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**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.

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}

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

NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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