Practical Geometry Ex-10.4

NCERT solutions for Maths Practical Geometry 

NCERT Solutions for Class 7 Maths Practical Geometry

Class –VII Mathematics (Ex. 10.4)

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

Answer:

To construct: $\Delta$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 compass.

(c) At point B, draw $\angle$XBA = ${30^ \circ }$ 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 $\Delta$PQR if PQ = 5 cm, $m\angle$PQR = $105^\circ$ and $m\angle$QRP = $40^\circ .$

Answer:

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

We know that sum of angles of a triangle is $180^\circ .$

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

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

$\Rightarrow$ $145^\circ$ + $m\angle$QPR = $180^\circ$

$\Rightarrow$ $m\angle$QPR = $180^\circ$ – $145^\circ$

$\Rightarrow$ $m\angle$QPR = $35^\circ$

NCERT Solutions for Class 7 Maths Exercise 10.4

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

Answer:

To construct: $\Delta$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 protractor.

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

(d) XP and YQ intersect at point R.

It is the required triangle PQR.