Mathematics Solutions Solutions for Class 7 Maths Chapter 1 Geometrical Constructions are provided here with simple step-by-step explanations. These solutions for Geometrical Constructions are extremely popular among Class 7 students for Maths Geometrical Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Maths Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 2:

#### Question 1:

#### 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.

#### Page No 2:

#### Question 2:

^{° }(2) 55

^{°}(3) 90

^{°}

#### 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.

#### Page No 2:

#### Question 3:

#### 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.

#### Page No 2:

#### Question 4:

#### 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.

#### Page No 2:

#### Question 5:

#### 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.

#### Page No 4:

#### Question 1:

*l*(AB) = 5.5 cm,

*l*(BC) = 4.2 cm,

*l*(AC) = 3.5 cm

*l*(ST) = 7 cm,

*l*(TU) = 4 cm,

*l*(SU) = 5 cm

*l*(PQ) = 6 cm,

*l*(QR) = 3.8 cm,

*l*(PR) = 4.5 cm

#### Answer:

(a) In $\u2206$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.

*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.

*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.

#### Page No 4:

#### Question 2:

Draw an isosceles triangle with base 5 cm and the other sides 3.5 cm each.

#### 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. $\u25b3$RPQ is the required isosceles triangle.

#### Page No 4:

#### Question 3:

Draw an equilateral triangle with side 6.5 cm.

#### 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. $\u25b3$ABC is the required equilateral triangle.

#### Page No 4:

#### Question 4:

Choose the lengths of the sides yourself and draw one equilateral, one isosceles and one scalene 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. $\u25b3$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. $\u25b3$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. $\u25b3$ZXY is the required isosceles triangle.

#### Page No 5:

#### Question 1:

Draw triangles with the measures given below.

*l*(MA) = 5.2 cm,

*m*∠A = 80

^{°},

*l*(AT) = 6 cm

#### Answer:

Steps of construction

1. Draw a line AM = 5.2 cm.

2. With A as centre, draw an angle of $80\xb0$ 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.

#### Page No 5:

#### Question 2:

Draw triangles with the measures given below.

*m*∠T = 40

^{°},

*l*(NT) =

*l*(TS) = 5 cm

#### Answer:

Steps of construction

1. Draw a line TS = 5 cm.

2. With T as centre, draw an angle of 40$\xb0$ 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.

#### Page No 5:

#### Question 3:

Draw triangles with the measures given below.

*l*(FU) = 5 cm,

*l*(UN) = 4.6 cm,

*m*∠U = 110

^{°}

#### Answer:

Steps of Construction:

1. Draw a line UN = 4.6 cm

2. With U as centre, draw an angle of 110$\xb0$ 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.

#### Page No 5:

#### Question 4:

Draw triangles with the measures given below.

*l*(RS) = 5.5 cm,

*l*(RP) = 4.2 cm,

*m*∠R = 90

^{°}

#### Answer:

Steps of construction:

1. Draw a line RS = 5.5 cm.

2. With R as centre, draw an angle of 90$\xb0$ 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.

#### Page No 6:

#### Question 1:

Construct triangles of the measures given below.

In ∆SAT,* l*(AT) = 6.4 cm, *m *∠A = 45^{°}, *m *∠T = 105^{°}

#### Answer:

Steps of construction:

1. Draw a line AT = 6.4 cm.

2. With A as centre, draw ∠XAT = 45$\xb0$.

3. With T as centre, draw ∠YTA = 105º.

4. Let YT and XA meet at point S.

∆SAT is the required triangle.

#### Page No 6:

#### Question 2:

Construct triangles of the measures given below.

*l*(NP) = 5.2 cm,

*m*∠N = 70

^{°},

*m*∠P = 40

^{°}

#### Answer:

Steps of construction:

1. Draw a line NP = 5.2 cm.

2. With N as centre, draw ∠XNP = 70$\xb0$.

3. With P as centre, draw ∠YPN = 40º.

4. Let YP and XN meet at point M.

∆MNP is the required triangle.

#### Page No 6:

#### Question 3:

Construct triangles of the measures given below.

In ∆EFG, *l*(EG) = 6 cm, *m*∠F = 65^{°},* m*∠G = 45^{°}

#### Answer:

Using the angle sum property, we can find the third angle of the triangle FEG.

$\angle \mathrm{F}+\angle \mathrm{E}+\angle \mathrm{G}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 65\xb0+\angle \mathrm{E}+45\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{E}=70\xb0$

Steps of construction:

1. Draw a line EG = 6 cm.

2. With E as centre, draw ∠XEG = 70$\xb0$.

3. With G as centre, draw ∠YGE = 45º.

4. Let YG and XE meet at point F.

∆FEG is the required triangle.

#### Page No 6:

#### Question 4:

Construct triangles of the measures given below.

*l*(XY) = 7.3 cm,

*m*∠X = 34

^{°},

*m*∠Y = 95

^{°}

#### Answer:

Steps of construction:

1. Draw a line XY = 7.3 cm.

2. With X as centre, draw ∠BXY = 34$\xb0$.

3. With Y as centre, draw ∠AYX = 95º.

4. Let BX and AY meet at point Z.

∆XYZ is the required triangle.

#### Page No 6:

#### Question 1:

Construct triangles of the measures given below.

*m*∠MAN = 90

^{°},

*l*(AN) = 8 cm,

*l*(MN) = 10 cm.

#### 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.

#### Page No 6:

#### Question 2:

Construct triangles of the measures given below.

In the right-angled ∆STU, hypotenuse SU = 5 cm and *l*(ST) = 4 cm.

#### 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.

#### Page No 6:

#### Question 3:

Construct triangles of the measures given below.

*l*(AC) = 7.5 cm,

*m*∠ABC = 90

^{°},

*l*(BC) = 5.5 cm.

#### 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.

#### Page No 6:

#### Question 4:

Construct triangles of the measures given below.

*l*(PQ) = 4.5 cm,

*l*(PR) = 11.7 cm,

*m*∠PQR = 90

^{°}.

#### 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.

#### Page No 6:

#### Question 5:

Students should take examples of their own and practise construction of triangles.

#### Answer:

*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.

#### Page No 8:

#### Question 1:

Write the names of pairs of congruent line segments. (Use a divider to find them.)

#### 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)

#### Page No 8:

#### Question 2:

On the line below, the distance between any two adjoining points shown on it is equal. Hence, fill in the blanks.

#### Answer:

(i) seg AB ≅ seg WA

(ii) seg AP ≅ seg YC

(iii) seg AC ≅ seg PY

(v) seg YA ≅ seg YQ

(vi) seg BW ≅ seg ZQ

#### Page No 10:

#### Question 1:

Some angles are given below. Using the symbol of congruence write the names of the pairs of congruent angles in these figures.

#### Answer:

OB is the angle bisector of $\angle $AOC.

So, $\angle $AOB = $\angle $BOC = 45$\xb0$

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 $PQR = 90$\xb0$

$\angle $AOC$\cong $$\angle $PQR

$\angle $DOC = $\angle $LMN = 30º

So, $\angle $DOC$\cong $$\angle $LMN

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