Class 9th

Construction

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Construct the following angles with the help of protector.

(01) Construct angles of 0°,15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180° with the help of protector.

निम्न कोणों को चाँद की सहायता से बनाऔं।

0°,15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180°

Construct the following angles with the help of scale and compass.

(02) Construct angles of 0°,15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180° with the help of scale and compass.

निम्न कोणों का परकार तथा पैमाना (स्केल) की सहायता से बनाओं।

0°,15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180°

Construct the following angles with the help of scale and compass and bisect each of them:

(03) Construct angles of 15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180° and bisect each of them with the help of scale and compass.

निम्न कोणों को बनाकर उनको परकार तथा पैमाना (स्केल) की सहायता से समद्विभाजित करो। 15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180°

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Construct the following segment with the help of scale and compass and bisect each of them:

(04) Construct the line segments of 5cm, 3cm, 4.5cm, 6cm, 7.5cm, 9cm, 10.5cm, 4.4cm, 5.5cm, 5.0cm, 6.6°, 8.2° and bisect each of them with the help of scale and compass.

निम्न रेखा खंडों को बनाकर उनको परकार तथा पैमाना (स्केल) की सहायता से समद्विभाजित करो। 5cm, 3cm, 4.5cm, 6cm, 7.5cm, 9cm, 10.5cm, 4.4cm, 5.5cm, 5.0cm, 6.6cm, 8.2cm।

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Rules and Properties of Triangles

★ A triangle with vertices P,Q,R

is denoted as ΔPQR.

★ By angle sum property, the sum of all angles in a triangle is 180∘.

★ In the triangle, the sum of any two sides is always greater than the third side.

★ The difference between any two sides of the triangle is always less than the third side.

★ An exterior angle of the triangle is the sum of opposite interior angles.

★ The side opposite to the largest side of the angle is greater in a triangle.

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Type 01

Construct the triangles with SSS condition:

(i) Equilateral Triangle

(01) Construct an equilateral triangle with side 4 cm.

4 सेमी भुजा वाले समबाहू त्रिभुज की रचना करो।

(02) Construct an equilateral triangle with side 5 cm.

5 सेमी भुजा वाले समबाहू त्रिभुज की रचना करो।

(03) Construct an equilateral triangle with side 6 cm.

6 सेमी भुजा वाले समबाहू त्रिभुज की रचना करो।

(04) Construct an equilateral triangle with side AB = BC = CA = 6 cm.

भुजा AB = BC = CA = 6 cm वाले समबाहू त्रिभुज की रचना करो।

(05) Construct an equilateral triangle with side KL = LM = MK = 5 cm.

भुजा KL = LM = MK = 5 cm वाले समबाहू त्रिभुज की रचना करो।

(06) Construct an equilateral triangle with side XY = YZ = ZX = 4.5 cm.

भुजा XY = YZ = ZX = 4.5 cm वाले समबाहू त्रिभुज की रचना करो।

(ii) Scaline Triangle

(01) Construct an Scaline triangle with side 4 cm, 3cm 5cm.

4 cm, 3cm 5cm भुजा वाले विषमबाहू त्रिभुज की रचना करो।

(02) Construct an Scaline triangle with side 4 cm, 6cm 10cm cm.

4 cm, 6cm 10cm भुजा वाले विषमबाहू त्रिभुज की रचना करो।

(03) Construct an Scaline triangle with side 5 cm, 6cm 7cm cm.

5 cm, 6cm 7cm भुजा वाले विषमबाहू त्रिभुज की रचना करो।

(04) Construct an Scaline triangle with side AB = 3cm, BC = 4cm, CA = 5 cm.

भुजा AB = 3cm, BC = 4cm, CA = 5 cm वाले विषमबाहू त्रिभुज की रचना करो।

(05) Construct an Scaline triangle with side KL = 5cm, LM = 6cm, MK = 7 cm.

भुजा KL = 5cm, LM = 6cm, MK = 7 cm वाले विषमबाहू त्रिभुज की रचना करो।

(06) Construct an Scaline triangle with side XY = 6cm, YZ = 7cm, ZX = 8 cm.

भुजा XY = 6cm, YZ = 7cm, ZX = 8 cm वाले विषमबाहू त्रिभुज की रचना करो।

(iii) Isoscales Triangle

(01) Construct an Scaline triangle with side 3 cm, 3cm, 5cm.

4 cm, 3cm 5cm भुजा वाले विषमबाहू त्रिभुज की रचना करो।

(02) Construct an Scaline triangle with side 6 cm, 6cm, 8cm.

6 cm, 6cm, 8cm भुजा वाले विषमबाहू त्रिभुज की रचना करो।

(03) Construct an Scaline triangle with side 5 cm, 5cm, 7cm.

5 cm, 5cm, 7cm भुजा वाले विषमबाहू त्रिभुज की रचना करो।

(04) Construct an Scaline triangle with side AB = 3cm, BC = 3cm, CA = 5 cm.

भुजा AB = 3cm, BC = 4cm, CA = 5 cm वाले विषमबाहू त्रिभुज की रचना करो।

(05) Construct an Scaline triangle with side KL = 5cm, LM = 6cm, MK = 7 cm.

भुजा KL = 5cm, LM = 6cm, MK = 7 cm वाले विषमबाहू त्रिभुज की रचना करो।

(06) Construct an Scaline triangle with side XY = 6cm, YZ = 7cm, ZX = 8 cm.

भुजा XY = 6cm, YZ = 7cm, ZX = 8 cm वाले विषमबाहू त्रिभुज की रचना करो।

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Type 02

Construct the triangles with SAS condition:

Construction of triangle if two sides and included angle is given.

$\begin{aligned} △ A B C \end{aligned}$ is the required triangle.

(01) Construct a triangle with sides AB=3cm, AC=4cm and ∠A=30°

(02) Construct a triangle with sides AB=4cm, AC=4cm and ∠A=40°

(03) Construct a triangle with sides AB=3cm, BC=4cm and ∠B=45°

Practice

(i) PR=6.5cm, ∠P=120°, PQ=5.2cm

(ii) AB=7.0cm, ∠B=60°, BC=7cm

(iii) MN=4cm, LM=5cm, ∠M=90°

(iv) AB=5.0cm, ∠B=45°, BC=6cm

(v) MN=5cm, LM=4cm, ∠M=90°

(vi) XY=7cm, ∠X=65°, YZ=7cm

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(i) PR=6.5cm, ∠P=100°, ∠Q=30°

(ii) AB=7.0cm, ∠B=60°, ∠A=7cm

(iii) MN=4cm, ∠M=5cm, ∠N=90°

(iv) XY=8cm, ∠X=65°, YZ=7cm

(v) XY=8cm, ∠X=65°, YZ=7cm

(i) AB=5.0cm, ∠B=45°, BC=6cm

(ii) MN=5cm, LM=4cm, ∠M=90°

(iii) XY=7cm, ∠X=65°, YZ=7cm

(iv) AB=6.0cm, ∠B=45°, BC=6cm

(v) MN=7cm, LM=4cm, ∠M=90°

(vi) XY=8cm, ∠X=65°, YZ=7cm

Type 03

Construction of Triangles – ASA

Construction of triangle, when two angles and one side are given, is explained by using an example:

Example: Construct a triangle with side $A B = 4 c m$ and $∠ A = 30^{\circ}, ∠ B = 60^{\circ}$

$∠ A = 30^{\circ}, ∠ B = 60^{\circ}$

Construction of Triangles – RHS

Let us discuss the construction of the triangle when the measurements of the side and hypotenuse of the right-angled triangle are given.

Example: Construct a triangle with sides AB=3cm,BC=5cmand∠A=90∘.

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(07) Constrict a triangle ∆ABC in which base BC=4.6cm. ∠B=45° and AB+CA=8.2cm.

एक ∆ABC की रचना करो जिसका आधार BC=4.6cm, आधार ∠B=45° और AB+CA=8.2cm.

(08) Constrict a triangle ∆ABC in which base RC=6 cm , ∠B=75° and AB+CA=14cm.

एक ∆ABC की रचना करो जिसका आधार BC=6 cm, और आधार कोण ∠B = 75° तथा AB+CA=14cm. है।

(09) Constrict a triangle triangle ∆PQR in which base PQ=4.6cm, ∠Q=45° and PQ+PR=7.9cm.

एक ∆PQR की रचना करो जिसका आधार PQ=4.6cm और आधार कोण ∠Q=45° तथा PQ+PR=7.9cm. है।

(10) Constrict a triangle ∆ABC in which base BC=5cm , ∠B=75° and AB +CA=9cm

एक ∆ABC की रचना जिसका आधार BC=5cm और आधार कोण ∠B = 75° तथा AB+CA=9cm. है।

(11) Construct a triangle whose base is 5 cm. The sum of the other two sides 7 cm, and one base angle of 60°.

एक त्रिभुज की रचना करो जिसका आधार 5 cm तथा जिसमे अन्य दो भुजाओं का योग 7 cm. और आधार कोण 60° है।

(12) Construct a right angle triangle when one side is 12 cm and the sum of the other side and hypotenuse is 18 cm.

एक समकोण त्रिभुज की रचना करो जिसका आधार 12 cm तथा जिसमे कर्ण व अन्य भुजा का योग 18 cm है।

(13) Construct a right angle triangle when one side is 3.5 cm and the sum of the other side and hypotenuse is 5.5 cm.

एक समकोण त्रिभुज की रचना करो जिसका आधार 3.5 cm तथा जिसमे कर्ण व अन्य भुजा का योग 5.5 cm है।

(14) Construct a triangle ∆ABC where BC=4.5cm, ∠B=45° and AB – AC=2.5cm.

एक ∆ABC की रचना करो जिसका आधार BC=4.5~cm आधार कोण ∠B=45° तथा AB–CA=2.5 cm. है।

(15) Construct a triangle ∆ABC where BC=8cm , ∠B=45° and AB–AC =3.5cm

एक ∆ABC की रचना करो जिसका आधार BC=8~cm आधार कोण ∠B=45° तथा AB–CA=3.5 cm. है।

(16) Construct a triangle ∆PQR where QR=6cm. , ∠Q=60° and PR–PO = 2 cm.

एक ∆PQR की रचना करो जिसका आधार QR=6cm. आधार कोण ∠Q=60° PR–PQ=2 cm है।

17) Construct a triangle ∆PQR where OR=3.4cm , ∠Q=75° and PR–PQ=1cm.

एक ∆PQR की रचना करो जिसका आधार OR=3.4 cm, आधार कोण ∠Q=75° तथा PR–PQ=1.2.

(18) Construct a triangle with base of length 7.5 cm, the difference of the other two sides 2.5 cm and one base angle of 45°.

एक त्रिभुज की रचना करो जिसका आधार 7.5 cm तथा जिसमे अन्य दो भुजाओं का अंतर 2.5 cm. और आधार कोण 45° ।

(19) Construct a triangle with base of length 5 cm, the difference of the other two sides 2 cm and one base angle of 30°.

एक त्रिभुज की रचना करो जिसका आधार 5 cm तथा जिसमे अन्य दो भुजाओं का अंतर 2 cm. और आधार कोण 30° है।

(20) Construct a ∆ABC where ∠B=90°, ∠C=45° and AB+BC+CA = 13 cm.

एक ∆ABC की रचना करो जिसका आधार कोण क्रमशः ∠B=90° , ∠C=45° तथा AB +BC+CA=13cm है।

(21) Construct a ∆ABC where ∠B=60°, ∠C=45° and AB+BC+CA= 12 cm.

एक ∆ABC की रचना करो जिसका आधार कोण क्रमशः ∠B=60°, ∠C=45° तथा AB +BC+CA=12cm है।

(22) Construct a ∆ABC where ∠B=60°, ∠C=45° and AB+BC+CA=

10 cm. एक ∆ABC की रचना करो जिसका आधार कोण क्रमशः ∠B=60° ∠C=45° तथा AB +BC+CA=10 cm है।

(23) Construct a ∆XYZ where ∠Z=90° , ∠Y=30° and XY+YZ+ZX= 11 cm.

एक ∆XYZ की रचना करो जिसका आधार कोण क्रमशः ∠Z=90° , ∠Y=30° तथा XY +YZ+ZX=11cm. है।

(24) Construct a triangle whose perimeter is 12 cm, and base angles are 60° and 45°.

एक त्रिभुज की रचना करो जिसका परिमाप 12 cm तथा जिसमे अन्य आधार कोण कमशः 60° और 45° ।

(25) Construct a triangle with perimeter 10 cm and base angles 60° and 45°.

एक त्रिभुज की रचना करो जिसका परिमाप 10 cm तथा जिसमे अन्य आधार कोण क्रमशः 60° और 45° है।

CHAPTER 11 CONSTRUCTIONS

EXERCISE 11.1

1. Construct an angle of 90° at the initial point of a given ray and justify the construction.

2. Construct an angle of 45° at the initial point of a given ray and justify the construction.

3. Construct the angles of the following measurements:

(i) 15°

(ii) 22½°

(iii) 30°

4. Construct the following angles and verify by measuring them by a protractor:

(i) 75°

(ii) 105°

(iii) 135°

5. Construct an equilateral triangle, given its side and justify the construction.

Example 1: Construct a triangle ∆ABC, in which ∠B=60°, ∠C=45° and AB+BC+CA=11 cm.

EXERCISE 11.2

1. Construct a triangle ∆ABC in which BC=7cm ∠B=75° and AB+AC=13

cm. 2. Construct a triangle ∆ABC in which BC=8cm, ∠B=45° and AB–AC= 3.5 cm.

3. Construct a triangle ∆PQR in which QR=6cm, ∠O=60° and PR–PQ= 2cm.

4. Construct a triangle ∆XYZ in which ∠Y=30° ∠Z=90° and XY+YZ+ ZX=11 cm.

5. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Practice Worksheet

Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 3 cm.

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

1. A triangle ABC can be constructed in which AB = 6 cm, ∠A=45° and BC + AC = 6 cm.

2. A triangle ABC can be constructed in which BC = 5 cm, ∠C=30° and AC - AB = 4 cm.

3. A triangle ABC can be constructed in which ∠B=105°, ∠C=90° and AB + BC + AC = 12 cm.

4. A triangle ABC can be constructed in which ∠B=60°, ∠C=45° and AB + BC + AC =10cm.

5. With the help of a ruler and a compass, it is possible to construct an angle of :

(A) 55° (B) 40° (C) 37.5° (D) 27.5°

6. The construction of a triangle ABC in which AB = 4 cm, <A = 60° is not possible when dif erence of BC and AC is equal to:

(A) 3.5 cm (B) 4.5 cm (C) 3 cm(D) 2.5 cm

7. The construction of a triangle ABC, given that BC = 6 cm, <B = 45° is not possible when dif erence of AB and AC is equal to:

(A) 6.9 cm (B) 5.2 cm (C) 5.0 cm (D

Example 1:

Construct a triangle ABC whose side lengths are 3 cm, 5 cm and 6 cm.

Construction:

Step 1: Draw the longest side using ruler. (i.e) AB = 6 cm.

Step 2: Take a compass, and draw an arc above line AB from point A, whose measurement is 5 cm.

Step 3: Similarly from Point B, draw an arc whose measurement is 3 cm. (Note: Draw an arc in such a way that both the arcs intersect at a point)

Step 4: Mark the intersection point as C and join CA and CB using a ruler.

Hence, ABC is the required triangle.

Construction of Triangle – SAS Criteria

In the construction of the SAS triangle, we need to know the side lengths of two sides and the angle between them are required.

Example 2:

Construct a triangle ABC, whose side lengths are 4 cm and 6 cm and the angle between them is 40°.

Construction:

Step 1: Draw the longest side of the triangle using a ruler. (i.e AC = 6 cm)

Step 2: Place the centre point of the protractor on point A and measure 40°. (i.e) Use inner reading and count 0° from the horizontal line to 40°. Mark this point as B.

Step 3: Using a ruler, draw a line AB, such that AB = 4 cm.

Step 4: Now draw the third side of the triangle by joining points B and C.

Hence, ABC is the required triangle.

Construction of Triangle – ASA Criteria

In the construction of a triangle based on ASA criteria, we need to know the measurement of two angles and the side length between them is required.

Example 3:

Construct a triangle whose two angle measurements are 40° and 70° and the side length between them is 8 cm.

Construction:

Step 1: Draw the line of length 8 cm using a ruler. (i.e) AB = 8 cm.

Step 2: Place the centre of the protractor on point A and measure 40° [Use the inner reading]. Now, put the construction mark at 40°.

Step 3: Using the ruler, draw a long line from A through the construction mark.

Step 4: Again, place the centre of the protractor on point B and measure 70° [Use the outer reading]. Now, put a mark at 70° and name the intersection point as C.

Step 5: Now, draw a line by joining points B and C.

Hence, ABC is the required triangle.

9.3

Construction of SAS Triangle

Last Modified: Aug 04, 2022

Lesson

SAS stands for "side-angle-side". Note that SAS is not the same as SSA because in SAS, the angle is included between the given sides.

Steps to Construct SAS Triangle

If two sides and included angle is given, then we follow the following steps for construction.

Step 0: Draw a rough sketch of the triangle ABC with required measures $\begin{aligned} A B C \end{aligned}$

Step 1: Draw a line segment AB of the given length.

Step 3: Draw a ray with measurement of given angle.

Step 4: With $\begin{aligned} A \end{aligned}$ as the centre and radius $\begin{aligned} A C, \end{aligned}$ draw an arc intersecting ray $\begin{aligned} A X \end{aligned}$ at a point $\begin{aligned} C . \end{aligned}$

Step 5: Join point $\begin{aligned} B \end{aligned}$ and point $\begin{aligned} C . \end{aligned}$

$\begin{aligned} △ A B C \end{aligned}$ is the required triangle.

Construction of SAS Triangle - Examples

Example 1

Construct $\begin{aligned} △ X Y Z \end{aligned}$ such that $\begin{aligned} ∠ Y = 60^{\circ}, X Y = 6 cm \end{aligned}$ and $\begin{aligned} Y Z = 8 cm . \end{aligned}$

Step 1: Draw a rough sketch of $\begin{aligned} △ X Y Z . \end{aligned}$

Step 2: Draw a line segment $\begin{aligned} Y Z \end{aligned}$ of length 8 cm.

Step 3: Construct a ray $\begin{aligned} Y A \end{aligned}$ such that $\begin{aligned} ∠ Z Y A = 60^{\circ} . \end{aligned}$

Step 4: With $\begin{aligned} Y \end{aligned}$ as centre and radius 6 cm, draw an arc intersecting ray $\begin{aligned} Y A \end{aligned}$ at a point $\begin{aligned} X . \end{aligned}$

Step 5: Join point $\begin{aligned} Z \end{aligned}$ with point $\begin{aligned} X . \end{aligned}$

$\begin{aligned} △ X Y Z \end{aligned}$ is the required triangle.

Example 2

Construct a $\begin{aligned} △ A B C \end{aligned}$ such that $\begin{aligned} A B = 11 cm, B C = 8 cm \end{aligned}$ and $\begin{aligned} ∠ A B C = 70^{\circ} . \end{aligned}$

i. Measure and write down the length of $\begin{aligned} A C . \end{aligned}$

ii. Construct the perpendicular bisector of $\begin{aligned} A B \end{aligned}$ such that it cuts $\begin{aligned} A C \end{aligned}$ at $\begin{aligned} S . \end{aligned}$ Measure and write down the length of $\begin{aligned} B S . \end{aligned}$

i. Steps of Construction:

Step 1: Draw a rough sketch of $\begin{aligned} △ A B C . \end{aligned}$

Step 2: Draw a line segment $\begin{aligned} A B \end{aligned}$ of length 11 cm.

Step 3: Draw a ray $\begin{aligned} B Z \end{aligned}$ such that $\begin{aligned} ∠ A B Z = 70^{\circ} . \end{aligned}$

Step 4: With $\begin{aligned} B \end{aligned}$ as centre and radius 8 cm, draw an arc intersecting ray $\begin{aligned} B Z \end{aligned}$ at a point $\begin{aligned} C . \end{aligned}$

Step 5: Join $\begin{aligned} A \end{aligned}$ with $\begin{aligned} C . \end{aligned}$

ii. Steps of Construction:

Step 1: With $\begin{aligned} A \end{aligned}$ as centre and radius more than half the length of $\begin{aligned} A B, \end{aligned}$ draw an arc above $\begin{aligned} A B \end{aligned}$ and another arc below $\begin{aligned} A B . \end{aligned}$

Step 2: Keeping the width of the compass the same, with $\begin{aligned} B \end{aligned}$ as the centre draw arcs intersecting the previous arcs at $\begin{aligned} X \end{aligned}$ and $\begin{aligned} Y . \end{aligned}$

Step 3: Join $\begin{aligned} X Y \end{aligned}$ to obtain the perpendicular bisector of $\begin{aligned} A B . \end{aligned}$

Step 4: Extend the perpendicular bisector of $\begin{aligned} A B \end{aligned}$ such that it cuts $\begin{aligned} A C \end{aligned}$ at $\begin{aligned} S . \end{aligned}$ Join $\begin{aligned} B S \end{aligned}$ using a dotted line.

Length of $\begin{aligned} B S = 7.3 cm . \end{aligned}$

Construction of SAS Triangle - Review Questions

1. Construct a $\begin{aligned} △ A B C \end{aligned}$ such that $\begin{aligned} A B = 6 cm, B C = 8 cm \end{aligned}$ and $\begin{aligned} ∠ A B C = 45^{\circ} . \end{aligned}$

2. Construct an isosceles triangle in which the length of equal sides is 5.2 cm and the vertex angle is $\begin{aligned} 72^{\circ} . \end{aligned}$ Measure the other two angles.

3. Construct $\begin{aligned} △ P Q R \end{aligned}$ such that $\begin{aligned} P Q = 8 cm, P R = 5 cm \end{aligned}$ and $\begin{aligned} ∠ Q P R = 54^{\circ} . \end{aligned}$ Measure and write down the length of $\begin{aligned} Q R . \end{aligned}$

4. Construct $\begin{aligned} △ A B C \end{aligned}$ such that $\begin{aligned} A B = 10 cm, B C = 7 cm \end{aligned}$ and $\begin{aligned} ∠ A B C = 72^{\circ} . \end{aligned}$

Measure the length of $\begin{aligned} A C . \end{aligned}$
Construct the perpendicular bisector of $\begin{aligned} A B \end{aligned}$ such that it cuts $\begin{aligned} A C . \end{aligned}$ Measure the length of $\begin{aligned} B S \end{aligned}$ such that $\begin{aligned} S \end{aligned}$ is the point where the perpendicular bisector of $\begin{aligned} A B \end{aligned}$ cuts $\begin{aligned} A C . \end{aligned}$

5. Construct $\begin{aligned} △ A B C \end{aligned}$ such that $\begin{aligned} A B = 9 cm, A C = 8 cm \end{aligned}$ and $\begin{aligned} ∠ A B C = 60^{\circ} . \end{aligned}$

Measure the size of the angle facing the shortest side.
Construct the angle bisector of $\begin{aligned} ∠ B A C \end{aligned}$ such that it cuts $\begin{aligned} B C . \end{aligned}$ Measure the length of $\begin{aligned} C S, \end{aligned}$ such that $\begin{aligned} S \end{aligned}$ is the point where the angle bisector of $\begin{aligned} ∠ B A C \end{aligned}$ cuts $\begin{aligned} B C . \end{aligned}$