Chapter-10 PRACTICAL GEOMETRY

Solution-

Explanation

Construction step-

1. Produce a segment of line and name it as AB.

2. Mark any point on this line and name it as Q now mark any point outside this line AB and name it as P then join the points P and Q.

3. Taking Q as the centre, produce arc of any radius, intersect AB at a point name it as E and intersect PQ at a point and name it as F.

4. Taking point P let it be the centre, produce an arc GH of equal radius and intersect QP at G.

5. Place your compass at the point marked as E and expand the aperture in such a way that the pencil point is at the marked point F.

6. Repeat the process with taking marked point G as center, produce another arc that intersects the arc GH at point H.

7. Finally add the PH pull cord CD.

Explanation

Solution-

Construction step are-

1. Produce the line named as L.

2. Consider a point on the line named L, let it be P.

3. Produce a vertical line N at point P.

4. Position the point of compass at point P, expand the compass to a segment of 4 cm and produce an arc that intersects the vertical line to point X.

5. Along X, produce a horizontal line named M which is parallel to line named L.

Solution-

Construction steps-

1. Produce the line named L.

2. Pick any point on the line named L, let it be P.

3. Produce a vertical line N at point P.

4. Position the point of compass at point P, expand the compass with the help of ruler to a length marked as 4 cm and produce an arc that intersects this vertical at a point named X.

5. At the point named P and at the point named X, produce a vertical line M.

Explanation

Solution-

Construction step are-

1. Produce a line of length YZ=5cm.

2. Consider Z as the centre and a length of compass as 6 cm, produce the marked arc.

3. Consider center Y and length of compass as 4.5 cm, produce one more arc that intersects the old arc at point named X.

4. Connect XY then connect XZ.

Hence, ΔXYZ the desired triangle is formed.

Explanation

Solution-

Construction step are-

1. Produce a line of length AB = 5.5 cm.

2. Taking point named A as center and length of compass as of 5.5 cm, produce the marked arc.

3. Produce arc taking point named B as center and length of compass as 5.5 cm at same point.

4. Connect CA then connect CB.

Hence, ΔABC the desired equilateral triangle is formed.

Solution-

Explanation

Construction step are-

1. Produce a line of length QR which is equal to 3.5 cm.

2. Taking point named Q as the centre and length of compass as 4 cm, produce the marked arc.

3. Taking point named R as the centre and length of compass as 4 cm, produce one more arc that intersects the old marked arc at point named P.

4. Add points named P & Q then add points named P & R.

Hence, with these points ΔPQR the desired triangle named isosceles is formed.

Explanation

Solution-

Construction steps-

1. Produce a line of length BC = 6 cm.

2. Taking point named B as centre, produce an arc with help of compass of 2.5 cm.

3. With C as the center and length of compass as 6.5 cm, produce one more arc that intersects the old marked arc at point named A.

4. Mark the point as A. Connect AB then connect AC.

Hence, ΔABC is the desired triangle.

Use a protractor to find the desired ∠B of the triangle, the value is equal to 90o

Explanation

Solution-

Construction step are-

1. Produce a line of length DF=3cm.

2. Draw a ray at point D to form a 90° angle, that is value of ∠XDF is = 90 o.

3. In this ray ray, cut DE equal to 5cm.

4. Point E and F are to be joined

Hence, ΔEDF the desired right angle triangle is formed.

Explanation

Solution-

Construction step are-

1. Produce a line of length AB is equal to 6.5 cm.

2. Draw a ray at point named A with an included angle of 110o. In other words, ∠XAB = 110o.

3. Along the ray, cut AC = 6.5 cm.

4. Point C and B are to be joined.

Hence, ΔABC the desired isosceles triangle is formed.

Explanation

Solution-

Construction steps-

1. Produce a line of length BC=7.5cm.

2. Produce a ray at point C with an included angle of 60°, that is ∠XCB = 60°.

3. Along the ray, mark AC = 5cm.

4. Point A and B are to be joined.

Hence, ΔABC the desired triangle is formed.

Explanation

Solution:-

Construction steps-

1. Produce a line of length AB is equal to 5.8 cm.

2. Produce a ray that makes angle of 60 o at point A. In other words, ∠PAB = 60 o.

3. Produce a ray at point B with an included angle of measure 30 o, that is ∠QBA is equal to 30 o.

4. Intersect these two lines/rays at point C.

Hence, ΔABC the desired triangle is formed.

Expalanation

Solution-

As we remember, the summation of interior angles in any triangle is equal to 180o.

Hence, ∠PQR + ∠QRP + ∠RPQ = 180o

⇨ 105o + 40o + ∠RPQ = 180° (put all the values)

⇨ 145° + ∠RPQ = 180° (transfer 145° from LHS to RHS)

⇨ ∠RPQ = 180o – 145° (solve)

⇨ ∠RPQ = 35°.

So, the value of ∠RPQ is 35o.

Construction process-

1. Produce a line of length PQ = 5 cm.

2. At P, produce ray that makes angle of measure 105o, that is ∠LPQ is equal to 35o.

3. At Q, produce ray that makes angle of measure 40o, that is ∠MQP is equal to 105o.

4. Here, these two lines/rays should cross each other at point named R.

Hence, ΔPQR the desired triangle is formed.

Explanation

Solution-

In the above question it is given that

EF = 7.2 cm, ∠E = 110°, ∠F = 80°

So, we need to check if we can construct ΔDEF from the above given data.

We remember that the summation of all the interior angles in any triangle is equal to 180°.

after that,

∠F + ∠D + ∠E = 180°

⇨ ∠D + 80° + 110° = 180° (put all the values)

⇨ ∠D + 190° = 180° (transfer 190° from LHS to RHS)

⇨ ∠D = 180° – 190° (simplify)

⇨ ∠D = –10°

We can see that the summation of the two angles is 190°, which is larger than the angle 180°. Therefore, it is impossible to create any triangle. Value of interior angle of a triangle is always positive.

Explanation

Solution-

Construction procedure-

1. Produce a line segment named QR of length = 8 cm.

2. At Q, produce ray that makes angle with measure of 90°. That is, ∠YQR = 90°.

3. Produce an arc with taking centre at R and compass length of 10 cm that intersects ray QY at point named P.

4. Add point P with Point R.

Hence, ΔPQR the desired right angle triangle is formed.

Explanation

Solution-

Assume, ∠B = 90° in the ΔABC which is a right angle triangle.

after that

Hypotenuse AC = 6 cm is ... [specified in question]

Also, its legs, BC = 4 cm

Now, we need to create a right angle triangle with help of above given data

Construction procedure-

1. Produce a line of length BC which is equal to 4 cm.

2. On point named B, produce a line/ray with an angle of measure 90°. That is, ∠XBC is equal to 90°

3. Produce arc with center C of radius 6 cm that intersects ray at A.

4. Connect point A and C.

Hence, ΔABC the desired right angle triangle is formed.

Explanation

Solution-

Construction steps

1. Produce a segment BC of length equal to 6 cm.

2. On point C produce a ray so that it makes an angle of measure 90°, that is ∠XCB is equal to 90°.

3. With center C and compass length of 6 cm, produce arc that intersects ray at point named A.

4. Connect point A and B.

Hence, ΔABC the desired right angle triangle is formed.