NCERT Solutions for Class 7 Maths Exercise 10.4 myCBSEguide App

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NCERT solutions for Maths Practical Geometry NCERT Solutions for Class 7 Maths Practical Geometry

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.

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.

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}

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} 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.

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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