius and P as centre, draw an arc CD, intersecting PQ at E.
- With the pointed tip of the compasses at A, adjust the opening such that the tip of the pencil touches the point B.
- With the same opening and E as centre, draw an arc intersecting the arc CD at F.
- Make a line m passing through P and F.
- Take a point R on m and draw an arc GH of radius PE intersecting the line m at I
- With I as centre and FE as radius, draw an arc intersecting the arc GH at J and draw a line passing through R and J.

The figure enclosed by the two pairs of parallel lines is a parallelogram PQRS.

### NCERT Solutions for Class 7 Maths Chapter 10 Exercise 10.2

**1. Construct ∆XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.**

**Solution:**

**Steps of construction:**

**Step 1.**Draw a line XY = 4.5 cm.**Step 2.**With X as centre and 6 cm (= ZX) as radius, draw an arc on one side of XY.**Step 3.**With Y as centre and 5 cm (= YZ) as radius, draw an arc intersecting the previous arc at Z.**Step 4.**Join XZ and YZ.

XYZ is the required triangle.

**2. Construct an equilateral triangle of side 5.5 cm.**

**Solution:**

**Steps of construction:**

**Step 1.**Draw a line AB of length 5.5 cm.**Step 2.**With A and B as centres and 5.5 cm as radius, draw arcs intersecting each other at C.**Step 3.**Join AC and BC.

ABC is the required triangle.

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

**Solution:**

**Steps of construction:**

**Step 1.**A line QR = 3.5 cm.**Step 2.**With Q and R as centres and 4 cm (= PQ = PR) as radius, draw arcs intersecting each other at P.**Step 3.**Join PQ and PR.

PQR is the isosceles triangle because its two sides (PQ and PR) are equal.

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

**Solution:**

**Steps of construction:**

**Step 1.**Draw a line BC = 6 cm.**Step 2.**With B as centre and 2.5 cm (= AB) as radius, draw an arc on one side of BC.**Step 3.**With C as centre and 6.5 cm (= AC) as radius, draw an arc intersecting the previous arc at A.**Step 4.**Join AC and AB.

The measure of ∠B is 90°.

ABC is the required triangle.

### NCERT Solutions for Class 7 Maths Chapter 10 Exercise 10.3

**1. Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.**

**Solution:**

**Steps of construction:**

**Step 1.**Draw a line DE = 5 cm.**Step 2.**At D, draw DX, making angle 90° with DE.**Step 3.**With D as centre and 3 cm (= DF) as radius, draw and arc intersecting DX at F.**Step 4.**Join EF

DEF is the required triangle.

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

**Solution:**

**Steps of construction**

- Draw a line BC = 6.5 cm.
- At B, draw BX, making angle 110° with BC.
- With B as centre and 6.5 cm as radius, make an arc intersecting BX at A.
- Join AC.

ABC is the required triangle.

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

**Solution:**

**Steps of construction **

**Step 1.**Draw a line AC = 5 cm.**Step 2.**At C, draw CX, making angle 60° with AC.**Step 3.**With C as centre and 7.5 cm (= BC) as radius, draw and arc intersecting CX at B.**Step 4.**Join BA.

ABC is the required triangle.

### NCERT Solutions for Class 7 Maths Chapter 10 Exercise 10.4

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

**Solution:**

**Steps of construction:**

**Step 1.**Draw a line AB = 5.8 cm.**Step 2.**At A, draw AX, making angle 60° with AB.**Step 3.**At B, draw BY, making angle 30° with AB.**Step 4.**Mark the point of intersection of AX and BY as C.

ABC is the required triangle.

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

**Solution:**

Angle sum of a triangle is 180°

∠PQR + ∠QRP + ∠RPQ = 180°

105° + 40° + ∠RPQ = 180°

145° + ∠RPQ = 180°

∠RPQ = 35°

** ****Steps of construction **

**Step 1.**Draw a line PQ = 5 cm.**Step 2.**At P, draw PX, making angle 35° with PQ.**Step 3.**At Q, draw QY, making angle 105° with PQ.**Step 4.**Mark the point of intersection of PX and QY as R.

PQR is the required triangle.

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

**Solution:**

∠E + ∠F = 110° + 80° = 190°

It is known that the angle sum of a triangle is always 180°, but in this case, it is not.

Hence, ∆DEF cannot be constructed.

### NCERT Solutions for Class 7 Maths Chapter 10 Exercise 10.5

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

**Solution:**

**Steps of construction:**

**Step 1.**Draw a line QR = 8 cm.**Step 2.**At Q, draw QX making angle 90° with QR.**Step 3.**With R as centre and 10 cm (= PR) as radius, draw and arc intersecting QX at P.**Step 4.**Join PR

PQR is the required triangle.

**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**

**Step 1 –**Draw a line BC = 4 cm.**Step 2 –**At B, draw BX, making angle 90° with BC.**Step 3 –**With C as centre and 6 cm as radius, make an arc intersecting BX at A.**Step 4 –**Join AC.

ABC is the required triangle.

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

**Solution:**

**Steps of construction **

**Step 1.**Draw a line AC = 6 cm.**Step 2.**At C, draw CX, making angle 90° with AC.**Step 3.**With C as centre and 6 cm (= CB) as radius, draw and arc intersecting CX at B.**Step 4.**Join BA.

ABC is the required triangle.

**Download NCERT Solutions for Class 7 Maths Chapter 10 – Practical Geometry**

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