Important Questions Class 10 Maths Chapter 11

Important Questions Class 10 Mathematics Chapter 11 – Constructions

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Chapter 11 Construction is a part of the Geometry section. You will learn about the construction of different shapes in this chapter. The important topics covered in the NCERT textbook include

• Division of a Line Segment
  • Construction 
• Construction of Tangents to a Circle
  • Construction

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Important Questions Class 10 Mathematics Chapter 11 – With Solutions

Below are a few Class 10 Mathematics Chapter 11 Important Questions and their solutions:

Question 1. To divide the line segment AB in the ratio 5:7, the ray AX is drawn such that BAX is the acute angle, and at the equal distances, the points are marked on the ray AX so that the minimum number of these points is

(A) 8 

(B) 10 

(C) 11 

(D) 12

Answer 1: (D) 12

As per the question,

The line segment AB is in the ratio of 5:7

Thus, A: B = 5:7

Draw the ray AX making the acute angle ∠BAX,

Mark A+B points at the equal distance.

Thus, we have A=5 as well as B=7.

Therefore, the minimum number of these points = A+B = 5+7 =12

Question 2. To divide the line segment AB in the ratio 4:7, the ray AX is drawn so that ∠BAX is the acute angle and the points A1, A2, A3,…. are located at the equal distances on the ray AX and the point B is joined to

(A) A12 

(B) A11 

(C) A10 

(D) A9

Answer 2. (B) A11

As per the question,

The line segment AB is in the ratio of 4:7

Thus, A: B = 4:7

Draw the ray AX making the acute angle BAX

The minimum number of the points located at the equal distances on the ray,

AX = A+B = 4+7= 11

A1, A2, A3, ………. are located at the same distances on the ray AX.

Point B has been joined to the last point to A11.

Question 3. For dividing the line segment AB in the ratio 5: 6, draw the ray AX so that ∠BAX is the acute angle. Now draw the ray BY parallel to the AX, and the points A1, A2, A3, … and B1, B2, B3, … are located at the equal distances on the ray AX and BY, respectively. So, the points joined are

(A) A5 and B6 

(B) A6 and B5 

(C) A4 and B5 

(D) A5 and B4

Answer 3. (A) A5 and B6

As per the question,

The line segment AB is in the ratio of 5:7

Thus, A: B = 5:7

Steps for the construction:

1. Draw the ray AX and the acute angle BAX.
2. Draw the ray BY ||AX, the angle ABY = angle BAX.
3. Then,locate the points A1,A2,A3,A4 and A5 on AX as well as B1,B2,B3,B4,B5 and B6

( As A:B = 5:7)

4. Join A5B6.

A5B6 cuts AB at the point C.

AC: BC= 5:6

Question 4. By the geometrical construction, it is possible to divide the line segment in the ratio √3:(1/√3)

Answer 4. True

Explanation:

As per the question,

Ratio= √3 : ( 1/√3)

Simplifying we have,

√3/ (1/√3) = (√3 x √3)/1 = 3:1

The required ratio is = 3:1

Therefore,

The geometrical construction is possible to divide the line segment in the ratio of 3:1.

Question 5. To construct the triangle similar to the given △ABC with the sides 7/3 of the corresponding sides of the△ABC. Draw the ray BX making the acute angle with BC, and X lies on the opposite side of the angle A with respect to the side BC. The points B1, B2, …., and B7 are located at the equal distances on BX, B3 is joined to the angle C, and the line segment B6C‘ is drawn parallel to the B3C where C‘ lies on the side BC produced. Finally, the line segment A‘C‘ is drawn parallel to the side AC.

Answer 5. False

Explanation

Steps for the construction,

1. Draw the line segment BC.
2. With B and C as the centres, draw two arcs of the suitable radius intersecting each other at the angle A.
3. Joining the side BA and CA, we have the required triangle ∆ABC.
4. Draw the ray BX from B downwards to make the acute angle ∠CBX.
5. Mark the seven points B1, B2, B3 …B7 on BX, such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
6. Join B3C and draw the line B7C’|| B3C from B7 so that it intersects the extended line segment BC at C’.
7. Draw the C’A’ ||CA so that it intersects the extended line segment BA at the angle A’.

Now, the ∆A’BC’ is the required triangle having sides as 7/3 of the corresponding sides of the ∆ABC.

As per the question,

We find,

The segment B6C’ || B3C. But it is clear from our construction that it is never possible that the segment B6C’||B3C as the similar triangle A’BC’ has the sides 7/3 of the corresponding sides of the triangle ABC.

Thus, B7C’ is parallel to B3C.

Question 6. Draw the line segment of the length of 7 cm. Find the point P on it that divides it in the ratio of 3:5.

Steps for the construction:

1. Draw the line segment, AB = 7 cm.
2. Draw the ray. AX is making the acute angle downward with AB.
3. Mark the points to A1, A2, A3 … A8 on the side AX.
4. Mark the points so that AA1 = A1A2 = A2A3 = ….., A7A8.
5. Join the side BA8.
6. Draw the line parallel to the side BA8 passing through the point A3 to meet the side AB on P.

Therefore, AP: PB = 3: 5

Question 7: Draw the line segment of the length of 7 cm. Find the point P on it that divides it in the ratio of 3:5.

Answer 7:

Steps for the construction:

Step 1: Draw the line segment, AB = 7 cm.

Step 2: Draw the ray, AX making the acute angle downward with AB.

Step 3: Mark the points to A1, A2, A3 … A8 on the side AX.

Step 4: Mark the points so that AA1 = A1A2 = A2A3 = ….., A7A8.

Step 5: Join the side BA8.

Step 6: Draw the line parallel to the side BA8 passing through the point A3 to meet the side AB on C.

Hence AC: CB = 3: 5

Question 8: Construct the triangle with the sides of 5 cm, 6 cm and 7 cm and the other triangle having the sides as 7/5 for the corresponding sides of the first triangle.

Answer 8:

Steps for the Construction:

Step 1: Draw the line segment AB =5 cm.

Step 2: Take A as well as B as the centre, and draw the arcs of the radius of 6 cm and 7 cm, respectively.

Step 3: These arcs will cut each other at the point C, and thus the ΔABC is the required the triangle with the length of the sides as 5 cm, 6 cm, and 7 cm, respectively.

Step 4: Draw the ray AX that makes an acute angle with the line segment AB at the opposite side of the vertex C.

Step 5: Locate the 7 points as A1, A2, A3, A4, A5, A6, A7 (since 7 is greater between 5 and 7), on the line AX so that you get AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.

Step 6: Join the points BA5 and then draw the line from A7 to BA5 that is parallel to the line BA5, here it cuts the extended line segment AB at the point B’.

Step 7: Draw the line from B’ and then extend the line segment AC at C’ that is parallel to the line BC, and it cuts to make the triangle.

So, ΔAB’C’ is the required triangle.

Question 9: Draw the circle with a radius of 3 cm. Take the two points P and Q such that one of the extended diameters, each at the distance of 7 cm from the centre. Draw the tangents towards the circle from the two points, P and Q.

Answer 9:

Steps for the construction:

Step 1: Draw the circle with the radius of 3 cm with a centre “O”.

Step 2: Draw the diameter of the circle with the endpoints P and Q, and extend it 7 cm from the centre.

Step 3: Draw the perpendicular bisector from the line PO and then mark the midpoint as M.

Step 4: Draw the circle with M as the centre and MO as the radius

Step 5: Join the points PA and PB, for which the circle with the radius MO cuts the circle at the points A and B.

Step 6: We obtain PA and PB as the required tangents.

Step 7: In the same way, from point Q, we could draw the tangents.

Step 8: We obtain QC and QD are the tangents required.

Question 10: Draw the circle with the help of the bangle. Take the point outside the circle. Construct the pair of the tangents from the point to the circle.

Answer 10:

Steps for the construction:

Step 1: Draw the circle with the help of the bangle.

Step 2: Draw the two non-parallel chords having AB and CD

Step 3: Draw the perpendicular bisector for the chords AB and CD

Step 4: Take the centre as O, here the perpendicular bisector cuts.

Step 5:  After drawing the tangents, take the point P outside the circle.

Step 6: Join the points as O and P.

Step 7: Draw the perpendicular bisector for the line PO and take the midpoint as M.

Step 8: Take M as the centre and MO as the radius, and draw the circle.

Step 9: Let it cut the circle at the points Q and R.

Step 10:  Join the chord PQ and PR.

Thus, PQ and PR are the required tangents.

Question 11: Draw the two concentric circles of the radii 3 cm and 5 cm. Taking the point on the outer circle, construct the pair of tangents to the other. Measure the length of the tangent and verify it by the actual calculation.

Answer 11:

Steps for the construction:

Step 1: Draw the circle with the centre O and the radius 3 cm.

Step 2: Draw the other circle with the centre O and the radius 5 cm.

Step 3: Take the point P on the circumference of the larger circle and join OP.

Step 4: Draw other circles so as it cuts the smallest circle at the point A and B.

Step 5: Join the points A to P and the points B to P.

So, AP and BP are the required tangents.

Question 12: Draw the line segment AB of the length of 7 cm. Taking A as the centre, draw the circle of the radius of 3 cm and taking B as the centre, draw other circles of the radius of 2 cm. Construct the tangents to each circle from the centre towards the other circle.

Steps for the Construction:

Step 1: Draw the line segment having AB as 7 cm.

Step 2: Taking A and B as the centres, draw the two circles of 3 cm and 2 cm radius, respectively.

Step 3: Bisect the line AB. Let assume the midpoint of AB be C.

Step 4: Taking C as the centre, and draw the circle of radius AC that intersects the two circles at the points P, Q, R and S. 

Step 5: Join the chord BP, BQ, AS and AR. 

PB, QB, RA and SA are the required tangents. 

Question 13:  Construct the equilateral ΔABC with each side having 5 cm. Construct another triangle whose sides are as 2/3 times the corresponding sides of the ΔABC.

Answer 13:

Steps for the construction:

Step 1: Draw the line segment BC = 5 cm.

Step 2: Taking B as the centre and the radius of 5 cm, draw the arc.

Step 3: Taking C as the centre and the radius of 5 cm, draw another arc meeting the previous arc at the point A.

Step 4: Join point AC and BC. So, the ΔABC is the required triangle.

Step 5: Draw the line BX so that ∠CBX is the acute angle and is opposite to the vertex A.

Step 6: Along BX, mark the 3 points B1, B2, and B3 so as BB1 = B1B2 = B2B3.

Step 7: Join B3 to C. 

Step 8: Draw the line B2C’ || B3C

Step 9: Draw the line A’C’ parallel to the side AC.

So, the ΔBA’C’ is the required triangle.

Question 14: Construct the triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Now construct other triangles whose sides are as 3/4 times the corresponding sides of the ΔABC.

Answer 14:

In the triangle ABC,

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° – 150° = 30°

Steps for the construction:

Step 1: Draw the line segment BC = 7 cm.

Step 2: At B, construct the right angle and cut it so as the angle ∠B = 45°.

Step 3: Construct the angle 30 degrees at C so as this line cuts the previous angle at A. So, the ΔABC is the required triangle.

Step 4: Draw the line BX so as the angle ∠CBX is the acute angle and is opposite to the vertex A.

Step 5: Along BX, mark the 4 points B1, B2, B3, and B4 such that BB1 = B1B2 = B2B3 = B3B4

Step 6: Join B3 to C. 

Step 7: Draw the line through B4 that is parallel to B3C so that it cuts the extended BC at C’.

Step 8: Draw the line A’C’ parallel to AC so that it meets the extended AB at A’.

Therefore, the ΔBA’C’ is the required triangle, similar to the triangle ABC.

Question 15: Construct the ΔABC for which AB = 6 cm, ∠A = 30° and ∠B = 60°. Construct the other ΔAB’C’ similar to the ΔABC with the base AB’ = 8 cm.

Answer 15:

Given that, AB = 6 cm as well as AB’ = 8 cm

Scale factor = AB’/AB = 8/6 = 4/3

Steps for the construction:

Step 1: Draw the line segment AB = 6 cm.

Step 2: At A and B, construct the angles 30° and 60°, respectively and let these lines cut each other at C. So, the ΔABC is the required triangle.

Step 3: Draw the line AX so that ∠BAX is the acute angle and is opposite to the vertex C.

Step 4: Along with AX, mark the 4 points A1, A2, A3, and A4 such that AA1 = A1A2 = A2A3 = A3A4

Step 5: Then join A3B. 

Step 6: Draw the line through A4, which is parallel to A3B so that it cuts the extended AB at B’.

Step 7: Draw the line B’C’ parallel to BC so that it meets the extended AC at the point C’.

Therefore, the ΔAB’C’  is the required triangle, similar to the triangle ABC.

Question 16. Draw the right triangle ABC for which BC = 12 cm, AB = 5 cm and ∠B = 90°. Construct the triangle similar to it and of the scale factor 2/3. Is the new triangle also the right triangle?

Answer 16:

Steps for the construction:

1. Draw the line segment AB = 5 cm. Construct the right-angle SAB at the point A.
2. Draw the arc of the radius of 12 cm with the point B as the centre to intersect SA at the point C.
3. Join the side BC to obtain ∆ABC.
4. Draw the ray AX making the acute angle with the side AB, opposite to the vertex C.
5. Locate the 3 points, A1, A2, A3 on the line segment AX so that AA1 = A1A2 = A2A3.
6. Join the side A3B.
7. Draw the line through A2 parallel to the side A3B cutting AB at the point B’.
8. From B’, draw the line parallel to the side BC intersecting AC at the vertex C’.
9. Triangle AB’C’ is the required triangle.

Question 17. Two line segments, AB and AC, include the angle of the angle 60°, where AB = 5 cm and AC = 7 cm. Locate the points P and Q on AB and AC, respectively, so that AP = ¾ AB and AQ = ¼ AC. Join the points P and Q and measure the length of PQ.

Answer 17:

Steps for the construction:

• Draw the line segment AB = 5 cm.
• Draw the angle ∠BAZ = 60°.
• With the centre A and the radius 7 cm, draw the arc cutting the line AZ at the vertex C.
• Draw the ray AX making the acute ∠BAX.
• Divide the side AX into four equal parts, as AA1 = A1A2 = A2A3 = A3A4. 
• Join the side A4B.
• Draw the side A3P || A4B meeting AB at the point P.
• Thus, we get that P is the point on AB so that AP = ¾ AB.
• Then, draw the ray AY so as it makes the acute ∠CAY.
• Divide the side AY into four parts as AB1 = B1B2 = B2B3 = B3B4.
• Join the side B4C.
• Draw the B1Q || B4C meeting the side AC at the point Q. We have, Q is the point on the side AC so as AQ = ¼ AC.
•   Join the side PQ and then measure it.
•   PQ = 3.25 cm.

Explanation:

The construction of the given problem can be clarified by showing that

AC/CB = 5/ 8

By the construction, we have the side A5C || A13B. From the Basic proportionality theorem for the triangle AA13B, we have,

AC/CB =AA5/A5A13….. (1)

From the given figure constructed, it is observed that the sides AA5 and A5A13 contain 5 and 8 as equal divisions of the line segments, respectively.

Hence, we get,

AA5/A5A13=5/8… (2)

Comparing the equations (1) and (2), we obtain

AC/CB = 5/ 8

Thus, Justified.

Question 19. Construct the triangle of the sides of 4 cm, 5 cm and 6 cm, then the triangle similar to it which has sides as 2/3 of the corresponding sides of the first triangle.

Answer 19. Construction Procedure:

1. Draw the line segment AB which measures 4 cm, that is AB = 4 cm.
2. Take the point A as the centre, and draw the arc of the radius of 5 cm. 
3. In the same way, take point B as the centre, and draw the arc of the radius of 6 cm.
4. The arcs drawn will cut each other at the point C.
5. We get the side AC = 5 cm and BC = 6 cm, and thus, the ΔABC is the required triangle.
6. Draw the ray AX, which makes the acute angle with the line segment AB on the opposite side of the vertex C.
7. Locate 3 points such that the A1, A2, and A3 as 3 is greater between 2 and 3 on the line AX so as it becomes AA1= A1A2 = A2A3.
8. Join the point BA3 and draw the line passing through A2, which is parallel to the line BA3, so that it cuts AB at point B’.
9. From the point B’, draw the line parallel to the line BC that cuts the line AC at the vertex C’.
10. Hence, the ΔAB’C’ is the required triangle.

Explanation:

The construction of the given problem could be justified by showing as,

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

From the construction given, we get B’C’ || BC

∠AB’C’ = ∠ABC (Corresponding angles)

In the triangle ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Hence, AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAAB’ and ΔAAB,

∠A2AB’ =∠A3AB (Common)

From the corresponding angles, we have,

∠AA2B’ =∠AA3B

Hence, from the AA similarity criterion, we get,

ΔAA2B’ and AA3B

Hence, AB’/AB = AA2/AA3

Thus, AB’/AB = 2/3 ……. (2)

From the equations (1) and (2), we have,

AB’/AB=B’C’/BC = AC’/ AC = 2/3

This could be rewritten as

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

Thus, justified.

Question 20. Construct the triangle with the sides of 5 cm, 6 cm and 7 cm and the other triangle which have sides as 7/5 of the corresponding sides of the first triangle

Answer 20. 

Construction Procedure:

1. Draw the line segment AB =5 cm.
2. Take A and B as the centre, and draw the arcs of the radius of 6 cm as well as 7 cm, respectively.
3. These arcs will have to intersect each other at the point C, and thus, the ΔABC is the needed triangle with the length of the sides as 5 cm, 6 cm, and 7 cm, respectively.
4. Draw the ray AX, which makes the acute angle with the line segment AB on the opposite side of the vertex C.
5. Locate the 7 points so that A1, A2, A3, A4, A5, A6, A7 as 7 is greater between 5 and 7 on the line AX so as it becomes AA1 =  A1 A2 = A2 A3 = A3 A4 = A4 A5 = A5 A6 =  A6 A7 
6. Join the points BA5 and draw the line from the A7 to BA5, which is parallel to the line BA5 in which it cuts the extended line segment AB at the point B’.
7. Then, draw the line from B’ the extended line segment AC at the vertex C’, that is parallel to the line BC, and it intersects to make the triangle.
8. Thus, the ΔAB’C’ is the required triangle.

Explanation:

The construction of the given problem could be justified by showing that

AB’ = (7/5)AB

B’C’ = (7/5)BC

AC’= (7/5)AC

From the construction given, we get B’C’ || BC

∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

∴ ΔAB’C’ ∼ ΔABC (From the AA similarity criterion)

Hence, AB’/AB = B’C’/BC= AC’/AC …. (1)

In the ΔAA7B’ and ΔAA5B,

∠A7AB’=∠A5AB (Common)

From the corresponding angles, we have,

∠A A7B’=∠A A5B

Thus, from the AA similarity criterion, we obtain

ΔA A2B’ and A A3B

Hence, AB’/AB = AA5/AA7

Thus, AB /AB’ = 5/7 ……. (2)

From the equations (1) and (2), we have,

AB’/AB = B’C’/BC = AC’/ AC = 7/5

This could be written as

AB’ = (7/5)AB

B’C’ = (7/5)BC

AC’= (7/5)AC

So, justified.

Question 21. Construct the isosceles triangle, which has the base as 8cm, and the altitude of 4cm and then another triangle which has sides are (1+1/2) time the corresponding sides of the isosceles triangle.

Answer 21. Construction Procedure:

1. Draw the line segment BC with the measure of 8 cm.
2. Draw the perpendicular bisector of the line segment BC and intersect at the point D
3. Take the point D as the centre and draw the arc with the radius of 4 cm, which cuts the perpendicular bisector at the point A
4. Join the lines AB and AC, and the triangle is the required triangle.
5. Draw the ray BX, which makes the acute angle with the line BC on the side opposite to the vertex A.
6. Locate the 3 points such that  B1, B2 and B3 on the ray BX such that   BB1 = B1B2 = B2B3
7. Join the points B2C and draw the line from B3, which is parallel to the line B2C, which cuts the extended line segment BC at the point C’.
8. Draw the line from the vertex C’ to the extended line segment AC at the point A’, which is parallel to the line AC as well as it intersects to make the triangle.
9. Hence, the ΔA’BC’ is the required triangle.

Explanation:

The construction of the given problem could be justified by showing that

A’B = (3/2)AB

BC’ = (3/2)BC

A’C’= (3/2)AC

From the construction given, we have A’C’ || AC

∠ A’C’B = ∠ACB (Corresponding angles)

In the ΔA’BC’ and ΔABC,

∠B = ∠B (common)

∠A’BC’ = ∠ACB

ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Hence, A’B/AB = BC’/BC= A’C’/AC

As the corresponding sides of the similar triangle are in the same ratio, we get,

A’B/AB = BC’/BC= A’C’/AC = 3/2

Thus, justified.

Question 22. Draw the triangle ABC with side BC = 6 cm, AB = 5 cm and the angle ∠ABC = 60°. Construct the triangle, which has sides that are 3/4 of the corresponding sides of the triangle ABC.

Answer 22. Construction Procedure:

1. Draw the ΔABC with the base side BC = 6 cm, and AB = 5 cm and the angle ∠ABC = 60°.
2. Draw the ray BX, which makes the acute angle with BC on the opposite side of the vertex A.
3. Locate the 4 points as 4 is greater in 3 and 4 so as B1, B2, B3, and B4, on the line segment BX.
4. Join the points B4C. Also, draw the line through B3, parallel to B4C, intersecting the line segment BC at the vertex C’.
5. Draw the line through C’ parallel to the line AC, which cuts the line AB at the point A’.
6. Thus, the ΔA’BC’ is the required triangle.

Explanation:

The construction of the given problem could be justified by showing that

As the scale factor is 3/4, we just prove

A’B = (3/4)AB

BC’ = (3/4)BC

A’C’= (3/4)AC

From the construction given, we find A’C’ || AC

In the ΔA’BC’ and ΔABC,

The ∠ A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

As the corresponding sides of the similar triangle are in the same ratio, we get,

Hence, A’B/AB = BC’/BC= A’C’/AC

Thus, it becomes A’B/AB = BC’/BC= A’C’/AC = ¾

Therefore, justified.

Question 23. Draw the triangle ABC with the side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Construct the triangle, which has sides as 4/3 times the corresponding sides of the ∆ ABC.

Answer 23. To find the angle ∠C:

Given that:

∠B = 45°, ∠A = 105°

We have,

The Sum of all the interior angles in the triangle is 180°.

∠A+∠B +∠C = 180°

105°+45°+∠C = 180°

∠C = 180° − 150°

∠C = 30°

Thus, from the property of the triangle, we have the angle ∠C = 30°

Construction Procedure:

The required triangle could be drawn as,

1. Draw the ΔABC with the side measures of the base BC = 7 cm, ∠B = 45°, and ∠C = 30°.
2. Draw the ray BX makes the acute angle with the side BC on the opposite side of the vertex A.
3. Locate 4 points as 4 is the greater in 4 and 3 so that B1, B2, B3, and B4, are on the ray BX.
4. Join the points B3C.
5. Draw the line through B4 parallel to B3C in which it cuts the extended line BC at C’.
6. Through point C’, draw a line parallel to the line AC which intersects the extended line segment at C’.
7. Therefore, ΔA’BC’ is the required triangle.

Explanation:

The construction of the given problem can be justified by proving that

hence the scale factor is 4/3, we need to prove

A’B = (4/3)AB

BC’ = (4/3)BC

A’C’= (4/3)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

therefore, ∠A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

therefore, ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

hence the corresponding sides of the similar triangle are in the same ratio, it becomes,

Therefore, A’B/AB = BC’/BC= A’C’/AC

thus, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3

Hence, justified.

Question 24. Draw the right triangle in which the sides other than hypotenuse are of the lengths 4 cm and 3 cm. Construct the other triangle, which has sides that are 5/3 times the corresponding sides of the given triangle.

Answer 24. Given that:

The sides other than the hypotenuse are having the lengths of 4cm and 3cm. It defines that the given sides are perpendicular to each other.

Construction Procedure:

The required triangle can be drawn as,

1. Draw the line segment BC =3 cm.
2. Measure and draw the angle of 90°
3. Taking B as the centre, draw the arc with a radius of 4 cm and cut the ray at point B.
4. On joining the lines AC, the triangle ABC is the required triangle.
5. Draw the ray BX makes the acute angle with the side BC on the opposite side of the vertex A.
6. Locate 5 so that B1, B2, B3, B4 on the ray BX so that BB1 = B1B2 = B2B3 = B3B4 = B4B5
7. Join the points B3C.
8. Draw the line through B5 parallel to B3C, which cuts the extended line BC at C’.
9. Through C’, draw the line parallel to the line AC that cuts the extended line AB at A’.
10. Hence, ΔA’BC’ is the required triangle.

Explanation:

The construction of the given problem could be justified by showing that

As the scale factor is 5/3, we just have to prove

A’B = (5/3)AB

BC’ = (5/3)BC

A’C’= (5/3)AC

From the construction given, we find A’C’ || AC

In the ΔA’BC’ and ΔABC,

∠ A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

ΔA’BC’ ∼ ΔABC (From the AA similarity criterion)

As the corresponding sides of the similar triangle are in the same ratio, we get,

Thus, A’B/AB = BC’/BC= A’C’/AC

Therefore, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3

So, justified.

Question 25. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Answer 25. Construction Procedure:

The construction to draw the pair of tangents to the given circle is as follows.

1. Draw a circle with a radius = 6 cm with centre O.
2. Locate point P, which is 10 cm away from point O.
3. Join points O and point P through the line
4. Draw the perpendicular bisector having the line OP.
5. Let’s assume M be the mid-point of the line PO.
6. Take M as a centre and measure the length of MO
7. The length MO is taken as a radius and draws a circle.
8. The circle drawn by the radius of MO intersects the previous circle at point Q and R.
9. Join PQ and PR.
10. hence, PQ and PR are the required tangents.

Explanation:

The construction of the given problem can be clarified by proving that PQ and PR are the tangents passing through the circle of radius 6cm with the centre O.

To prove the following, join OQ and OR represented in dotted lines.

From the construction,

∠PQO is the angle in the semicircle.

We get that the angle in a semicircle is a right angle, so it becomes,

∴ ∠PQO = 90°

Such that

⇒ OQ ⊥ PQ

Since OQ is the radius of the circle with a radius of 6 cm, PQ might be a tangent of the circle. same as, we can prove that PR is a tangent of the circle.

thus, justified.

Hence, the length of the tangents to a circle of radius 6 cm, from a point 10 cm away from the centre of the circle, is  8 cm.

Question 26. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.

Answer 26. Construction Procedure:

For the given circle, the tangent could be drawn as follows.

1. Draw the circle of 4 cm radius with a centre “O”.
2. Again, take O as the centre and draw a circle of radius 6 cm.
3. Locate the point P on this circle
4. Join the points O and P passing through lines so that it becomes OP.
5. Draw the perpendicular bisector towards the line OP
6. Let assume M be the mid-point of PO.
7. Draw the circle with M as its centre and MO as its radius
8. The circle drawn having the radius OM intersects the given circle at the points Q and R.
9. Join PQ and PR.
10. PQ and PR are the needed tangents.

From the construction, we have seen that PQ and PR are of the length of 4.47 cm each.

It can be calculated manually as follows

In ∆PQO,

hence PQ is a tangent,

∠PQO = 90°. PO = 6cm and QO = 4 cm

Applying the Pythagoras theorem in ∆PQO, we obtain PQ2+QO2 = PQ2

PQ2+(4)2 = (6)2

PQ2 +16 =36

PQ2 = 36−16

PQ2 = 20

PQ = 2√5

PQ = 4.47 cm,

thus, the tangent length PQ = 4.47

Explanation:

The construction of the given problem can be clarified by proving that PQ and PR are the tangents for the circle of radius 4 cm with centre O.

To prove this, join OQ as well as OR represented in dotted lines.

From the construction,

∠PQO is an angle in the semicircle.

We know that the angle in a semicircle is a right angle, so it becomes,

∴ ∠PQO = 90°

Such that

⇒ OQ ⊥ PQ

Since OQ is the radius of the circle with a radius of 4 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle.

Hence, justified.

Question 27. Draw a circle of radius 3 cm. Take two points, P and Q, on one of its extended diameters, each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points, P and Q

Answer 27. Construction Procedure:

The tangent for the given circle can be constructed as follows.

1. Draw the circle with a radius of 3cm with a centre “O”.
2. Draw a diameter of the circle and it extends 7 cm from the centre and marks it as P and Q.
3. Draw the perpendicular bisector for the line PO and mark the midpoint as M.
4. Draw a circle with M as the centre and MO as the radius
5. then join the points PA and PB in which the circle with radius MO intersects the circle of circle 3cm.
6. then, PA and PB are the required tangents.
7. same as, from the point Q, we could draw the tangents.
8. From that required, QC and QD are the needed tangents.

Explanation:

The construction of the given problem can be clarified by proving that PQ as well as PR are the tangents for the circle of radius 3 cm with centre O.

To prove this, join the line OA and OB.

From the construction,

∠PAO is an angle in the semicircle.

We know that the angle in a semicircle is a right angle, so it becomes,

therefore, ∠PAO = 90°

Such that

⇒ OA ⊥ PA

Since OA is the radius of the circle with a radius of 3 cm, PA must be a tangent of the circle. same as, we can prove that PB, QC and QD are the tangents of the circle.

Hence, justified

Question 28. Draw a pair of tangents to a circle of radius 5 cm, which is inclined to each other at an angle of 60°

Answer 28. Construction Procedure:

The tangents could be constructed in the following manner:

1. Draw a circle of radius 5 cm and with a centre as O.
2. Take the point Q on the circumference of the circle and join OQ.
3. Draw a perpendicular to QP at point Q.
4. Draw the radius OR, making an angle of 120°, i.e. (180°−60°) with OQ.
5. Draw a perpendicular to RP at point R.
6. then, both the perpendiculars intersect at point P.
7. hence, PQ and PR are the required tangents at an angle of 60°.

Explanation:

The construction can be clarified by proving that ∠QPR = 60°

By our construction

∠QOP = 90°

∠ORP = 90°

And ∠QOR = 120°

We get that the sum of all interior angles of a quadrilateral = 360°

∠OQP+∠QOR + ∠ORP +∠QPR = 360o

90°+120°+90°+∠QPR = 360°

Therefore, ∠QPR = 60°

Hence Justified

Question 29. Draw the line segment AB of the length of 8 cm. Taking the point A as the centre, draw the circle of the radius of 4 cm and take B as the centre, draw the other circle of the radius of 3 cm. Construct tangents to each of the circles from the centre of the other circle.

Answer 29. Construction Procedure:

The tangent for the given circle could be constructed as,

1. Draw the line segment AB = 8 cm.
2. Taking the point A as the centre, draw the circle of the radius of 4 cm
3. Taking the point B as the centre, draw the circle of the radius of 3 cm
4. Draw the perpendicular bisector having the line AB and take the midpoint as M.
5. Taking the point M as the centre, draw the circle with the radius of MA or MB that cuts the circle at the points P, Q, R and S.
6. Join AR, AS, BP and BQ
7. Thus, the required tangents are AR, AS, BP and BQ

Explanation:

The construction could be justified by showing that the chords AS and AR are the tangents of the circle, which has the centre as B with a radius of 3 cm. BP aswell as BQ are the tangents of the circle, which has the centre as A and the radius 4 cm.

From the given construction, 

We can prove this by joining AP, AQ, BS, and BR.

∠ASB is the angle in the semicircle. We have the angle in the semicircle as the right angle.

∠ASB = 90°

 BS ⊥ AS

As BS is the radius of the circle, AS must be the tangent of the circle.

In the same way, AR, BP, and BQ are the required tangents to the given circle.

Question 30. Let ABC be the right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from the point B on AC. The circle passing through the point B, C, and D is drawn. Construct the tangents from the point A to the circle.

Answer 30. Construction Procedure:

The tangent for the given circle could be constructed as,

1. Draw the line segment with the base BC = 8cm
2. Measure the angle 90° at the having point B, so as ∠ B = 90°.
3. Take B as the centre and draw the arc with the measure of 6cm.
4. Let the point be A ,here the arc cuts the ray.
5. Join the line AC.
6. Hence, the ∆ABC be the required triangle.
7. Draw the perpendicular bisector to the line BC, and the midpoint is marked as E.
8. Take E as the centre. BE and taking EC is the measure of the radius; draw the circle.
9. Join the point A to the midpoint E of the circle
10. Then, draw the perpendicular bisector to the line AE and the midpoint is taken as the point M
11. Take M as the centre. AM and ME is the measure of the radius; draw the circle.
12. This circle cuts the previous circle at the points B and Q
13. Join the points A and Q
14. Hence, AB and AQ are the required tangents

Explanation:

The construction could be justified by proving that the sides AG and AB are the tangents to the circle.

From the given construction, join the side EQ.

∠AQE is an angle of the semicircle. We have that the angle in the semicircle is the right angle.

∠AQE = 90°

 EQ⊥ AQ

As the side EQ is the radius of the circle, the side AQ has to be the tangent of the circle. 

In the same way, ∠B = 90°

AB ⊥ BE

As BE is the radius of the circle, the side AB has to be the tangent of the circle.

Thus, justified.

Question 31. Draw the circle with using of the bangle. Take the point outside the circle. Construct the pair of the tangents from the point to the circle.

Answer 31. Construction Procedure:

The required tangents could be constructed on the given circle as,

1. Draw the circle with the help of the bangle.
2. Draw two non-parallel chords so as AB and CD
3. Draw the perpendicular bisector to the AB and CD
4. Take the centre as O here the perpendicular bisector cuts.
5. To draw the tangents, take the point P outside the circle.
6. Join the points O and P.
7. Then draw the perpendicular bisector of the line PO, and the midpoint is taken as M
8. Take the point M as the centre and MO as the radius to draw the circle.
9. Let the circle cut the circle at the points Q and R
10. Then join the side PQ and PR
11. Hence, PQ and PR are the required tangents.

Explanation:

The construction could be justified by proving that the side PQ and PR are the tangents to the circle.

As O is the centre of the circle, we have that the perpendicular bisector of the chords passes through the centre.

Join the points OQ and OR.

We get that the perpendicular bisector of the chord passing through the centre.

It is clear that the cutting point of the perpendicular bisectors is the centre of the circle.

As the angle ∠PQO is the angle in the semicircle. We have that the angle in the semicircle is the right angle.

 ∠PQO = 90°

 OQ⊥ PQ

As OQ is the radius of the circle, the side PQ has to be the tangent of the circle. 

In the same way,

 ∠PRO = 90°

OR ⊥ PO

As OR is the radius of the circle, the side PR has to be the tangent of the circle.

Hence, PQ and PR are the required tangents of the circle.

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