NCERT Solutions For Class 7 Maths Chapter 10

NCERT Solutions Class 7 Maths Practical Geometry

NCERT Solutions For Class 7 Maths Chapter 10 Exercises

Exercise 10.1

Q1:

Draw a line and name it PQ, take a point R outside it. Through R, draw a line parallel to PQ using the ruler and compass only.

Solution:

To construct: A line, parallel to the given line by using ruler and compass.

Construction:

(i) Draw a line segment PQ and take a point R outside PQ.

(ii) Take any point X on PQ and join R to X.

(iii) With X as the center and take convenient radius, draw an arc cutting PQ at Y and RX at Z.

(iv) With R as center and the same radius as in step (iii), draw an arc AB cutting PX at O

(v) With the same arc YZ, draw the equal arc cutting AB at M.

(vi) Join MR to draw a line N.

This the required line PQ ǁ N

1.1

 

Q2 :

Draw a line N. Draw a perpendicular to N at any point on N. On this perpendicular choose a point P, 4cm away from N. Through A, draw a line P parallel to N.

Solution:

To construct: A line parallel to given line when perpendicular line is also given.

Construction:

(i) Draw a line N and take a point O on it.

(ii) At point O, draw a perpendicular line Q

(iii) Take OA = 4 cm on line Q.

(iv) At point A again draw a perpendicular line P

It is the required construction.

1.2

 

Q3:

Let X be a line and A be a point not on X. Through A, draw a line Y parallel to X. Now join A to any point B on X. Choose any point C on Y. Through C, draw a line parallel to AB. Let this meet X at P. What shape do the two sets of parallel lines enclose?

Solution:

To construct: A pair of parallel lines intersecting other part of parallel lines.

Construction:

(i) Draw a line X and take a point A outside of X

(ii) Take point B on line X and join AB

(iii) Make equal angle at point A such that Ð B = Ð A

(iv) Extend line at A to get line Y

(v) Similarly, take a point C which intersects at P on line X, draw line PC

Thus, we get parallelogram ABCP

 1.3

 

Exercise 10.2

 

Q1 :

Construct Δ PQR in which PQ = 4.5 cm , QR = 5 cm and RP = 6cm.

Solution:

Construct: Δ PQR, where PQ = 4.5 cm, QR = 5 cm, RP = 6 cm

Construction:

(i) Draw a line segment QR = 5 cm.

(ii) Taking R as center and radius 6cm, draw arc.

(iii) Similarly, taking Q as center and radius 4.5 cm  draw another arc which intersects the first arc at point P

(iv) Join PR and PR

It is the required Δ PQR

 2.1

 

Q2:

Construct an equilateral triangle of side 5.5 cm

Solution:

To construct: A Δ PQR where PQ = QR = QP = 5.5 cm

Construction :

(i) Draw a line segment BC = 5.5 cm

(ii) Taking points Q and R as centers and radius 5.5 cm , draw arcs which intersect at point P.

(iii) Join PQ and PR

It is the required Δ PQR

 2.2

 

Q3:

Draw Δ ABC with AB = 4cm, BC = 3.5 and AC = 4cm. What type of triangle is this?

Solution:

To construct: Δ ABC, in which AB = 4cm, BC = 3.5 and AC = 4 cm

Construction:

(i) draw a line segment BC = 3.5 cm

(ii) Taking B as center and radius 4 cm, draw an arc.

(iii) Similarly, taking C as center and radius 4 cm, draw another arc which intersects first arc at A.

(iv) Join AB and AC.

It is the required isosceles Δ ABC

2.3

 

 Q4:

Construct Δ PQR such that PQ = 2.5 cm , QR = 6 cm and PR = 6.5 cm. Measure Q

Solution:

To Construct:

Δ PQR such that PQ = 2.5 cm, QR = 6cm and PR = 6 cm and PR = 6.5 cm

Construction :

(i) Draw a line segment QR = 6cm

(ii) Taking B as center and radius 2.5 cm, draw an arc.

(iii) Similarly, taking C as center and radius 6.5 cm , draw another arc which intersects first arc at point P

(iv) Join PQ and PR

(v) Measure angle Q with the help of protractor

It is the required Δ PQR where Q = 80o

2.4

 

 

Exercise 10.3

 

Q1: Construct   a Δ  ABC so that AC = 5 cm, AB = 3 cm and CAB = 90O .

Solution:

Constructing a Δ DEF where DE = 5 cm, DF = 3 cm and m EDF = 90O .

Steps of construction:

(a) Draw a line segment AB = 3 cm.

(b) At point A, draw an angle of 90O with the help of compass i.e., XAB = 900.

(c) Taking A as the center, draw an arc of radius 5 cm, which cuts AX at the point E.

(d)Connect BC.

This  is the required right angled triangle ACB:

3.1

 

Q2: Construct an isosceles triangle which has equal sides of length  6.5 cm and with a angle of 110 ° between them.

Solution:

Constructing an isosceles triangle where AC = AB = 6.5 cm and A = 110 ° .

Steps of construction:

(a) Make a line segment AB= 6.5 cm.

(b) At  A, make a  110° with the help of protractor, i.e., XAB = 110 ° .

(c) Take A as the  center, and draw an arc of radius 6.5 cm, which cuts AX at point  C.

(d) Connect BC

This is the required isosceles Δ ABC:

3.2

 

Q3:

Construct a Δ ZXC with XC = 7.5 cm, ZC = 5 cm and C = 60O .

Solution:

Constructing a  Δ ZXC where XC = 7.5 cm, ZC = 5 cm and C = 60O.

Steps of construction:

(a) Make a line segment XC = 7.5 cm.

(b) At C, make an angle of 600 with the help of protractor, i.e., XCB = 60O .

(c) Take C as the center  and draw an arc of 5cm, which cuts the line YC at the point Z.

(d)Connect  AB

Thus, this is the required Δ ZXC:

3.3

 

Exercise 10.4

 

Q1:

Construct a triangle ZXC, where Z = 60O , X = 30O and ZX = 5.8 cm.

Solution:

Now, constructing a Δ ZXC where Z = 60O , X = 300 and ZX = 5.8 cm.

Steps for construction:

(a) Make a line segment ZX = 5.8 cm.

(b) At  Z, make an angle XZC = 60O  with a compass.

(c) At X, draw QXZ = 300  with  a compass.

(d) ZP and XQ intersect at the point C.

Thus we have the required triangle ZXC:

4.1

 

Q2:

Construct a triangle ABC if AB = 5 cm, ABC = 105° and BCA = 40°.

Solution:

Given:

ABC  = 105° and BAC = 40°

We know ,sum of the angles of a triangle = 180 .°

ABC + BCA + CAB = 180°

105. 40 °+ 40°+BCA = 180°

145° + BCA = 180°

BCA = 180° – 145°

BCA = 35°

Constructing a triangle ABC  whose A = 35°, B = 105° and AB = 5 cm.

Steps of construction:

(a) Make a line segment PQ = 5 cm.

(b) At  A, make an angle  QAB = 35° with the help of a protractor.

(c) At  B, make an angle ABP = 105° with the help of a protractor.

(d) QA and BP intersect at point C.

Thus, we have the required triangle ABC.

4.2

 

Q3:

Can you construct  a Δ ABC such that AB = 4 cm, B = 120° and C= 70 °. Explain your answer.

Solution:

Given:

In Δ ABC, A = 120° and m F = 70 °

We know the angles of a triangle sum up to 180o

Thus, we have

A + C+ B= 180°

A + 120° + 70° = 180°

A + 190° = 180°

 

A = 180° − 190° = −10°

Since the sum of the angles is coming more than 180o, this triangle construction is not possible.

 

Exercise 10.5

 

Q1:

Construct a right angled Δ ABC, where B = 90o  BC = 8 cm and AC = 10 cm

Solution:

To construct:

A right angled triangle ABC where B = 90o BC = 8cm and AC = 10cm

Construction:

(i) Draw a line segment BC = 8cm

(ii) At point B, draw BP    BC

(iii) taking C as the center, draw an arc of radius 10cm

(iv) this arc cuts BP at point X

(V) join XB

It is the required right angled triangle ABC.

5.1

 

Q2 :

Construct a right angled triangle whose hypotenuse is 6cm long and one of the legs is 4cm long.

Solution:

TO construct:

A right angled triangle ABC where AC = 6cm and BC = 4cm

Construction:

(i) Draw a line segment BC = 4cm

(ii) At point X, draw BX    BC

(iii) Taking C as the center and radius 6cm, draw an arc. (Hypotenuse)

(iv) This arc cuts the BX at point A

(v) Join AC

It is the required right angled triangle ABC.

5.2

 

Q3 :

Construct an isosceles right angled triangle PQR, where PQR = 90o and PR = 6cm.

Solution:

To construct:

An isosceles right triangle PQR where R = 90o, PR = QR = 6cm

Construction:

(i) Draw a line segment PR = 6cm

(ii) At point R, draw XR    RP

(iii) Taking R as the center and radius 6 cm, draw an arc

(iv) This arc cuts RX at point Q

(v) Join QP

It is the required isosceles right angled triangle PQR

5.3