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Exercise 11.1 - Chapter 11 Constructions class 10 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. Give the justification of the construction.

Answer:

A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.

Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.

Step 2 Locate 13 (= 5 + 8) points, A1, A2, A3, A…….. A13, on AX such that AA1 = A1A= A2A3 and so on.

Step 3 Join BA13.

Step 4 Through the point A5, draw a line parallel to BA13 (by making an angle equal to ∠AA13B) at A5 intersecting AB at point C.

C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8.

The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2159/Chapter%2011_html_54401b0c.jpg

Justification

The construction can be justified by proving that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2159/Chapter%2011_html_m2620e40b.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2159/Chapter%2011_html_4562af32.gif  … (1)

From the figure, it can be observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments respectively.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2159/Chapter%2011_html_420b2a9c.gif  … (2)

On comparing equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2159/Chapter%2011_html_m2620e40b.gif

This justifies the construction.

 

Question 2:

Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides arehttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m64d8549d.gif of the corresponding sides of the first triangle.

Give the justification of the construction.

Answer:

Step 1

Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ΔABC is the required triangle.

Step 2

Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.

Step 3

Locate 3 points A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that AA= A1A2 = A2A3.

Step 4

Join BA3 and draw a line through Aparallel to BA3 to intersect AB at point B’.

Step 5

Draw a line through B’ parallel to the line BC to intersect AC at C’.

ΔAB’C’ is the required triangle.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m3e2a42f9.jpg

Justification

The construction can be justified by proving that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_64ea3df2.gif

By construction, we have B’C’ || BC

∴ ∠Ahttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m918f803.gif  = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m400c0479.gif  = ∠ABC (Proved above)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m5eb5263.gif  = ∠BAC (Common)

∴ Δhttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m400c0479.gif  ∼ ΔABC (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m40f68e8f.gif  … (1)

In ΔAA2B’ and ΔAA3B,

∠A2AB’ = ∠A3AB (Common)

∠AA2B’ = ∠AA3B (Corresponding angles)

∴ ΔAA2B’ ∼ ΔAA3B (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m4a2631f.gif

From equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2161/Chapter%2011_html_m191785cb.gif

This justifies the construction.

 

Question 3:

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2172/Chapter%2011_html_48ccc641.gif of the corresponding sides of the first triangle.

Give the justification of the construction.

Answer:

Step 1

Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 2

Draw a ray AX making acute angle with line AB on the opposite side of vertex C.

Step 3

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

Step 4

Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B’.

Step 5

Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C’ is the required triangle.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2172/Chapter%2011_html_45681385.jpg

Justification

The construction can be justified by proving that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2172/Chapter%2011_html_3e12d3ac.gif

In ΔABC and ΔAB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (Common)

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2172/Chapter%2011_html_m1e58dcdb.gif  … (1)

In ΔAA5B and ΔAA7B’,

∠A5AB = ∠A7AB’ (Common)

∠AA5B = ∠AA7B’ (Corresponding angles)

∴ ΔAA5B ∼ ΔAA7B’ (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2172/Chapter%2011_html_240ad0b3.gif

On comparing equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2172/Chapter%2011_html_35fca366.gif

⇒ https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2172/Chapter%2011_html_3e12d3ac.gif

This justifies the construction.

 

Question 4:

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2215/Chapter%2011_html_75d58f26.gif  times the corresponding sides of the isosceles triangle.

Give the justification of the construction.

Answer:

Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.

A ΔAB’C’ whose sides are https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2215/Chapter%2011_html_m47d40970.gif times of ΔABC can be drawn as follows.

Step 1

Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O’. Join OO’. Let OO’ intersect AB at D.

Step 2

Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO’ at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.

Step 3

Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.

Step 4

Locate 3 points (as 3 is greater between 3 and 2) A1, A2, and A3 on AX such that AA1 = A1A2 = A2A3.

Step 5

Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment AB at point B’.

Step 6

Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C’ is the required triangle.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2215/Chapter%2011_html_15caa8b9.jpg

Justification

The construction can be justified by proving that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2215/Chapter%2011_html_37cc5a80.gif

In ΔABC and ΔAB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (Common)

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2215/Chapter%2011_html_m1e58dcdb.gif  … (1)

In ΔAA2B and ΔAA3B’,

∠A2AB = ∠A3AB’ (Common)

∠AA2B = ∠AA3B’ (Corresponding angles)

∴ ΔAA2B ∼ ΔAA3B’ (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2215/Chapter%2011_html_3b8f8f03.gif

On comparing equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2215/Chapter%2011_html_m15587be1.gif

⇒ https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2215/Chapter%2011_html_37cc5a80.gif

This justifies the construction.

 

Question 5:

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides arehttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2216/Chapter%2011_html_m2111158f.gif of the corresponding sides of the triangle ABC.

Give the justification of the construction.

Answer:

A ΔA’BC’ whose sides are https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2216/Chapter%2011_html_m2111158f.gif of the corresponding sides of ΔABC can be drawn as follows.

Step 1

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on line segment BX.

Step 4

Join B4C and draw a line through B3, parallel to B4C intersecting BC at C’.

Step 5

Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2216/Chapter%2011_html_167a4fd8.jpg

Justification

The construction can be justified by proving

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2216/Chapter%2011_html_7096acec.gif

In ΔA’BC’ and ΔABC,

∠A’C’B = ∠ACB (Corresponding angles)

∠A’BC’ = ∠ABC (Common)

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2216/Chapter%2011_html_f7bc0e0.gif  … (1)

In ΔBB3C’ and ΔBB4C,

∠B3BC’ = ∠B4BC (Common)

∠BB3C’ = ∠BB4C (Corresponding angles)

∴ ΔBB3C’ ∼ ΔBB4C (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2216/Chapter%2011_html_451475b8.gif

From equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2216/Chapter%2011_html_m272b4ae0.gif

⇒ https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2216/Chapter%2011_html_7096acec.gif

This justifies the construction.

 

Question 6:

Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2217/Chapter%2011_html_37cb7ba8.gif times the corresponding side of ΔABC. Give the justification of the construction.

Answer:

∠B = 45°, ∠A = 105°

Sum of all interior angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

The required triangle can be drawn as follows.

Step 1

Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.

Step 4

Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C’.

Step 5

Through C’, draw a line parallel to AC intersecting extended line segment at C’. ΔA’BC’ is the required triangle.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2217/Chapter%2011_html_3306b5f4.jpg

Justification

The construction can be justified by proving that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2217/Chapter%2011_html_12149ef3.gif

In ΔABC and ΔA’BC’,

∠ABC = ∠A’BC’ (Common)

∠ACB = ∠A’C’B (Corresponding angles)

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2217/Chapter%2011_html_2d04f55b.gif  … (1)

In ΔBB3C and ΔBB4C’,

∠B3BC = ∠B4BC’ (Common)

∠BB3C = ∠BB4C’ (Corresponding angles)

∴ ΔBB3C ∼ ΔBB4C’ (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2217/Chapter%2011_html_m215b76f6.gif

On comparing equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2217/Chapter%2011_html_m759d3972.gif

⇒ https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2217/Chapter%2011_html_12149ef3.gif

This justifies the construction.

 

Question 7:

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. the construct another triangle whose sides are https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2218/Chapter%2011_html_45970558.gif times the corresponding sides of the given triangle. Give the justification of the construction.

Answer:

It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other.

The required triangle can be drawn as follows.

Step 1

Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.

Step 2

Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.

Step 3

Draw a ray AX making an acute angle with AB, opposite to vertex C.

Step 4

Locate 5 points (as 5 is greater in 5 and 3), A1, A2, A3, A4, A5, on line segment AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.

Step 5

Join A3B. Draw a line through A5 parallel to A3B intersecting extended line segment AB at B’.

Step 6

Through B’, draw a line parallel to BC intersecting extended line segment AC at C’. ΔAB’C’ is the required triangle.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2218/Chapter%2011_html_1e91658b.jpg

Justification

The construction can be justified by proving that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2218/Chapter%2011_html_m437b215b.gif

In ΔABC and ΔAB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (Common)

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2218/Chapter%2011_html_m1e58dcdb.gif  … (1)

In ΔAA3B and ΔAA5B’,

∠A3AB = ∠A5AB’ (Common)

∠AA3B = ∠AA5B’ (Corresponding angles)

∴ ΔAA3B ∼ ΔAA5B’ (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2218/Chapter%2011_html_m77f09f8d.gif

On comparing equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2218/Chapter%2011_html_m1fa2c516.gif

⇒ https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/138/2218/Chapter%2011_html_m437b215b.gif

This justifies the construction.

Also Read : Exercise-11.2-Chapter-11-Constructions-class-10-ncert-solutions-Maths

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