{"id":150477,"date":"2021-02-22T06:48:11","date_gmt":"2021-02-22T06:48:11","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=150477"},"modified":"2023-09-15T06:56:01","modified_gmt":"2023-09-15T06:56:01","slug":"ncert-solutions-for-7th-class-maths-chapter-10-practical-geometry","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-10-practical-geometry\/","title":{"rendered":"NCERT Solutions for 7th Class Maths: Chapter 10-Practical Geometry"},"content":{"rendered":"\n

Class 7: Maths Chapter 10 solutions. Complete Class 7 Maths Chapter 10 Notes.<\/p>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 10-Practical Geometry<\/strong><\/h2>\n\n\n\n

NCERT 7th Maths Chapter 10, class 7 Maths Chapter 10 solutions<\/p>\n\n\n\n

1. Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: A line, parallel to given line by using ruler and compasses.<\/p>\n\n\n\n

Steps of construction:<\/p>\n\n\n\n

(a) Draw a line-segment AB and take a point C outside AB.<\/p>\n\n\n\n

(b) Take any point D on AB and join C to D.<\/p>\n\n\n\n

(c) With D as centre and take convenient radius, draw an arc cutting AB at E and CD at F.<\/p>\n\n\n\n

(d) With C as centre and same radius as in step 3, draw an arc GH cutting CD at I.<\/p>\n\n\n\n

(e) With the same arc EF, draw the equal arc cutting GH at J.<\/p>\n\n\n\n

(f) Join JC to draw a line l. This the required line.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

2. Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: A line parallel to given line when perpendicular line is also given.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line l and take a point P on it.<\/p>\n\n\n\n

(b) At point P, draw a perpendicular line n.<\/p>\n\n\n\n

(c) Take PX = 4 cm on line n.<\/p>\n\n\n\n

(d) At point X, again draw a perpendicular line m.<\/p>\n\n\n\n

Given figure is the required construction.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

3. Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: A pair of parallel lines intersecting other part of parallel lines.<\/p>\n\n\n\n

Steps of construction:<\/p>\n\n\n\n

(a) Draw a line l and take a point P outside of l .<\/p>\n\n\n\n

(b) Take point Q on line l and join PQ.<\/p>\n\n\n\n

(c) Make equal angle at point P such that \u2220Q = \u2220P.<\/p>\n\n\n\n

(d) Extend line at P to get line m.<\/p>\n\n\n\n

(e) Similarly, take a point R online m, at point R, draw angles such that \u2220P = \u2220R.<\/p>\n\n\n\n

(f) Extended line at R which intersects at S online l. Draw line RS.<\/p>\n\n\n\n

Thus, we get parallelogram PQRS.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

NCERT 7th Maths Chapter 10, class 7 Maths Chapter 10 solutions<\/p>\n\n\n\n

Exercise 10.2<\/h4>\n\n\n\n

1. Construct \u0394XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct:  \u0394XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment YZ = 5 cm.<\/p>\n\n\n\n

(b) Taking Z as centre and radius 6 cm, draw an arc.<\/p>\n\n\n\n

(c) Similarly, taking Y as centre and radius 4.5 cm, draw another arc which intersects first arc at point X.<\/p>\n\n\n\n

(d) Join XY and XZ. It is the required \u0394XYZ.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

2. Construct an equilateral triangle of side 5.5 cm. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct:  A \u0394ABC where AB = BC = CA = 5.5 cm<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment BC = 5.5 cm<\/p>\n\n\n\n

(b) Taking points B and C as centers and radius 5.5 cm, draw arcs which intersect at point A.<\/p>\n\n\n\n

(c) Join AB and AC. It is the required \u0394ABC.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

3. Draw \u0394PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construction: \u0394PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm.<\/p>\n\n\n\n

Steps of construction:<\/p>\n\n\n\n

(a) Draw a line segment QR = 3.5 cm.<\/p>\n\n\n\n

(b) Taking Q as centre and radius 4 cm, draw an arc.<\/p>\n\n\n\n

(c) Similarly, taking R as centre and radius 4 cm, draw an another arc which intersects first arc at P.<\/p>\n\n\n\n

(d) Join PQ and PR. It is the required isosceles \u0394PQR.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

4. Construct \u0394ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure \u2220B. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct:  \u0394ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment BC = 6 cm.<\/p>\n\n\n\n

(b) Taking B as centre and radius 2.5 cm, draw an arc.<\/p>\n\n\n\n

(c) Similarly, taking C as centre and radius 6.5 cm, draw another arc which intersects first arc at point A.<\/p>\n\n\n\n

(d) Join AB and AC.<\/p>\n\n\n\n

(e) Measure angle B with the help of protractor. It is the required \u0394ABC where \u2220B = 80\u00b0.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

NCERT 7th Maths Chapter 10, class 7 Maths Chapter 10 solutions<\/p>\n\n\n\n

Exercise 10.3<\/h4>\n\n\n\n

1. Construct \u0394DEF such that DE = 5 cm, DF = 3 cm and  \u2220EDF = 90\u00b0.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: \u0394DEF where DE = 5 cm, DF = 3 cm and m \u2220EDF = 90\u00b0.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment DF = 3 cm.<\/p>\n\n\n\n

(b) At point D, draw an angle of 90o with the help of compass i.e., \u2220XDF = 90\u00b0.<\/p>\n\n\n\n

(c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E.<\/p>\n\n\n\n

(d) Join EF. It is the required right angled triangle DEF.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

2. Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110\u00b0.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and \u2220Q = 110\u00b0<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment QR = 6.5 cm.<\/p>\n\n\n\n

(b) At point Q, draw an angle of 110\u00b0 with the help of protractor, i.e., \u2220YQR = 110\u00b0<\/p>\n\n\n\n

(c) Taking Q as centre, draw an arc with radius 6.5 cm, which cuts QY at point P.<\/p>\n\n\n\n

(d) Join PR<\/p>\n\n\n\n

Given figure is the required isosceles triangle PQR.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

3. Construct \u0394ABC with BC = 7.5 cm, AC = 5 cm and m \u2220C = 60\u00b0 . <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: \u0394ABC where BC = 7.5 cm, AC = 5 cm and m \u2220C = 60\u00b0.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment BC = 7.5 cm.<\/p>\n\n\n\n

(b) At point C, draw an angle of 60 with the help of protractor, i.e., \u2220XCB = 60\u00b0 .<\/p>\n\n\n\n

(c) Taking C as centre and radius 5 cm, draw an arc, which cuts XC at the point A.<\/p>\n\n\n\n

(d) Join AB It is the required triangle ABC.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

NCERT 7th Maths Chapter 10, class 7 Maths Chapter 10 solutions<\/p>\n\n\n\n

Exercise 10.4<\/h4>\n\n\n\n

1. Construct \u0394ABC, given m\u2220A = 60\u00b0, m\u2220B = 30\u00b0 and AB = 5,8 cm.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: \u0394ABC, given m\u2220A = 60\u00b0, m\u2220B = 30\u00b0 and AB = 5,8 cm.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment AB = 5.8 cm.<\/p>\n\n\n\n

(b) At point A, draw an angle Z YAB = 60\u00b0= with the help of a compass.<\/p>\n\n\n\n

(c) At point B, draw Z XBA = 30\u00b0 with the help of a compass.<\/p>\n\n\n\n

(d) AY and BX intersect at the point C.<\/p>\n\n\n\n

Given figure is the required triangle ABC.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

2. Construct \u0394 PQR if PQ = 5 cm, m\u2220PQR = 105\u00b0 and m\u2220.QRP = 40\u00b0.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Given: <\/p>\n\n\n\n

m\u2220 PQR = 105\u00b0 and m\u2220 QRP = 40\u00b0<\/p>\n\n\n\n

We know that sum of angles of a triangle is 180\u00b0.<\/p>\n\n\n\n

\u2234 m\u2220PQR + m\u2220QRP + m\u2220 QPR = 180\u00b0<\/p>\n\n\n\n

\u21d2 105\u00b0 + 40\u00b0 + m\u2220QPR= 180\u00b0<\/p>\n\n\n\n

\u21d2 145\u00b0 + m\u2220QPR= 180\u00b0<\/p>\n\n\n\n

\u21d2  m\u2220QPR= 180\u00b0 – 145\u00b0<\/p>\n\n\n\n

\u21d2 m\u2220QPR = 35\u00b0<\/p>\n\n\n\n

To construct: \u0394PQR where m\u2220P = 35\u00b0, m\u2220Q = 105\u00b0 and PQ = 5 cm.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment PQ = 5 cm,<\/p>\n\n\n\n

(b) At point P, draw \u2220 XPQ = 35\u00b0 with the help of protractor.<\/p>\n\n\n\n

(c) At point Q, draw \u2220 YQP = 105\u00b0 with the help of protractor.<\/p>\n\n\n\n

(d) XP and YQ intersect at point R.<\/p>\n\n\n\n

It is the required triangle PQR.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

3. Examine whether you can construct \u0394DEF such that EF = 7.2 cm, m\u2220 E= 110\u00b0 and m\u2220F= 80\u00b0. Justify your answer.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Given: In \u0394DEF, m\u2220E = 110\u00b0 and m\u2220T = 80\u00b0<\/p>\n\n\n\n

Using angle sum property of triangle<\/p>\n\n\n\n

\u2220D + \u2220E + \u2220F = 180\u00b0<\/p>\n\n\n\n

\u21d2\u2220D + 110\u00b0+ 80\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u21d2 \u2220D + 190\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u21d2 \u2220D = 180\u00b0 – 190\u00b0 = -10\u00b0<\/p>\n\n\n\n

Which is not possible.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 10, class 7 Maths Chapter 10 solutions<\/p>\n\n\n\n

Exercise 10.5<\/h4>\n\n\n\n

1. A right angled triangle PQR where m\u2220Q = 9ff, QR= 8 cm and PQ = 10 cm.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: Construct the right angled \u0394 PQR, where m\u2220Q = 90″, QR = 8 cm and PR = 10 cm.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a)  Draw a line segment QR = 8 cm,<\/p>\n\n\n\n

(b) At point Q, draw QX \u22a5 QR,<\/p>\n\n\n\n

(c) Taking R as centre, draw an arc of radius 10 cm.<\/p>\n\n\n\n

(d) This arc cuts QX at point P.<\/p>\n\n\n\n

(e) foin PQ.<\/p>\n\n\n\n

Given figure is the required right-angled triangle PQR,<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

2. Construct a right-angled triangle whose hypotenuse is 6 cm long and one the legs is 4 cm long. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct: A right-angled triangle DEF where DF = 6 cm and EF = 4 cm.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment EF = 4 cm.<\/p>\n\n\n\n

(b) At point Q, draw EX \u22a5 EF.<\/p>\n\n\n\n

(c) Taking F as centre and radius 6 cm, draw an arc. (Hypotenuse)<\/p>\n\n\n\n

(d) This arc cuts the EX at point D.<\/p>\n\n\n\n

(e) Join DF.<\/p>\n\n\n\n

It is the required right-angled triangle DEF.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

3. Construct an isosceles right angled triangle ABC, where m\u2220ACB = 90\u00b0 and AC = 6 cm.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To construct:  An isosceles right angled triangle ABC where m\u2220 C = 90% AC = BC = 6 cm.<\/p>\n\n\n\n

Steps of construction: <\/p>\n\n\n\n

(a) Draw a line segment AC = 6 cm.<\/p>\n\n\n\n

(b) At point C, draw XC \u22a5 CA.<\/p>\n\n\n\n

(c) Taking C as centre and radius 6 cm, draw an arc.<\/p>\n\n\n\n

(d) This arc cuts CX at point B.<\/p>\n\n\n\n

(e) Join BA.<\/p>\n\n\n\n

It is the required isosceles right-angled triangle ABC.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 10: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 10-Practical Geometry<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 7th Class Maths: Chapter 10-Practical Geometry PDF<\/a><\/p>\n\n\n\n

Chapter-wise NCERT Solutions Class 7 Maths<\/h2>\n\n\n\n
Chapter 1 Integers<\/a><\/td><\/tr>
Chapter 2 Fractions and Decimals<\/a><\/td><\/tr>
Chapter 3 Data Handling<\/a><\/td><\/tr>
Chapter 4 Simple Equations<\/a><\/td><\/tr>
Chapter 5 Lines and Angles<\/a><\/td><\/tr>
Chapter 6 The Triangle and its Properties<\/a><\/td><\/tr>
Chapter 7 Congruence of Triangles<\/a><\/td><\/tr>
Chapter 8 Comparing Quantities<\/a><\/td><\/tr>
Chapter 9 Rational Numbers<\/a><\/td><\/tr>
Chapter 10 Practical Geometry<\/a><\/td><\/tr>
Chapter 11 Perimeter and Area<\/a><\/td><\/tr>
Chapter 12 Algebraic Expressions<\/a><\/td><\/tr>
Chapter 13 Exponents and Powers<\/a><\/td><\/tr>
Chapter 14 Symmetry<\/a><\/td><\/tr>
Chapter 15 Visualising Solid Shapes<\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n


<\/strong>About NCERT<\/h2>\n\n\n\n


<\/strong>About NCERT<\/h2>\n\n\n\n

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n

\n
NCERT Solutions<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Class 7: Maths Chapter 10 solutions. Complete Class 7 Maths Chapter 10 Notes. NCERT Solutions for 7th Class Maths: Chapter 10-Practical Geometry NCERT 7th Maths Chapter 10, class 7 Maths Chapter 10 solutions 1. Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler […]<\/p>\n","protected":false},"author":302,"featured_media":627323,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,907],"tags":[1550],"boards":[1180],"class_list":["post-150477","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-7","tag-ncert-maths-class-7","boards-ncert","entry"],"yoast_head":"\nNCERT Solutions Class 7, Maths Chapter 10 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 7th Class Maths: Chapter 10-Practical Geometry | Browse all Class 7 Maths Solutions of NCERT books- IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-10-practical-geometry\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for 7th Class Maths: Chapter 10-Practical Geometry\" \/>\n<meta property=\"og:description\" content=\"Class 7: Maths Chapter 10 solutions. Complete Class 7 Maths Chapter 10 Notes. 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