Mathematics Solutions Solutions for Class 7 Maths Chapter 1 - Geometrical Constructions

Answer:

(1) 5.3 cm
Steps of construction:
1. Draw line segment AB = 5.3 cm.
2. With A as centre and radius more than half of AB, mark two arcs, one above and other below the line AB.
3. With B as centre and same radius, draw two arcs cutting the previous drawn arcs and name the point of intersection as X and Y.
4. Join XY and name the point where this line cuts AB as point P. 
XY is the perpendicular bisector of AB. 



(2) 6.7 cm
Steps of construction:
1. Draw line segment CD = 6.7 cm.
2. With C as centre and radius more than half of CD, mark two arcs, one above and other below the line CD.
3. With D as centre and same radius, draw two arcs cutting the previous drawn arcs and name the point of intersection as A and B.
4. Join AB and name the point where this line cuts CD as point Q. 
AB is the perpendicular bisector of CD. 


(3) 3.8 cm
Steps of construction:
1. Draw line segment XY = 3.8 cm.
2. With X as centre and radius more than half of XY, mark two arcs, one above and other below the line XY.
3. With Y as centre and same radius, draw two arcs cutting the previous drawn arcs and name the point of intersection as A and B.
4. Join AB and name the point where this line cuts XY as point R. 
AB is the perpendicular bisector of XY. 

Answer:

(1) 105°
Steps of Construction:
1. Draw a ray BC.
2. With B as centre, use a protractor to make an angle of 105°. Thus, $\angle$ABC = 105°. 
3. With X and Y as centre, draw arcs intersecting each other at point M. 
BM is the required angle bisector of $\angle$ABC.


(2) 55^°
Steps of Construction:
1. Draw a ray OX.
2. With O as centre, use a protractor to make an angle of 55°. Thus, $\angle$YOX = 55°. 
3. With P and Q as centre, draw arcs intersecting each other at point A. 
OA is the required angle bisector of $\angle$YOX.


(3) 90°
Steps of Construction:
1. Draw a ray OB.
2. With O as centre, use a protractor to make an angle of 90°. Thus, $\angle$AOB = 90°. 
3. With X and Y as centre, draw arcs intersecting each other at point S. 
OS is the required angle bisector of $\angle$AOB.

Answer:

I. Right angled triangle
Steps of construction
1. Draw a right angled triangle ABC, right angled at B. 
2. Make the angle bisectors of the angles A, B and C. 
The angle bisectors meet at the point O. This point of concurrence of the angle bisectors lies inside the triangle ABC. 


II. Obtuse-angled triangle
Steps of construction
1. Draw an obtuse angled triangle XYZ. 
2. Make the angle bisectors of angles X, Y and Z. 
The angle bisectors meet at point S. This point of concurrence of the angle bisectors lies inside the obtuse angled triangle XYZ. 

Answer:

Steps of construction
1. Draw the right angle triangle ABC. 
2. Draw the perpendicular bisectors of the sides AB, BC and CA. 
The perpendicular bisectors meet at the point D which lies on the hypotenuse AC. 

Answer:

Maithili, Shaila and Ajay be the three vertices of a triangle. 
A toy shop equidistant from these three points will be the point of concurrence of the perpendicular bisectors 
of the lines joining the three vertices of the triangle. Thus, the geometrical construction representing this will be the 
circumcircle. 
 

Answer:

(a)  In  $∆$ABC , l(AB) = 5.5 cm,l(BC) = 4.2 cm, l(AC) = 3.5 cm
Steps of construction
1. Draw a line AB = 5.5 cm
2. With A as centre and 3.5 cm as the radius, draw an arc above the line AB.
3. With B as the centre and 4.2 cm as the radius, draw an arc cutting the previous drawn arc at point C. 
4. Join CA and CB. 

$∆$ABC is thus the required triangle. 

 

(b)  In ∆ STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm
Steps of construction
1. Draw a line ST = 7 cm
2. With S as centre and 5 cm as the radius, draw an arc above the line ST.
3. With T as the centre and 4 cm as the radius, draw an arc cutting the previous drawn arc at point U. 
4. Join US and UT. 

$∆$STU is thus the required triangle. 

 

(c)  In ∆ PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm
Steps of construction
1. Draw a line PQ = 6 cm
2. With P as centre and 4.5 cm as the radius, draw an arc above the line PQ.
3. With Q as the centre and 3.8 cm as the radius, draw an arc cutting the previous drawn arc at point R. 
4. Join RP and RQ. 

$∆$PQR is thus the required triangle. 

Answer:

Steps of construction:
1. Draw a line PQ = 5 cm.
2. With P as centre and 3.5 cm as radius, draw an arc above the line PQ. 
3. With Q as centre and 3.5 cm as radius, draw an arc cutting the previous drawn arc. Name the point of intersection as point R.
Join RP and RQ. $△$RPQ is the required isosceles triangle. 

Answer:

Steps of construction:
1. Draw a line BC = 6.5 cm.
2. With B as centre and 6.5 cm as radius, draw an arc above the line BC. 
3. With C as centre and 6.5 cm as radius, draw an arc cutting the previous drawn arc. Name the point of intersection as point A.
Join AB and AC. $△$ABC is the required equilateral triangle. 

Answer:

1. Equilateral triangle
Steps of construction:
1. Draw a line BC = 4 cm.
2. With B as centre and 4 cm as radius, draw an arc above the line BC. 
3. With C as centre and 4 cm as radius, draw an arc cutting the previous drawn arc. Name the point of intersection as point A.
Join AB and AC. $△$ABC is the required equilateral triangle.


2. Isosceles triangle
Steps of construction:
1. Draw a line QR = 6 cm.
2. With Q as centre and 4 cm as radius, draw an arc above the line QR. 
3. With R as centre and 4 cm as radius, draw an arc cutting the previous drawn arc. Name the point of intersection as point P.
Join PQ and RP. $△$RPQ is the required isosceles triangle. 


3. Scalene triangle
Steps of construction
1. Draw a line XY = 5 cm.
2. With X as centre and 3 cm as radius, draw an arc above the line XY. 
3. With Y as centre and 4.5 cm as radius, draw an arc cutting the previous drawn arc. Name the point of intersection as point Z.
Join ZX and ZY. $△$ZXY is the required isosceles triangle. 

Answer:

Steps of construction
1. Draw a line AM = 5.2 cm.
2. With A as centre, draw an angle of $80°$ using protractor. Name this angle as $\angle$XAM.
3. With A as centre and 6 cm as radius, cut an arc on XA and name it as point T. 
4. Join TM. 
∆MAT is the required triangle.

 

Answer:

Steps of construction
1. Draw a line TS = 5 cm.
2. With T as centre, draw an angle of 40$°$ using protractor. Name this angle as $\angle$XTS.
3. With T as centre and 5 cm as radius, cut an arc on XT and name it as point N. 
4. Join NS. 
∆NTS is the required triangle.

Answer:

Steps of Construction:
1. Draw a line UN = 4.6 cm
2. With U as centre, draw an angle of 110$°$ using the protractor. Name the angle thus formed as $\angle$XUN. 
3. With U as centre and 5 cm radius, cut an arc on XU and name it as point F. 
4. Join FN. 
∆FUN is the required triangle. 

Answer:

Steps of construction:
1. Draw a line RS = 5.5 cm.
2. With R as centre, draw an angle of 90$°$ and name it as $\angle$XRS.
3. With R as centre and 4.2 cm radius in compass, cut an arc on XR and name it P.
Join PS. 
∆PRS is the required triangle. 

Answer:

Steps of construction:
1. Draw a line AT = 6.4 cm.
2. With A as centre, draw ∠XAT = 45$°$. 
3.  With T as centre, draw ∠YTA = 105º.
4. Let YT and XA meet at point S. 
∆SAT is the required triangle. 

Answer:

Steps of construction:
1. Draw a line NP = 5.2 cm.
2. With N as centre, draw ∠XNP = 70$°$. 
3.  With P as centre, draw ∠YPN = 40º.
4. Let YP and XN meet at point M. 
∆MNP is the required triangle. 

Answer:

Using the angle sum property, we can find the third angle of the triangle FEG. 
$$\begin{gathered} \angle \mathrm{F}+\angle \mathrm{E}+\angle \mathrm{G}=180° \\ \Rightarrow 65°+\angle \mathrm{E}+45°=180° \\ \Rightarrow \angle \mathrm{E}=70° \end{gathered}$$
Steps of construction:
1. Draw a line EG = 6 cm.
2. With E as centre, draw ∠XEG = 70$°$. 
3.  With G as centre, draw ∠YGE = 45º.
4. Let YG and XE meet at point F. 
∆FEG is the required triangle. 

Answer:

Steps of construction:
1. Draw a line XY = 7.3 cm.
2. With X as centre, draw ∠BXY = 34$°$. 
3. With Y as centre, draw ∠AYX = 95º.
4. Let BX and AY meet at point Z. 
∆XYZ is the required triangle. 

Answer:

Steps of construction
1. Draw a line AN = 8 cm.
2. With A as centre, draw ∠XAN = 90º.
3. With N as centre and 10 cm as radius, cut an arc on XA and name it as point M.
4. Join MN.
∆MAN is thus formed. 

Answer:

Steps of construction
1. Draw a line ST = 4 cm.
2. With T as centre, draw ∠XTS = 90º.
3. With S as centre and 5 cm as radius, cut an arc on XT and name it as point U.
4. Join US.
∆UST is thus formed. 

Answer:

Steps of construction
1. Draw a line BC = 5.5 cm.
2. With B as centre, draw ∠XBC = 90º.
3. With C as centre and 7.5 cm as radius, cut an arc on XB and name it as point A.
4. Join AC.
∆ABC is thus formed. 

Answer:

Steps of construction
1. Draw a line PQ = 4.5 cm.
2. With Q as centre, draw ∠XQP = 90º.
3. With P as centre and 11.7 cm as radius, cut an arc on XQ and name it as point R.
4. Join RP.
∆PQR is thus formed. 

Answer:

In ∆PQR, l(PR) = 8 cm, m ∠PQR = 90^° , l(QR) = 6 cm.
Steps of construction
1. Draw a line QR = 6 cm.
2. With Q as centre, draw ∠XQR = 90º.
3. With R as centre and 8 cm as radius, cut an arc on XQ and name it as point P.
4. Join PR.
∆PQR is thus formed. 

Answer:

(i) seg(MG) $\cong$ seg(GR)
(ii) seg(MG) $\cong$ seg(NG)
(iii) seg(GC) $\cong$ seg(GB)
(iv) seg(GE) $\cong$ seg(GR)

Answer:

Answer:

OB is the angle bisector of $\angle$AOC.
So, $\angle$AOB = $\angle$BOC = 45$°$
Thus, $\angle \mathrm{AOB}\cong \angle \mathrm{BOC}$
Also, $\angle \mathrm{AOB}\cong \angle \mathrm{SRT}$ and $\angle$BOC $\cong$$\angle$RST. 
 $\angle$AOC = $\angle$PQR = 90$°$
$\angle$AOC$\cong$$\angle$PQR
$\angle$DOC = $\angle$LMN = 30º
So, $\angle$DOC$\cong$$\angle$LMN