Class 6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.4

1. Draw a circle of the following radius:

Question (i)
3.5 cm
Solution:
Steps of construction.

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measure OA = 3.5 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (ii)
4 cm
Solution:
Steps of construction.

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measure OA = 4 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob.

Question (iii)
2.8 cm
Solution:
Steps of construction

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measures OA = 2.8 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (iv)
4.7 cm
Solution:
Steps of construction

1. Mark a point O on the page of your note book.
2. Take compasses fixed with sharp pencil and measures OA = 4.7 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (v)
5.2 cm.
Solution:
Steps of construction

1. Mark a point O on the page of your note book.
2. Take compasses fixed with sharp pencil and measures OA = 5.2 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

2. Draw a circle of diameter 6 cm.
Solution:
Steps of construction

1. Draw a line segment PQ = 6 cm.
2. Draw the perpendicular bisector of PQ intersecting PQ at O.
3. With O as centre and radius = OQ = 3 cm (= OP), draw a circle.
The circle thus drawn is the required circle.

3. With the same centre O, draw two concentric circles of radii 3.2 cm and 4.5 cm.
Solution:
Steps of Construction

1. Mark a point O on the page of your note book, where a circle is drawn.
2. Take compasses fixed with sharp pencil measuring OA = 4.5 cm using scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw complete arc by holding the compasses from its knob.
After completing one round, we get circle I.
4. Again with the same centre O and new radius = 3.2 cm draw another circle II following the same step 3.

4. Draw a circle of radius 4.2 cm with centre at O. Mark three points A, B and C such that point A is on the circle, B is in the interior and C is in the exterior of the circle.
Solution:
Steps of Construction

1. Mark a point O on the page of your note book, where a circle is to be drawn
2. Take compasses fixed with sharp pencil and measure OA = 4.2 cm using scale (∴ A is on the circle).
3. Without changing the opening of the compasses, keep the needle at point O and draw complete arc by rotating the compasses from the knob. After completing one round, we get required circle.
4. Mark point B in the interior of the circle and point C in the exterior of the circle.

5. Draw a circle of radius 3 cm and draw any chord. Draw the perpendicular bisector of the chord. Does the perpendicular bisector passes through the centre?
Solution:
Steps of Construction.

1. Draw a circle with C as centre and radius 3 cm.
2. Draw AB the chord of the circle.
3. Draw PQ the perpendicular bisector of chord AB.
4. We see that the perpendicular bisector of chord AB passes through the centre C of the circle.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.3

1. Draw a line r and mark a point P on it. Construct a line perpendicular to r at point P.

Question (i)
Using a ruler and compasses.
Solution:
Using ruler and compasses

Steps of Construction.

1. Draw a line r and mark a point P on it.

2. Draw an arc from P to the line r of any suitable radius which intersects line r at A and B.

3. Draw arcs of any radius which is more than half of arc made in step (2) from A and B which intersect at Q.

4. Join PQ.

Thus PQ is perpendicular to AB or line l or PQ ⊥ A.
Here P is called foot of perpendicular.

Question (ii)
Using a ruler and a set square.
Solution:
Using a ruler and a set square

Steps of Construction

1. Draw a line r and a point P on it.
2. Place one of the edges of a ruler along the line l and hold if firmly.

3. Place the set square in such a way that one of its edges contaning the right angle coincides with the ruler.
4. Holding the ruler, slide the set square along the line l till the vertical side reaches the point P.

5. Firmly hold the set square in this position. Draw PQ along its vertical edge. Now PQ is the required perpendicular to l ie. PQ ⊥ r.

2. Draw a line p and mark a point z above it. Construct a line perpendicular to p, from the point z.

Question (i)
Using a ruler and compasses.
Solution:
1. Draw a line p and mark a point z not lying on it.
2. From point z draw an arc which intersects line p at two points P and Q.


3. Using any radius and taking P and Q as centre, draw two arcs that intersect at point say B. On the other side (a shown in figure).
4. Join AB to obtain altitude to the line p.

Thus xz is altitude to line p.
i.e. xz ⊥ p.

Question (ii)
Using a ruler and set square
Solution:
Steps of constructions:
1. Draw a line p and mark a point z which is not lying on it.
2. Place one of the edge of a ruler along the line p and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Holding the ruler firmly, slide the set square along the line p till its vertical side reaches the point z.

5. Firmly hold the set square in this position, Draw xz along its vertical edge. Now xz is the required altitude to p i.e. xz ⊥ p.

3. Draw a line AB and mark two points P and Q on either side of line AB, Construct two lines perpendicular to AB, from P and Q using a ruler and compasses.
Solution:
1. Draw a line AB and Mark two points P and Q on either side of AB.

2. From point P draw an arc which intersect line AB at two points C and D.
3. Using any radius and taking C and D as centre draw two arcs that intersects at point say E on the other side as shown in figures.

4. Join PE to obtain perpendicular to AB.
5. From point Q draw an arc which intersects AB at two points X and Y.

6. Using any radius and taking X and Y as centre draw two arcs that intersects at point say R on the other side of line AB as shown in figures.
7. Join QR to obtain perpendicular to AB.
Thus, PE ⊥ AB and QR ⊥ AB

4. Draw a line segment of 7 cm and draw perpendicular bisector of this line segment.
Solution:
Steps of Construction:
1. Draw a line segment AB = 7 cm.
2. With A as centre and radius more than half of AB, draw an arc on both sides of AB.

3. With B as centre and the same radius as in step 2, draw an arc intersecting the first arc at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

5. Draw a line segment PQ = 6.8 cm and draw its perpendicular bisector XY which bisect PQ at M. Find the length of PM and QM. Is PM = QM ?
Solution:
Steps of Construction:

1. Draw a line segment PQ = 6.8 cm
2. With P as centre and radius more than half of PQ draw arcs on both sides of PQ.

3. Now with Q as centre and the same radius as in step 2 draw arcs intersecting the previous drawn arcs at A and B respectively.
4. Join AB intersecting PQ at M. Then M bisects the line segment.
5. Measure the length of PM and QM
PM = 3.4 cm and QM = 3.4 cm
∴ PM = QM.

6. Draw perpendicular bisector of line segment AB = 5.4 cm. Mark point X anywhere on perpendicular bisector Join X with A and B. Is AX = BX ?
Solution:
Steps of construction.
1. Draw a line segment AB = 5.4 cm.
2. With A as centre and radius more than half of AB, draw an arc in both sides of AB.

3. With B as centre and the same radius as in step 2, draw an arc intersecting the first arc at C and D.
4. Join CD intersecting AB at O.
Then CD is the perpendicular bisector of AB.
Mark any point X on the perpendicular bisector CD. Drawn. Then join AX and BX.
On examination, we find that AX = BX.

7. Draw perpendicular bisectors of line segment of the following lengths.

Question (i)
8.2 cm
Solution:
Steps of Construction.
1. Draw a line regment AB = 8.2 cm

2. With A as centre and radius more than half of AB, draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw an arcs intersecting the previous arc at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

Question (ii)
7.8 cm
Solution:
Steps of Construction.

1. Draw a line segment AB = 7.8 cm

2. With A as centre and radius more than half of AB draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw arcs intersecting the previous arcs at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

Question (iii)
6.5 cm.
Solution:
Steps of Construction.
1. Draw a line segment AB = 6.5 cm

2. With A as centre and radius more than half of AB draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2 draw arcs intersecting the previous arcs at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

8. Draw a line segment of length 8 cm and divide it into four equal parts Using compasses. Measure each part.
Solution:
Steps of construction.

1. Draw a line segment AB of length 8 cm
2. With A as centre and radius more than half of AB, draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw arcs intersecting the previous arcs at P and Q.
4. Join PQ intersecting AB at C then PQ is the perpendicular bisector of AB intersecting AB at C.
5. Similarly draw the perpendicular bisector of AC intersecting AC at D.
6. Draw the perpendicular bisector of CB intersecting CB at E.
By actual measurement, it can be verified that
AD = DC = CE = EB

Punjab State Board PSEB 6th Class Hindi Book Solutions Hindi Grammar Vakyansh Bodhak Shabd वाक्यांशों के लिए एक शब्द Exercise Questions and Answers, Notes.

PSEB 6th Class Hindi Grammar वाक्यांश बोधक शब्द

वाक्यांश बोधक शब्द

जो कभी जन्म न ले – अजन्मा, अज
जो दिखाई न दे – अदृश्य
जो पढ़ा हुआ न हो – अनपढ़
जिसका आदि न हो – अनादि
जिसका कोई शत्रु पैदा न हुआ हो – अजातशत्रु
जो परीक्षा में पास न हो – अनुत्तीर्ण
जो विद्या ग्रहण करे – विद्यार्थी
जो बिना वेतन के काम करे – अवैतनिक
जो दूर की न सोचे – अदूरदर्शी
जिसका अन्त न हो – अनन्त

जिसको कोई जीत न सके – अजेय
जिस पर विश्वास न किया जा सके – अविश्वसनीय
बिना सोचे-समझे किया जाने वाला विश्वास – अन्धविश्वास
जो कानून के प्रतिकूल हो – अवैध
जिसके समान दूसरा कोई, न हो – अद्वितीय
जो पहले कभी न हुआ हो – अभूतपूर्व
जो मनुष्यता के अनुकूल न हो – अमानवीय
जिसकी तुलना न की जा सके – अतुलनीय
जिसके माता-पिता न हों – अनाथ
जो ईश्वर को मानता हो – आस्तिक
जो सारे जन्म तक हो – आजन्म
जो अपने पर बीती हो – आपबीती
जिसने ऋण चुका दिया हो – उऋण
इतिहास से सम्बन्ध रखने वाला – ऐतिहासिक
जो काम करने से दिल चुराए – कामचोर
जो किए हुए उपकार को भूल जाए – कृतघ्न
जो काम करने वाला हो – कर्मठ
जिसकी चार भुजाएं हों – चतुर्भुज
जिसके मुँह से आग निकलती हो – ज्वालामुखी

जो जानने की इच्छा रखता हो – जिज्ञासु
जल में रहने वाला या घूमने वाला – जलचर
तीन महीने बाद आने वाला – त्रैमासिक
जहाँ कठिनता से पहुँचा जाए – दुर्गम
जिसमें दया हो – दयालु
जो दूर की बातें सोचता हो – दूरदर्शी
जो प्रतिदिन होता हो – दैनिक
पति और पत्नी का जोडा – दम्पति
जिसका कोई आकार न हो – निराकार
जो ईश्वर में विश्वास न करता हो – नास्तिक
जिसमें दया न हो – निर्दय
जिसमें कोई रस न हो – नीरस
जो (स्त्री) पति-भक्त हो – पतिव्रता
जो आँखों के सामने हो – प्रत्यक्ष
जो आँखों के पीछे हो – परोक्ष
जो दूसरों का भला करे – परोपकारी
केवल फल खाने वाला – फलाहारी

जो अनेक रूप धारण कर सकता हो – बहुरूपिया
जो मीठा बोलने वाला हो – मधुरभाषी
मांस खाने वाला – मांसाहारी
सोच-समझकर खर्च करने वाला – मितव्ययी
जो एक मास में होता हो – मासिक
जिससे कुछ लाभ प्राप्त हो – लाभदायक
जिसकी पत्नी मर गई हो – विधुर
जिसका पति मर गया हो – विधवा
जिस पर विश्वास किया जा सके – ‘विश्वसनीय
विष से भरा हुआ – विषैला
ऐश्वर्य में लीन रहने वाला – विलासी
जो व्यक्ति बहुत बोलने वाला हो – वाचाल
जो वर्ष में एक बार हो – वार्षिक
केवल सब्जी खाने वाला – शाकाहारी

जिसका पति जीवित हो – सधवा
जिसमें सहनशक्ति हो – सहनशील
जो सप्ताह में एक बार हो – साप्ताहिक
जो अच्छे आचरण वाला हो – सदाचारी
जो सब कुछ जानने वाला हो – सर्वज्ञ
जो साथ पढ़ता हो – सहपाठी
जो स्वयं सेवा करता हो – स्वयंसेवक

Punjab State Board PSEB 6th Class Hindi Book Solutions Hindi Grammar Shuddh-Ashuddh शुद्ध-अशुद्ध Exercise Questions and Answers, Notes.

PSEB 6th Class Hindi Grammar शुद्ध-अशुद्ध

अशुद्ध – शुद्ध
प्रयाप्त – पर्याप्त
आधीन – अधीन
शात्र – छात्र
श्रृंगार – शृंगार
बहीन – बहन
प्रगट – प्रकट
प्रमात्मा – परमात्मा
गियान -ज्ञान
पत्नीयों – पत्नियों
जाग्रति – जागृति
श्राप – शाप
परिक्षा – परीक्षा
सुचि – सूची
स्वास्थ – स्वास्थ्य
जन्ता – जनता
अनीवार्य – अनिवार्य

अवश्यकता – आवश्यकता
प्रमाणु – परमाणु
बुद्यिमान – बुद्धिमान्
मीठाई – मिठाई
प्रन्तु – परन्तु
शरन – शरण
मिठाईयां – मिठाइयाँ
अनन्द – आनन्द
जित – जीत
पेंसल – पेंसिल
कवी – कवि
आश्य – आशय
प्रमान – प्रमाण
दुश्ट – दुष्ट
शान्ती – शान्ति
पैत्रिक – पैतृक
स्मृद्धि – समृद्धि
स्वयंम्बर – स्वयंवर
कवित्री – कवयित्री
औषधी – औषधि

उज्जवल- उज्ज्वल
सुन्द्र – सुन्दर
परीश्रम – परिश्रम
नदि – नदी
दुरोपयोग – दुरुपयोग
न्यायधीश – न्यायाधीश
आग्या – आज्ञा
दूख – दुःख
परार्थना – प्रार्थना
कृश्ण – कृष्ण
महातमा – महात्मा
प्रतीख्श – प्रतीक्षा
लक्षमी – लक्ष्मी
पंडत – पंडित
मन्दर – मन्दिर
हिरदा – हृदय
सुरेन्दर – सुरेन्द्र
अमोद-परमोद – आमोद-प्रमोद
गुपत – गुप्त
त्यारी – तैयारी
गणेश – गणेश
सुतंतर – स्वतंत्र

Punjab State Board PSEB 6th Class Hindi Book Solutions Hindi Grammar Paryayvachi ya Samanarthak Shabd पर्यायवाची या समानार्थक शब्द Exercise Questions and Answers, Notes.

PSEB 6th Class Hindi Grammar पर्यायवाची या समानार्थक शब्द

अग्नि – अनल, पावक, आग, दहन, ज्वाला।
अपमान – अनादर, तिरस्कार, निरादर।
अमृत – सुधा, पीयूष, सोम।
असुर – राक्षस, दैत्य, दानव, दनुज ।
अन्धकार – अन्धेरा, तम, तिमिर।
अश्व – घोड़ा, वाजी, घोटक।
आंख – नेत्र, चक्षु, नयन, लोचन।
आकाश – गगन, अम्बर, नभ, आसमान, अन्तरिक्ष ।
इनाम – पुरस्कार, पारितोषिक, आनन्दकर।
इच्छा – अभिलाषा, चाह, मनोरथ, कामना, लालसा।
इन्द्र – देवेन्द्र, सुरेन्द्र, सुरपति, देवराज।
ईश्वर – ईश, परमात्मा, परमेश्वर, प्रभु।
उषा – प्रभात, सवेरा, निशान्त।

उद्देश्य – मकसद, लक्ष्य, ध्येय।
उपकार – हित, भलाई, नेकी, भला।
कपड़ा – वस्त्र, अम्बर, पट, वसन।
कमल – जलज, नलिन, पंकज, सरोज, राजीव, नीरज।
कान – कर्ण, श्रोता, श्रवण।
किनारा – तट, तीर, कूल।
किरण – रश्मि, अंशु, कर।
केश – बाल, अलक. कच।
कोयल – पिक, कोकिल।
क्रोध – गुस्सा, रोष, कोप।
गंगा – भागीरथी, देवनदी, सुरसरी, नदीश्वरी।
गृह – घर, सदन, निकेतन, भवन, वास, आलय, शाला।
चन्द्रमा – शशि, चन्द्र, राकेश, चाँद, सोम।
जल – नीर, पानी, पय, रस।
तलवार – खड्ग, कृपाण, करवाल।
तालाब – सर, सरोवर, जलाशय, ताल ।

दास – नौकर, सेवक, अनुचर।
देवता – सुर, अमर, देव।
दुनिया – विश्व, संसार, जगत्, जग, भूमण्डल।
दूध – दुग्ध, गोरस, पय, क्षीर।
नदी – सरिता, जलमाला, नद, तटनी।
नारी – स्त्री, महिला, अबला, वनिता।
नाव – नौका, तरिणी, जलयान, बेड़ा।
पवन – हवा, वायु, समीर, अनिल।
पत्नी – वधू, गृहिणी, स्त्री, प्राणप्रिया, भार्या ।
पति – स्वामी, नाथ, प्राणनाथ।
पर्वत – गिरि, पहाड़, शैल, नग, अचल।
पक्षी – खग, पतंग, नभचर। पुत्र-सुत,बेटा, लड़का, पूत।
पुत्री – सुता, बेटी, लड़की।
पुष्प – कुसुम, सुमन, फूल, प्रसून ।
पृथ्वी – भू, भूमि, धरती, वसुधा, धरा।

बाग – बगीचा, वाटिका, उपवन, उद्यान।
बादल – मेघ, घन, जलद, नीरद।
माता – जननी, माँ, मात, मैया, अम्ब ।
मृत्यु – निधन, देहान्त, अन्त, मौत।
मनुष्य – मनुज, नर, आदमी, मानव, पुरुष।
युवक – जवान, युवा, तरुण।
राजा – नृप, नरेश, सम्राट्, नरेन्द्र, नरपति।
रात – रजनी, निशा, रात्रि।

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.2

1.

Question (i)
(a) Which of the following are simple curves?
(b) Classify the following as open or closed curve.

Solution:
(a) Simple curves :
(i), (iii), (iv), (vi), (vii), (iii)

(b) Open curves :
(iii) , (vi), (viii)
Closed curves :
(i), (ii), (iv), (v), (vii)

2. Identify the polygons:

Question (i)

Solution:
(ii), (iii), (v) are polygons.

3. Draw any polygon and shade its interior.
Solution:

4. Name the points which are:

Question (i)
In the interior of the closed figure.
Solution:
Points in the interior of closed figure are :
A B, Q

Question (ii)
In the exterior of the closed figure,
Solution:
Points in the exterior of closed figure are :
R, N

Question (iii)
On the boundary of the closed figure.
Solution:
Points in the boundary of closed figure are :
P, M

5. In the given figure, name :

Question (i)
The vertices
Solution:
Vertices are :
D, E, A, B, C

Question (ii)
The sides
Solution:
Sides are :
AB, BC, CD, DE, EA

Question (iii)
The diagonals
Solution:
Diagonals are :
AC, AD, BE, BD, CE

Question (iv)
Adjacent sides of AB
Solution:
Adjacent sides of AB are :
AE and BC

Question (v)
Adjacent vertices of E.
Solution:
Adjacent vertices of E are :
A and D.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.1

1. Give the examples of:

Question (i)
A point
Solution:
A point. Point A •
Examples:
(i) A small dot marked by a sharp pencil on a sheet of paper.
(ii) A tiny prick made by a fine needle or pin on a paper.
(iii) Bindi.
(iv) A star in the sky.

Question (ii)
A line segment
Solution:
A line segment
Examples:
(i) An edge of a box.
(ii) A tube light.
(iii) The edge of a postcard.
(iii) Parallel lines.

Question (iii)
Parallel lines
Solution:
Examples:
(i) The opposite edges of ruler (scale).
(ii) The crossbars of window.
(iii) The opposite edges of blackboard.
(iv) Rail lines.
(iv) Interescting lines.

Question (iv)
Intersecting lines
Solution:
Examples:
(i) Two adjacent edges of your notebook.
(ii) The letter X of the English alphabet.
(iii) Crossing roads.

Question (v)
Concurrent lines.
Solution:
Concurrent lines.
(i) Three angle bisectors of a triangle.
(ii) Three medians of a triangle.
(iii) Three perpendiculars of a triangle.
(iv) The intersection of the three walls of a room.

2. Name the lines segments in given lines.

Solution:
AB, AC, AD, BC, BD, CD are line segments.

3. How many lines can pass through a point?
Solution:
Infinite lines can pass through a point.

4. How many points lie on line?
Solution:
Infinite points lie on a line.

5. How many lines pass through two points?
Solution:
One and only one line passes through two points.

6. Use the figure to name:

Question (i)
Five Points
Solution:
Five Points are :
O, A, B, C, D, or E

Question (ii)
A line
Solution:
BE is the line.

Question (iii)
Four rays
Solution:
Four rays are :
\(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}, \overrightarrow{\mathrm{OC}}, \overrightarrow{\mathrm{OD}} \text { or } \overrightarrow{\mathrm{OE}}\)

Question (iv)
Five line segments.
Solution:
Five line segments are :
OA, OB, OC, OD, OE, DE.

7. Name the given ray in all possible ways.

Solution:
The possible rays are:
\(\overrightarrow{\mathrm{PQ}}, \overrightarrow{\mathrm{PR}}, \overrightarrow{\mathrm{QR}}\)

8. Use the figure to name :

Question (i)
Pair of parallel lines.
Solution:
Pair of parallel lines are :
l and m.

Question (ii)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are:
p and n, n and l, n and m, p and l, p and m.

Question (iii)
Lines whose point of intersection is S.
Solution:
Lines whose point of intersection is S :
m and n.

Question (iv)
Collinear points.
Solution:
Collinear points are :
P, Q, S and P, R, T.

9. Use the figure to name:

Question (i)
All pairs of parallel lines.
Solution:
All pairs of parallel lines are : n and p, q and p, n and q.

Question (ii)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are :
m and l, m and n, m and p, m and q, l and n, l and p, l and q.

Question (iii)
Lines whose point of intersection is D.
Solution:
Lines whose point of intersection is D are :
p and l.

Question (iv)
Point of intersection of lines m and p.
Solution:
Point of intersection of lines m and p is E.

Question (v)
All sets of collinear points.
Solution:
All sets of collinear points are :
G, E, C, A and F, D, C, B.

10. Use the figure to name:

Question (i)
Line containing point P.
Solution:
Lines containing point are : l, n.

Question (ii)
Lines whose point of intersection is B.
Solution:
Lines whose point of intersection is B are : l and m.

Question (iii)
Point of intersection of lines m and l.
Solution:
Point of intersection of lines m and l is : B.

Question (iv)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are : m and l, n and l.

11. State which of the following statements are True (T) or False (F):

Question (i)
Two lines in a plane, always intersect at a point
Solution:
False

Question (ii)
If four lines intersect at a point, those are called concurrent lines.
Solution:
True

Question (iii)
Point has a size because we can see it as a thick dot on the paper.
Solution:
False

Question (iv)
Through a given point, only one line can be drawn.
Solution:
False

Question (v)
Rectangle is a part of the plane.
Solution:
True.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 7 Algebra MCQ Questions

Multiple Choice Questions

Question 1.
Each side of square is represented by ‘s’ then perimeter of square is:
(a) 4 + s
(b) s – 4
(c) 4s
(d) s.
Answer:
(c) 4s

Question 2.
Write commutative property of multiplication using variables x and y:
(a) xy = yx
(b) x + y = y + x
(c) x + y
(d) xy.
Answer:
(a) xy = yx

Question 3.
How many terms in expression 7l – 3l?
(a) 1
(b) 3
(c) 2
(d) 4.
Answer:
(c) 2

Question 4.
5 is subtracted from m = ……………….. .
(a) 5 – m
(b) m + 5
(c) 5 + m
(d) m – 5.
Answer:
(d) m – 5.

Question 5.
Multiply p by 3 then 2 is added = ……………. .
(a) 2p + 3
(b) 3p – 2
(c) 3p + 2
(d) 2p – 3.
Answer:
(c) 3p + 2

Question 6.
If Armaan’s present age is x years then what will be his age after 4 years?
(a) x – 4
(b) x + 4
(c) 4x
(d) 4 – x.
Answer:
(b) x + 4

Question 7.
Write as algebraic equation: 7 more than 4 times ofy gives 23 :
(a) 4 + 7y = 23
(b) 7 + y = 23
(c) 4y – 7 = 23
(d) 4y + 7 = 23.
Answer:
(d) 4y + 7 = 23.

Question 8.
Find x if x – 3 = 2 :
(a) 3
(b) 6
(c) 5
(d) 2.
Answer:
(c) 5

Question 9.
Solve:
4l – 3 = 5
(a) 3
(b) 4
(c) 1
(d) 2.
Answer:
(d) 2.

Question 10.
If \(\frac {a}{4}\) = 5 then a = ……………. .
(a) 5
(b) 20
(c) 4
(d) 18.
Answer:
(b) 20

Question 21.
What is algebraic expression for subtracting 7 from – m?
(a) m – 1
(b) m + 7
(c) 7 – m
(d) – m – 7.
Answer:
(d) – m – 7.

Question 22.
What is algebraic expression for subtracting 7 from p?
(a) p – 7
(b) p + 7
(c) 7 – p
(d) 7 × p.
Answer:
(a) p – 7

Question 23.
What is algebraic expression for multiplying p by 16?
(a) 16 p
(b) p + 6
(c) p – 16
(d) \(\frac {p}{16}\)
Answer:
(a) 16 p

Question (iv)
What is algebraic expression for first multiplying x by 3 and then adding 2 to the product?
(a) x + 6
(b) 3x + 2
(c) 3x – 2
(d) 6x
Answer:
(b) 3x + 2

Question (v)
What is algebraic expression for first multiplying y by 2 and then subtracting 5 from the product?
(a) 2y + 5
(b) y + 10
(c) 2y – 5
(d) 10y.
Answer:
(c) 2y – 5

Fill in the blanks:

Question (i)
The algebraic expression for first multiplying y by 10 and then adding 7 to the product is ………… .
Answer:
10y + 7

Question (ii)
The algebraic expression for first multiplying n by 2 and then subtracting l from the product ………… .
Answer:
2n – l

Question (iii)
7 × 20 – 82 is expression of only ………………. .
Answer:
Numbers

Question (iv)
Each side of a square is l, then perimeter of square is ……………. .
Answer:
4l

Question (v)
5 is added to x = …………….. .
Answer:
x + 5

Write True/False:

Question (i)
If \(\frac {a}{5}\) = 4, then a = 20. (True/False)
Answer:
True

Question (ii)
If x – 3 = 2, then x = l. (True/False)
Answer:
False

Question (iii)
If 4l – 3 = 5, then l = 2. (True/False)
Answer:
True

Question (iv)
Number of terms in expression 3p + 2 is two. (True/False)
Answer:
True

Question (v)
The letters which can take any numerical value are called variables. (True/False)
Answer:
True

Punjab State Board PSEB 6th Class Hindi Book Solutions Hindi Grammar Visheshan Rachana विशेषण रचना Exercise Questions and Answers, Notes.

PSEB 6th Class Hindi Grammar विशेषण रचना

शब्द – विशेषण
अज्ञान – अज्ञात
अन्तर – आन्तरिक
अपमान – अपमानित
अभिमान – अभिमानी
अवश्य – आवश्यक
अन्त – अन्तिम
आदर – आदरणीय
इच्छा – इच्छुक
इतिहास – ऐतिहासिक
ईर्ष्या – ईर्ष्यालु
ईश्वर – ईश्वरीय
उद्यम – उद्यमी
उदास – उदासी
घर – घरेलू
घना – घनिष्ठ
घाव – घायल

चमक – चमकीला, चमकदार
चाचा – चचेरा
चिकित्सा – चिकित्सक
चाह – चाहत, चहेता
जंगल – जंगली
जाति – जातीय
एक – अकेला, एकाकी
ऐश्वर्य – ऐश्वर्यवान्
केन्द्र – केन्द्रीय
कथन – कथित
कांटा – कंटीला
काल – कालीन
कोढ़ – कोढ़ी
कुल – कुलीन
कृपा – कृपालु
खोज – खोजी
खेलना – खिलाड़ी
गहराई – गहरा
गाना – गवैया

गाँव – गँवार
ग्राम – ग्रामीण
दुध – दुधारू
धन – धनी
दो – दूसरा
नगर – नागरिक
नमक – नमकीन
नोक – नुकीला
निन्दा – निन्दनीय
जहर – ज़हरीला
जागना – जागरूक
जोश – जोशीला
झूठ – झूठा
झगड़ा – झगड़ालू
डर – डरावना
तप – तपस्वी
तेज – तेजस्वी
दया – दयालु
दान – दानी
पाँच – पाँचवाँ
पिता – पैतृक
पुजा – पुजारी
दिन – दैनिक
द्रोह – द्रोही

देश – देशीय
यश – यशस्वी
रस – रसीला
रक्ष – रक्षक
रंग – रंगीन
सम्मान – सम्मानित
सभा – सभ्य
स्थान – स्थानीय
सन्तोष – सन्तुष्ट
सच्च – सच्चा
समाज – सामाजिक
सप्ताह – साप्ताहिक
सम्बन्ध – सम्बन्धी
स्वर्ग – स्वर्गीय
स्वदेश – स्वदेशी
स्वाद – स्वादिष्ट
लाख – लखपति
श्रीमान् – श्रद्धालु
शारीर – शारीरिक
नीचे – नीचा
नाटक – नाटकीय
नील – नीला
नशा – नशीला
पतन – पतित
पर्वत – पर्वतीय

परिश्रम – पारिश्रमिक
परिवार – पारिवारिक
पहरा – पहरेदार
पाप – पापी
पालन – पालक
प्यास – प्यासा
मास – मासिक
मौखिक – मौलिक
मूल्य – मूल्यवान्
भार – भारी
भारत – भारतीय
भागना – भगोड़ा
भीख – भिखारी
भूख – भूखा
भूल – भुलक्कड़
मधु – मधुर
मास – मासिक
सोना – सुनहरी

संसार – सांसारिक
साँप – सपेरा
सेना – सैनिक
स्वभाव – स्वाभाविक
सुर – सुरीला
विदेश – विदेशी
वन – वन्य
विज्ञान – वैज्ञानिक
शीत – शीतल
शाक्त – शक्ति
वन – वनैला
बात – बातूनी
बल – बलवान्
बाहर – बाहरी
बुद्धि – बुद्धिमान्
बुराई – बुरा
विष – विषैला
विरोध – विरोधी
वर्ष – वार्षिक
विष्णु – वैष्णव
लोभ – लोभी

लूट – लुटेरा
लाज – लजीला
लज्जा – लज्जालु
लड़ना – लडाक
हिंसक – हँसोड़
बर्फ – बर्फीला
हत्या – हत्यारा
संस्कृति – सांस्कृतिक
हिंसा – हँसी

Punjab State Board PSEB 6th Class Hindi Book Solutions Hindi Grammar Bhav vachak Sangyaen भाववाचक संज्ञाएँ Exercise Questions and Answers, Notes.

PSEB 6th Class Hindi Grammar भाववाचक संज्ञाएँ

शब्द – भाववाचक
अज्ञान – अज्ञानता
अपना – अपनापन
अच्छा – अच्छाई
अधिक – अधिकता
उजाला – उजालापन
उलझन – उलझाव
उठना – उठाव
ऊँचा – ऊँचाई
एक – एकता
कठिन – कठिनता
कठोर – कठोरता
अंगड़ाना – अंगड़ाई
आलसी – आलस्य
आवश्यक – आवश्यकता
आगे-पीछे – आगा-पीछा

इन्सान – इन्सानियत
ईश्वर – ऐश्वर्य
उदार – उदारता
गुरु – गौरव
गिरावट – गिरना
गर्म – गर्मी
गाना – गान, गीत
कड़ा – कड़ाई
कंजूस – कंजूसी
कायर – कायरता
शिशु – शैशव
कुशल – कुशलता
कृत्रिम – कृत्रिमता
कृपण – कृपणता
कमाना – कमाई
काटना – कटौती
खट्टा – खटाई, खटास
खुजलाना – खुजली
खेलना – खेल
खोदना – खुदाई
गहरा – गहराई

ठण्डा – ठण्डक
ठगना – ठगी
डाकू – डाका, डकैती
ढीठ – ढिठाई
तेज – तेजी
तीव्र – तीव्रता
दास – दासता
देव – देवत्व
दुर्बल – दुर्बलता
दुष्ट – दुष्टता
दौड़ाना – दौड़
दुहराना – दुहराई
धोना – धुलाई
नीच – नीचता
पण्डित – पण्डिताई, पाण्डित्य
प्यासा – प्यास
पूर्ण – पूर्णता
पकड़ना – पिसाई पहचान
पहचानना – पकड़

मीठा – मिठास
पीसना – पीसाई
गुलाम – गुलामी
घबराना – घबराहट
घेरना – घिराव
चतुर – चतुराई
चालाक – चलाकी
चौड़ा – चौड़ाई
चढ़ना – चढ़ाई
चलना – चाल
चिकना – चिकनाई
चमकना – चमक
चिल्लाना – चिल्लाहट
चुनना – चुनाव
जागना – जागरण
जलना – जलन
जीना – जीवन
परतन्त्र – परतन्त्रता
पालना – पालन
बच्चा – बचपन
बालक – बालकपन
बुद्धिमान् – बुद्धिमत्ता
बूढ़ा – बुढ़ापा
बनाना – बनावट

बैठना – बैठक
बहना – बहाव
बुरा – बुराई
बोलना – बोल
नम्र – नम्रता
बेचना – बेच
भला – भलाई
भूखा – भूख
भय – भयानक
भिड़ना – भिड़न्त
मित्र – मित्रता
मधुर – मधुरता
महान् – महत्ता
मुस्कराना – मुस्कराहट
मूर्ख – मूर्खता
मोटा – मुटापा, मोटापन
मिलना – मिलावट

सुन्दर – सुन्दरता
स्वतन्त्र – स्वतन्त्रता
सहायक – सहायता
सेवक – सेवा, सेवकाई
सज्जन – सज्जनता
स्वस्थ – स्वास्थ्य
मारना – मार
राजा – राज्य
लड़का – लड़कपन
लिखना – लिखाई