Chapter 10: Practical Geometry

Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

Steps of construction are as follows:

a.  Draw a line-segment AB and take a point C above AB.

b.  Mark any point D on AB and join C to D.

c.  Take a convenient radius using compass. With D as centre, draw an arc cutting AB at E and CD at F.

d.  Considering the same radius as in step $3$, with C as centre draw an arc GH cutting CD at I.

e.  Place the pointed tip of compass at E and adjust the opening so that the pencil tip is at F..

f.    With the same opening and centre at I, draw an arc cutting the arc GH at J.

g.  Join JC to draw a line $l$.

This the required line $\text{AB}\parallel l$.

 

 

Draw a line $l$. Draw a perpendicular to $l$ at any point on $l$. On this perpendicular choose a point X, $4\text{ cm}$ away from $l$. Through X, draw a line m parallel to $l$.

Steps of construction are as follows:

a.  Draw a line $l$ and take a point P on it.

b.  Place the compass point on P and draw an arc of any size below the line that crosses the line l twice.

c.  Place the compass point where the arc crossed the line l on one side and make a small arc below the line.

d.   Without changing the radius on the compass, place the compass point where the first arc crossed the line on the OTHER side and make another arc. Your two small arcs should be intersecting.

e.  Join the intersection of the two small arcs to point P. thus at point P a perpendicular line $n$ is drawn.

f.    Take $\text{PX}=4$ cm on line $n$.

g.  At point X, again draw a perpendicular line $m$.

Thus, line m is parallel to line l and the distance between these two lines is 4 cm  .

 

 

Let $l$ be a line and P be a point not on $l$. Through P, draw a line m parallel to $l$. Now join P to any point Q on $l$. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet $l$ at S. What shape do the two sets of parallel lines enclose?

Steps of construction are as follows:

a.  Draw a line $l$ and take a point P above line $l$.

b.  Take point Q on line $l$ and join PQ.

c.  Make equal angle at point P such that $\angle \text{Q}=\angle \text{P}.$

d.  Extend line at P to get line $m$.

e.  Similarly, take a point R on line $m$. At point R, draw angles such that $\angle \text{P}=\angle \text{R}.$

f.    Extend the line at R which intersects at S on line $l$. Draw line RS.

Thus, we get parallelogram PRSQ.

 

 

Construct $\Delta \text{XYZ}$ in which $\text{XY}=4.5\text{ cm},$ $\text{YZ}=5\text{ cm}$ and $\text{ZX}=6\text{ cm}$.

Steps of construction are as follows:

a.  Draw a line segment $\text{YZ}=5\text{ cm}$.

b.  Taking Z as centre and radius $6\text{ cm}$, draw an arc.

c.  Similarly, taking Y as centre and radius $4.5\text{ cm,}$ draw another arc which intersects the first arc at point X.

d.  Join X to Y and X to Z.

$\Delta \text{XYZ}$ is the required triangle.

 

 

Construct an equilateral triangle of side $5.5\text{ cm}$.

Steps of construction are as follows:

a.  Draw a line segment $\text{BC}=5.5\text{ cm}$

b.  Taking points B and C as centres and radius $5.5\text{ cm,}$ draw arcs which intersect at point A above the line BC.

c.  Join A to B and A to C.

$\Delta \text{ABC}$ is the required triangle.

 

 

Draw $\Delta \text{PQR}$ with $\text{PQ}=4\text{ cm}$, $\text{QR}=3.5\text{ cm}$ and $\text{PR}=4\text{ cm}$. What type of triangle is this?

Steps of construction are as follows:

a.  Draw a line segment $\text{QR}=3.5\text{ cm}\text{.}$

b.  Taking Q as centre and radius $4\text{ cm}$, draw an arc above the line QR.

c.  Similarly, taking R as centre and radius $4\text{ cm}$, draw another arc which intersects the first arc at P.

d.  Join P to Q and P to R.

$\Delta \text{PQR}$ is an isosceles triangle.

 

 

Construct $\Delta \text{ABC}$ such that $\text{AB}=2.5\text{ cm},$ $\text{BC}=6\text{ cm}$ and $\text{AC}=6.5\text{ cm}$. Measure $\angle \text{B}$.

To construct: $\Delta \text{ABC}$ in which $\text{AB}=2.5\text{ cm,}$ $\text{BC}=6\text{ cm}$ and $\text{AC}=6.5\text{ cm}$.

Steps of construction:

a.  Draw a line segment $\text{BC}=6\text{ cm}$.

b.  Taking B as centre and radius $2.5\text{ cm}$, draw an arc above the line BC.

c.  Similarly, taking C as centre and radius $6.5\text{ cm}$, draw another arc which intersects the first arc at point A.

d.  Join A to B and A to C.

e.  Measure angle B with the help of protractor.

This is the required $\Delta \text{ABC}$ where $\angle \text{B}=80°.$

 

 

 

 

Construct $\Delta \text{DEF}$ such that $\text{DE}=5\text{ cm},$ $\text{DF}=3\text{ cm}$ and $m\angle \text{EDF}=90°.$

To construct: $\Delta \text{DEF}$ where $\text{DE}=5\text{ cm,}$ $\text{DF}=3\text{ cm}$ and $m\angle \text{EDF}=90°.$

Steps of construction:

a.  Draw a line segment $\text{DF}=3\text{ cm}\text{.}$

b.  At point D, draw an angle of $90°$ with the help of compass i.e., $\angle \text{XDF}=90°.$

c.  Taking D as centre, draw an arc of radius $5$ cm, which cuts DX at the point E.

d.  Join E to F.

This is the required right angled triangle DEF.

 

 

Construct an isosceles triangle in which the lengths of each of its equal sides is $6.5\text{ cm}$ and the angle between them is $110°$.

To construct: An isosceles triangle PQR where $\text{PQ}=\text{RQ}=6.5\text{ cm}$ and $\angle \text{Q}=110°.$

Steps of construction:

a.  Draw a line segment $\text{QR}=6.5\text{ cm}\text{.}$

b.  At point Q, draw an angle of $110°$ with the help of protractor, i.e., $\angle \text{YQR}=110°.$

c.  Taking Q as centre, draw an arc with radius $6.5$ cm, which cuts QY at point P.

d.  Join P to R.

It is the required isosceles triangle PQR.

 

 

Construct $\Delta \text{ABC}$ with $\text{BC}=7.5\text{ cm},$ $\text{AC}=5\text{ cm}$ and $\text{m}\angle \text{C}=60°.$

To construct: $\Delta \text{ABC}$ where $\text{BC}=7.5\text{ cm,}$ $\text{AC}=5\text{ cm}$ and $m\angle \text{C}=60°.$

Steps of construction:

a.  Draw a line segment $\text{BC}=7.5\text{ cm}\text{.}$

b.  At point C, draw an angle of $60°$ with the help of protractor, i.e., $\angle \text{XCB}=60°.$

c.  Taking C as centre and radius $5\text{ cm}$, draw an arc, which cuts XC at the point A.

d.  Join A to B.

This is the required triangle $\text{ABC}\text{.}$

 

 

 

 

Construct $\Delta \text{ABC}$, given $\text{m}\angle \text{A}=60°,$ $\text{m}\angle \text{B}=30°$ and $\text{AB}=5.8\text{ cm}\text{.}$

To construct: $\Delta \text{ABC}$ where $\text{m}\angle \text{A}=60°,$ $\text{m}\angle \text{B}=30°$ and $\text{AB}=5.8\text{ cm}\text{.}$

Steps of construction:

a.  Draw a line segment $\text{AB}=5.8\text{ cm}\text{.}$

b.  At point A, measure and draw an angle $\angle \text{YAB}=60°$ with the help of compass.

c.  At point B, measure and draw $\angle \text{XBA}=30°$ with the help of compass.

d.  AY and BX intersect at a point. Name this point as C.

This is the required triangle ABC.

 

 

Construct $\Delta \text{PQR}$ if $\text{PQ}=5\text{ cm,}$ $m\angle \text{PQR}=105°$ and $m\angle \text{QRP}=40°.$

(Hint: Recall angle-sum property of a triangle).

Given: $m\angle \text{PQR}=105°$ and $m\angle \text{QRP}=40°.$

We know that sum of angles of a triangle is $180°.$

$\therefore m\angle \text{PQR}+m\angle \text{QRP}+m\angle \text{QPR}=180°$

$\Rightarrow 105°+40°+m\angle \text{QPR}=180°$

$\Rightarrow 145°+m\angle \text{QPR}=180°$

$\Rightarrow m\angle \text{QPR}=180°-145°$

$\Rightarrow m\angle \text{QPR}=35°$

To construct: $\Delta \text{PQR}$ where $m\angle \text{P}=35°,$ $m\angle \text{Q}=105°$ and $\text{PQ}=5\text{ cm}.$

Steps of construction:

a.  Draw a line segment $\text{PQ}=5\text{ cm}\text{.}$

b.  At point P, draw $\angle \text{XPQ}=35°$ with the help of protractor.

c.  At point Q, draw $\angle \text{YQP}=105°$ with the help of protractor.

d.  XP and YQ intersect at a point. Name the point as R.

This is the required triangle PQR.

 

 

Examine whether you can construct $\Delta \text{DEF}$ such that $\text{EF}=7.2\text{ cm},$ $\text{m}\angle \text{E}=110°$ and $\text{m}\angle \text{F}=80°$. Justify your answer.

Given: In $\Delta \text{DEF}$, $\text{m}\angle \text{E}=110°$ and $\text{m}\angle \text{F}=80°$.

Using the angle sum property of triangle, we get

$\angle \text{D}+\angle \text{E}+\angle \text{F}=180°$

$\Rightarrow \angle \text{D}+110°+80°=180°$

$\Rightarrow \angle \text{D}+190°=180°$

$\Rightarrow \angle \text{D}=180°-190°=-10°$

Which is not possible.

Therefore we cannot construct such a triangle.

 

Construct the right angled $\Delta \text{PQR}$, where $\text{m}\angle \text{Q}=90°$, $\text{QR}=8\text{ cm}$ and $\text{PR}=10\text{ cm}\text{.}$

To construct:

A right angled $\Delta \text{PQR}$ where $\text{m}\angle \text{Q}=90°$, $\text{QR}=8\text{ cm}$ and $\text{PR}=10\text{ cm}\text{.}$

Steps of construction:

a.  Draw a line segment $\text{QR}=8\text{ cm}\text{.}$

b.  At point Q, draw $\text{QX}\perp \text{QR}.$

c.  Taking R as centre, draw an arc of radius $10\text{ cm}.$

d.  This arc cuts QX at point P.

e.  Join P to Q.

This is the required right angled triangle PQR.

 

 

Construct a right-angled triangle whose hypotenuse is $6\text{ cm}$ long and one of the legs is $4\text{ cm}$ long.

To construct:

A right angled triangle DEF where hypotenuse, $\text{DF}=6$ cm and $\text{EF}=4\text{ cm}$

Steps of construction:

a.  Draw a line segment $\text{EF}=4\text{ cm}\text{.}$

b.  At point E, draw $\text{EX}\perp \text{EF}.$

c.  Taking F as centre and radius $6$ cm, draw an arc on line EX

d.  This arc cuts the line EX at point D.

e.  Join D to F.

This is the required right angled triangle DEF.

 

 

Construct an isosceles right-angled triangle ABC, where $\text{m}\angle \text{ACB}=90°$ and $\text{AC}=6\text{ cm}\text{.}$

To construct:

An isosceles right angled triangle ABC where $\text{m}\angle \text{C}=90°,$ $\text{AC}=\text{BC}=6\text{ cm}\text{.}$

Steps of construction:

a.  Draw a line segment $\text{AC}=6\text{ cm}\text{.}$

b.  At point C, draw $\text{XC}\perp \text{CA}.$

c.  Taking C as centre and radius $6$ cm, draw an arc on line CX.

d.  This arc cuts CX at point B.

e.  Join B to A.

This is the required isosceles right-angled triangle. ABC.