NCERT solutions for class 7 maths chapter 10 Practical Geometry

 

Steps for construction,

1. Draw a line AB.

2. Take any point Q on AB and a point P outside AB and join PQ.

3. With Q as the center and any radius, draw an arc to cut AB at E and PQ at F.

4. With P as the center and the same radius, draw an arc IJ to cut QP at G.

5. Place the pointed tip of the compass at E and adjust the opening so that the pencil tip is at F.

6. With the same opening as in step 5 and with G as the center, draw an arc cutting the arc IJ at H.

7. Now, join PH to draw a line CD.

 

 

Steps for construction,

1. Draw a line L.

2. Take any point P on line L.

3. At point P, draw a perpendicular line N.

4. Place the pointed tip of the compass at P and adjust the compass up to the length of 4 cm. Draw an arc to cut this perpendicular at point X.

5. At point X, again draw a perpendicular line M.

 

 

Steps for construction,

1. Draw a line L.

2. Take any point Q on L and a point P outside L and join PQ.

3. Make sure that angles at point P and point Q are equal, i.e. ∠Q = ∠P

4. At point P, extend the line to get line M which is parallel to L.

5. Then take any point R on line M.

6. At point R, draw an angle such that ∠P = ∠R

7. At point R, extend the line which intersects line L at S and draw a line RS.

 

Steps of construction

1. Draw a line segment YZ = 5 cm.

2. With Z as a centre and radius 6 cm, draw an arc.

3. With Y as a centre and radius 4.5 cm, draw another arc, cutting the previous arc at X.

4. Join XY and XZ.

Then, ΔXYZ is the required triangle.

 

Steps of construction

1. Draw a line segment AB = 5.5 cm.

2. With A as a centre and radius 5.5 cm, draw an arc.

3. With B as a centre and radius 5.5 cm, draw another arc, cutting the previous arc at C.

4. Join CA and CB.

Then, ΔABC is the required equilateral triangle.

 

Steps of construction:

1. Draw a line segment QR = 3.5 cm.

2. With Q as a centre and radius 4 cm, draw an arc.

3. With R as a centre and radius 4 cm, draw another arc, cutting the previous arc at P.

4. Join PQ and PR.

Then, ΔPQR is the required isosceles triangle.

 

1. Draw a line segment BC = 6 cm.

2. With B as a centre and radius 2.5 cm, draw an arc.

3. With C as a centre and radius 6.5 cm, draw another arc, cutting the previous arc at A.

4. Join AB and AC.

Then, ΔABC is the required triangle.

5. When we measure the angle B of the triangle with a protractor, then the angle is equal to ∠B = 90o

 

 

Steps of construction:

1. Draw a line segment DF = 3 cm.

2. At point D, draw a ray DX to make an angle of 90o, i.e., ∠XDF = 90o

3. Along DX, set off DE = 5cm

4. Join EF.

Then, ΔEDF is the required right-angled triangle.

 

 

Steps of construction:

1. Draw a line segment DF = 3 cm.

2. At point D, draw a ray DX to make an angle of 90o, i.e., ∠XDF = 90o

3. Along DX, set off DE = 5cm

4. Join EF.

Then, ΔEDF is the required right-angled triangle.

 

 

Steps of construction:

1. Draw a line segment AB = 6.5 cm.

2. At point A, draw a ray AX to make an angle of 110o, i.e., ∠XAB = 110o.

3. Along AX, set off AC = 6.5cm.

4. Join CB.

Then, ΔABC is the required isosceles triangle.

 

 

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Steps of construction:

1. Draw a line segment BC = 7.5 cm.

2. At point C, draw a ray CX to make an angle of 60o, i.e., ∠XCB = 60o.

3. Along CX, set off AC = 5cm.

4. Join AB.

Then, ΔABC is the required triangle.

 

Steps of construction:

1. Draw a line segment AB = 5.8 cm.

2. At point A, draw a ray P to making an angle of 60o i.e. ∠PAB = 60o.

3. At point B, draw a ray Q to making an angle of 30o i.e. ∠QBA = 30o.

4. Now the two rays AP and BQ intersect at the point C.

Then, ΔABC is the required triangle.

 

 

We know that the sum of the angles of a triangle is 180o.

∴ ∠PQR + ∠QRP + ∠RPQ = 180o

= 105o+ 40o+ ∠RPQ = 180o

= 145o + ∠RPQ = 180o

= ∠RPQ = 180o– 1450

= ∠RPQ = 35o

Hence, the measures of ∠RPQ is 35o.

Steps of construction:

1. Draw a line segment PQ = 5 cm.

2. At point P, draw a ray L to making an angle of 105o i.e. ∠LPQ = 35o.

3. At point Q, draw a ray M to making an angle of 40o i.e. ∠MQP = 105o.

4. Now the two rays PL and QM intersect at the point R.

Then, ΔPQR is the required triangle.

 

From the question it is given that,

EF = 7.2 cm

∠E = 110o

∠F = 80o

Now we have to check whether it is possible to construct ΔDEF from the given values.

We know that the sum of the angles of a triangle is 180o.

Then,

∠D + ∠E + ∠F = 180o

∠D + 110o+ 80o= 180o

∠D + 190o = 180o

∠D = 180o– 1900

∠D = -10o

We may observe that the sum of two angles is 190o is greater than 180o. So, it is not possible to construct a triangle.

 

 

Steps of construction

1. Draw a line segment QR = 8 cm.

2. At point Q, draw a ray QY to make an angle of 90o, i.e., ∠YQR = 90o.

3. With R as a centre and radius of 10 cm, draw an arc that cuts the ray QY at P.

4. Join PR.

Then, ΔPQR is the required right-angled triangle.

 

Let us consider ΔABC is a right-angled triangle at ∠B = 90o

Then,

AC is hypotenuse = 6 cm … [given in the question]

BC = 4 cm

Now, we have to construct the right-angled triangle by the above values.

Steps of construction

1. Draw a line segment BC = 4 cm.

2. At point B, draw a ray BX to make an angle of 90o, i.e., ∠XBC = 90o.

3. With C as a centre and radius of 6 cm, draw an arc that cuts the ray BX at A.

4. Join AC.

Then, ΔABC is the required right-angled triangle.

 

 

Steps of construction

1. Draw a line segment BC = 6 cm.

2. At point C, draw a ray CX to make an angle of 90o, i.e., ∠XCB = 90o.

3. With C as a centre and radius of 6 cm, draw an arc that cuts the ray CX at A.

4. Join AB.

Then, ΔABC is the required right-angled triangle.