NCERT Solutions for Class 7 Maths Chapter 10: Practical Geometry - Exercise 10.4

Exercise 10.4

1. Construct $\vartriangle ABC$, given $m\angle A = {60^ \circ }$, $m\angle B = {30^ \circ }$ and $AB = 5.8$cm.

Ans: To construct: $\vartriangle ABC$ where $m\angle A = {60^ \circ }$, $m\angle B = {30^ \circ }$ and $AB = 5.8$cm.

Steps of Construction:

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

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

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

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

This is the required triangle $ABC$.

2. Construct $\vartriangle PQR$ if $PQ = 5$cm, $m\angle PQR = {105^ \circ }$ and $m\angle QRP = {40^ \circ }$. 

Ans: 

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

The sum of measure of all the interior angles of a triangle is always ${180^ \circ }$.

$\therefore {\text{    }}m\angle PQR + m\angle QRP + m\angle QPR = {180^ \circ }$ 

Put the value of given angles and add then subtract ${145^ \circ }$ to obtain a measure of the unknown angle.

${105^ \circ } + {40^ \circ } + m\angle QPR = {180^ \circ }$

${145^ \circ } + m\angle QPR = {180^ \circ }$

$m\angle QPR = {180^ \circ } - {145^ \circ }$

$m\angle QPR = {35^ \circ }$ 

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

Steps of Construction:

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

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

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

(d) $XP$ and $YQ$ intersect at point $R$.

It is the required triangle $PQR$.

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

Ans: We know that the sum of all angles of a triangle is always ${180^ \circ }$.

And the sum of two given angles of triangle $\vartriangle DEF$ is ${110^ \circ } + {80^ \circ } = {190^ \circ }$ which exceeds${180^ \circ }$. 

Hence, the $\vartriangle DEF$ cannot be constructed.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4

Opting for the NCERT solutions for Ex 10.4 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.4 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Subject Practical Geometry textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 10 Exercise 10.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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