NCERT Solutions for Class 7 Maths Exercise 10.4

NCERT Solutions for Class 7 Maths Exercise 10.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 7 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Practical Geometry Download as PDF

NCERT Solutions for Class 7 Maths Practical Geometry

Class –VII Mathematics (Ex. 10.4)

Question 1.Construct {tex}\Delta {/tex}ABC, given {tex}m\angle {/tex}A = {tex}{60^ \circ },{/tex}{tex}m\angle {/tex}B = {tex}{30^ \circ }{/tex} and AB = 5.8 cm.

Answer:

To construct: {tex}\Delta {/tex}ABC where {tex}m\angle {/tex}A = {tex}{60^ \circ },{/tex}{tex}m\angle {/tex}B = {tex}{30^ \circ }{/tex} and AB = 5.8 cm.

Steps of construction:

(a) Draw a line segment AB = 5.8 cm.

(b) At point A, draw an angle {tex}\angle {/tex}YAB = {tex}{60^ \circ }{/tex} with the help of compass.

(c) At point B, draw {tex}\angle {/tex}XBA = {tex}{30^ \circ }{/tex} with the help of compass.

(d) AY and BX intersect at the point C.

It is the required triangle ABC.

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NCERT Solutions for Class 7 Maths Exercise 10.4

Question 2.Construct {tex}\Delta {/tex}PQR if PQ = 5 cm, {tex}m\angle {/tex}PQR = {tex}105^\circ {/tex} and {tex}m\angle {/tex}QRP = {tex}40^\circ .{/tex}

Answer:

Given: {tex}m\angle {/tex}PQR = {tex}105^\circ {/tex} and {tex}m\angle {/tex}QRP = {tex}40^\circ {/tex}

We know that sum of angles of a triangle is {tex}180^\circ .{/tex}

{tex}\therefore {/tex} {tex}m\angle {/tex}PQR + {tex}m\angle {/tex}QRP + {tex}m\angle {/tex}QPR = {tex}180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}105^\circ + 40^\circ + m\angle {/tex}QPR = {tex}180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}145^\circ {/tex} + {tex}m\angle {/tex}QPR = {tex}180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}m\angle {/tex}QPR = {tex}180^\circ {/tex} – {tex}145^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}m\angle {/tex}QPR = {tex}35^\circ {/tex}

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NCERT Solutions for Class 7 Maths Exercise 10.4

Question 3.Examine whether you can construct {tex}\Delta {/tex}DEF such that EF = 7.2 cm, {tex}m\angle {/tex}E = {tex}110^\circ {/tex} and {tex}m\angle {/tex}F = {tex}80^\circ .{/tex} Justify your answer.

Answer:

To construct: {tex}\Delta {/tex}PQR where {tex}m\angle {/tex}P = {tex}35^\circ {/tex}, {tex}m\angle {/tex}Q = {tex}105^\circ {/tex} and PQ = 5 cm.

Steps of construction:

(a) Draw a line segment PQ = 5 cm.

(b) At point P, draw {tex}\angle {/tex}XPQ = {tex}35^\circ {/tex} with the help of protractor.

(c) At point Q, draw {tex}\angle {/tex}YQP = {tex}105^\circ {/tex} with the help of protractor.

(d) XP and YQ intersect at point R.

It is the required triangle PQR.

NCERT Solutions for Class 7 Maths Exercise 10.4

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