﻿ Class 10 NCERT Math Solution
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Exercise - 11.1

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. Give the justification of the construction.

Solution :-
```  Given that :
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.

Steps of Construction :
1. Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.
2. Locate 13 (= 5 + 8) points, A₁, A₂, A₃, A₄ .... A₁₃, on AX such that AA₁ = A₁A₂ = A₂A₃ and so on.
3. Join BA₁₃.
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.

Justification :-
The construction can be justified by proving that AC/CB = 5/8
By construction, we have A₅C || A₁₃B.
By applying Basic proportionality theorem for the triangle AA₁₃B, we obtain AC/BC = AA₅/A₅A₁₃ --- (1)
From the figure, it can be observed that AA₅ and A₅A₁₃ contain 5 and 8 equal divisions of line segments respectively.
AA₅/A₅A₁₃ = 5/8 ----(2)
On comparing equations (1) and (2), we obtain
AC/CB = 5/8
```

Question-2 :-  Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it 2/3 of the corresponding sides of the first triangle. Give the justification of the construction.

Solution :-
```  Given that :
A triangle of sides 4 cm, 5cm and 6cm.

Steps of Construction :
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.
2. Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
3. Locate 3 points A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁ = A₁A₂ = A₂A₃.
4. Join BA₃ and draw a line through A₂ parallel to BA₃ to intersect AB at point B'.
5. Draw a line through B' parallel to the line BC to intersect AC at C'.
ΔAB'C' is the required triangle.

Justification :-
The construction can be justified by proving that AB' = 2AB/3, B'C' = 2BC/3, AC' = 2AC/3
By construction, we have B’C’ || BC
∠AB'C' = ∠ABC (Corresponding angles)

In ΔAB'C' and ΔABC,
∠AB'C' = ∠ABC (Proved above)
∠B'AC' = ∠BAC (Common)
∴ ΔAB'C'∼ΔABC (AA similarity criterion)
So, AB'/AB = B'C'/BC = AC'/AC -----(1)

In ΔAA₂B' and ΔAA₃B, ∠A₂AB' = ∠A₃AB (Common)
∠AA₂B' = ∠AA₃B (Corresponding angles)
∴ΔAA₂B' ∼ΔAA₃B (AA similarity criterion)
So, AB'/AB = AA₂/AA₃
AB'/AB = 2/3 -------(1)

From equations (1) and (2), we obtain
AB'/AB = B'C'/BC = AC'/AC = 2/3
Therefore, AB' = 2AB/3, B'C' = 2BC/3, AC' = 2AC/3
```

Question-3 :-  Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle. Give the justification of the construction.

Solution :-
```  Given that :
A triangle with sides 5 cm, 6 cm and 7 cm.

Steps of Construction :
1. Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5cm 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.
2. Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
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.
4. Join BA₅ and draw a line through A₇ parallel to BA₅ to intersect extended line segment AB at point B'.
5. Draw a line through B' parallel to BC intersecting the extended line segment AC at C'.
ΔAB'C' is the required triangle.

Justification :-
The construction can be justified by proving that AB' = 7AB/5, B'C' = 7BC/5, AC' = 7AC/5
By construction, we have B’C’ || BC
∠AB'C' = ∠ABC (Corresponding angles)

In ΔAB'C' and ΔABC,
∠AB'C' = ∠ABC (Proved above)
∠B'AC' = ∠BAC (Common)
∴ ΔAB'C'∼ΔABC (AA similarity criterion)
So, AB'/AB = B'C'/BC = AC'/AC
or AB/AB' = BC/B'C' = AC/AC' -----(1)

In ΔAA₅B and ΔAA₇B',
∠A₅AB = ∠A₇AB' (Common)
∠AA₅B = ∠AA₇B' (Corresponding angles)
∴ΔAA₅B ∼ΔAA₇B' (AA similarity criterion)
So, AB/AB' = AA₅/AA₇
AB'/AB = 7/5 -------(2)

From equations (1) and (2), we obtain
AB'/AB = B'C'/BC = AC'/AC = 7/5
Therefore, AB' = 7AB/5, B'C' = 7BC/5, AC' = 7AC/5
```

Question-4 :-  Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are 1 by 1/2 times the corresponding sides of the isosceles triangle. Give the justification of the construction.

Solution :-
```  Given that :
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 3/2 times of ΔABC can be drawn as follows.

Steps of Construction :
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.
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.
3. Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
4. Locate 3 points (as 3 is greater between 3 and 2) A₁, A₂, and A₃ on AX such that AA₁ = A₁A₂ = A₂A₃.
5. Join BA₂ and draw a line through A₃ parallel to BA₂ to intersect extended line segment AB at point B'.
6. Draw a line through B' parallel to BC intersecting the extended line segment AC at C'.
ΔAB'C' is the required triangle.

Justification :-
The construction can be justified by proving that AB' = 3AB/2, B'C' = 3BC/2, AC' = 3AC/2
By construction, we have B’C’ || BC
∠AB'C' = ∠ABC (Corresponding angles)

In ΔAB'C' and ΔABC,
∠AB'C' = ∠ABC (Proved above)
∠B'AC' = ∠BAC (Common)
∴ ΔAB'C'∼ΔABC (AA similarity criterion)
So, AB'/AB = B'C'/BC = AC'/AC
or AB/AB' = BC/B'C' = AC/AC' -----(1)

In ΔAA₂B and ΔAA₃B',
∠A₂AB = ∠A₃AB' (Common)
∠AA₂B = ∠AA₃B' (Corresponding angles)
∴ ΔAA₂B ∼ΔAA₃B' (AA similarity criterion)
So, AB/AB' = AA₂/AA₃
AB/AB' = 2/3
AB'/AB = 3/2

From equations (1) and (2), we obtain
AB'/AB = B'C'/BC = AC'/AC = 3/2
Therefore, AB' = 3AB/2, B'C' = 3BC/2, AC' = 3AC/2
```

Question-5 :-  Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC. Give the justification of the construction.

Solution :-
```  Given that :
A ΔA'BC' whose sides are 3/4 of the corresponding sides of ΔABC can be drawn as follows.

Steps of Construction :
1. Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
2. Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 3 and 4), B₁, B₂, B₃, B₄, on line segment BX.
4. Join B₄C and draw a line through B₃, parallel to B₄C intersecting BC at C'.
5. Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.
6. Through B', draw a line parallel to BC intersecting extended line segment AC at C'.
ΔAB'C' is the required triangle.

Justification :-
The construction can be justified by proving that AB' = 3AB/4, B'C' = 3BC/4, AC' = 3AC/4
By construction, we have B’C’ || BC
∠AB'C' = ∠ABC (Corresponding angles)

In ΔAB'C' and ΔABC,
∠AB'C' = ∠ABC (Proved above)
∠B'AC' = ∠BAC (Common)
∴ ΔAB'C'∼ΔABC (AA similarity criterion)
So, AB'/AB = B'C'/BC = AC'/AC
or AB/AB' = BC/B'C' = AC/AC' -----(1)

In ΔBB₃C' and ΔBB₄C,
∠B₃BC' = ∠B₄BC (Common)
∠BB₃C' = ∠BB₄C (Corresponding angles)
AB/AB' = 4/3
AB'/AB = 3/4 ------(2)

From equations (1) and (2), we obtain
AB'/AB = B'C'/BC = AC'/AC = 3/4
Therefore, AB' = 3AB/4, B'C' = 3BC/4, AC' = 3AC/4
```

Question-6 :-  Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding side of ΔABC. Give the justification of the construction.

Solution :-
```  Given that :
∠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°

Steps of Construction :
1. Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
2. Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 4 and 3), B₁, B₂, B₃, B₄, on BX.
4. Join B₃C. Draw a line through B₄ parallel to B₃C intersecting extended BC at C'.
5. Through C', draw a line parallel to AC intersecting extended line segment at C'.
ΔA'BC' is the required triangle.

Justification :-
The construction can be justified by proving that AB' = 4AB/3, B'C' = 4BC/3, AC' = 4AC/3
By construction, we have B’C’ || BC
∠AB'C' = ∠ABC (Corresponding angles)

In ΔAB'C' and ΔABC,
∠AB'C' = ∠ABC (Proved above)
∠B'AC' = ∠BAC (Common)
∴ ΔAB'C'∼ΔABC (AA similarity criterion)
So, AB'/AB = B'C'/BC = AC'/AC
or AB/AB' = BC/B'C' = AC/AC' -----(1)

In ΔBB₃C and ΔBB₄C',
∠B₃BC = ∠B₄BC' (Common)
∠BB₃C = ∠BB₄C' (Corresponding angles)
∴ΔBB₃C ∼ΔBB₄C' (AA similarity criterion)
BC/B'C' = BB₃/BB₄
BC/B'C' = 3/4
B'C'/BC = 4/3 ------(2)

From equations (1) and (2), we obtain
AB'/AB = B'C'/BC = AC'/AC = 4/3
Therefore, AB' = 4AB/3, B'C' = 4BC/3, AC' = 4AC/3
```

Question-7 :-  Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Give the justification of the construction.

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

Steps of Construction :
1. Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.
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.
3. Draw a ray AX making an acute angle with AB, opposite to vertex C.
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.
5. Join A₃B. Draw a line through A₅ parallel to A₃B intersecting extended line segment AB at B'.
6. Through B', draw a line parallel to BC intersecting extended line segment AC at C'.
ΔAB'C' is the required triangle.

Justification :-
The construction can be justified by proving that AB' = 5AB/3, B'C' = 5BC/3, AC' = 5AC/3
By construction, we have B’C’ || BC
∠AB'C' = ∠ABC (Corresponding angles)

In ΔAB'C' and ΔABC,
∠AB'C' = ∠ABC (Proved above)
∠B'AC' = ∠BAC (Common)
∴ ΔAB'C'∼ΔABC (AA similarity criterion)
So, AB'/AB = B'C'/BC = AC'/AC
or AB/AB' = BC/B'C' = AC/AC' -----(1)

In ΔAA₃B and ΔAA₅B',
∠A₃AB = ∠A₅AB' (Common)
∠AA₃B = ∠AA₅B' (Corresponding angles)
∴ΔAA₃B ∼ΔAA₅B' (AA similarity criterion)
So, AB/AB' = AA₃/AA₅
AB/AB' = 3/5
AB'/AB = 5/3 ------(2)

From equations (1) and (2), we obtain
AB'/AB = B'C'/BC = AC'/AC = 5/3
Therefore, AB' = 5AB/3, B'C' = 5BC/3, AC' = 5AC/3
```
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