Maths Chapter-10: Practical Geometry

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 ? N

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.

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

Exercise 10.2

Q1 :

Construct \(\Delta\) PQR in which PQ = 4.5 cm , QR = 5 cm and RP = 6cm.

Solution:

Construct: \(\Delta\) 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 \(\Delta\) PQR

Q2:

Construct an equilateral triangle of side 5.5 cm

Solution:

To construct: A \(\Delta\) 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 \(\Delta\) PQR

Q3:

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

Solution:

To construct: \(\Delta\) 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 \(\Delta\) ABC

Q4:

Construct \(\Delta\) PQR such that PQ = 2.5 cm , QR = 6 cm and PR = 6.5 cm. Measure \(\angle\) Q

Solution:

To Construct:

\(\Delta\) 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 \(\Delta\) PQR where \(\angle\) Q = 80^o

Exercise 10.3

Q1: Construct a \(\Delta\) ABC so that AC = 5 cm, AB = 3 cm and \(\angle\) CAB = 90^O .

Solution:

Constructing a \(\Delta\) DEF where DE = 5 cm, DF = 3 cm and m\(\angle\) EDF = 90^O .

Steps of construction:

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

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

(d)Connect BC.

This is the required right angled triangle ACB:

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 \(\angle\) A = 110 ° .

Steps of construction:

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

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

(d) Connect BC

This is the required isosceles \(\Delta\) ABC:

Q3:

Construct a \(\Delta\) ZXC with XC = 7.5 cm, ZC = 5 cm and \(\angle\) C = 60^O .

Solution:

Constructing a \(\Delta\) ZXC where XC = 7.5 cm, ZC = 5 cm and \(\angle\) C = 60^O.

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., \(\angle\) XCB = 60^O .

(d)Connect AB

Thus, this is the required \(\Delta\) ZXC:

Exercise 10.4

Q1:

Construct a triangle ZXC, where\(\angle\) Z = 60^O , \(\angle\) X = 30^O and ZX = 5.8 cm.

Solution:

Now, constructing a \(\Delta\) ZXC where \(\angle\) Z = 60^O , \(\angle\) 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 \(\angle\) XZC = 60^O with a compass.

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

Thus we have the required triangle ZXC:

Q2:

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

Solution:

Given:

\(\angle\)ABC = 105° and \(\angle\) BAC = 40°

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

\(\angle\) ABC + \(\angle\) BCA + \(\angle\) CAB = 180°

105. 40 °+ 40°+\(\angle\)BCA = 180°

145° + \(\angle\) BCA = 180°

\(\angle\)BCA = 180° – 145°

\(\angle\)BCA = 35°

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

Steps of construction:

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

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

(d) QA and BP intersect at point C.

Thus, we have the required triangle ABC.

Q3:

Can you construct a \(\Delta\) ABC such that AB = 4 cm, \(\angle\) B = 120° and \(\angle\) C= 70 °. Explain your answer.

Solution:

Given:

In \(\Delta\) ABC, \(\angle\) A = 120° and m\(\angle\) F = 70 °

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

Thus, we have

\(\angle\)A + \(\angle\)C+ \(\angle\)B= 180°

?\(\angle\)A + 120° + 70° = 180°

?\(\angle\)A + 190° = 180°

?\(\angle\)A = 180° ? 190° = ?10°

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

Exercise 10.5

Q1:

Construct a right angled \(\Delta\) ABC, where \(\angle\) B = 90^o BC = 8 cm and AC = 10 cm

Solution:

To construct:

A right angled triangle ABC where \(\angle\) B = 90^o BC = 8cm and AC = 10cm

Construction:

(i) Draw a line segment BC = 8cm

(ii) At point B, draw BP \(\perp\) 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.

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 \(\perp\) 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.

Q3 :

Construct an isosceles right angled triangle PQR, where \(\angle\)PQR = 90^o and PR = 6cm.

Solution:

To construct:

An isosceles right triangle PQR where \(\angle\)R = 90^o, PR = QR = 6cm

Construction:

(i) Draw a line segment PR = 6cm

(ii) At point R, draw XR \(\perp\) 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