Chapter 6: The Triangle and its Properties

# Question: 1

In $\Delta \text{PQR,}$ D is the mid-point of $\overline{\text{QR}}$.

$\overline{\text{PM}}$ is _______________

$\text{PD}$ is _________________.

Is $\text{QM}=\text{MR?}$ ## Solution

Given: $\text{QD}=\text{DR}$

$\therefore \overline{\text{PM}}$ is altitude.

$\text{PD}$ is median.

No, $\text{QM}\ne \text{MR}$ as $\text{D}$ is the mid-point of $\text{QR}$.

# Question: 2

Draw rough sketches for the following:

a.    In  is a median.

b.   In  and $\text{PR}$ are altitudes of the triangle.

c.    In $\Delta \text{XYZ,}\text{\hspace{0.17em}}\text{YL}$ is an altitude in the exterior of the triangle.

## Solution

a.    Here, $\text{BE}$ is a median in $\Delta \text{ABC,}$ and $\text{AE}=\text{EC}$. b.   Here, $\text{PQ}$ and $\text{PR}$ are the altitudes of the $\Delta \text{PQR}$ and $\text{RP}\perp \text{QP}$. c.    $\text{YL}$ is an altitude in the exterior of $\Delta \text{XYZ}$. # Question: 3

Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.

## Solution

Isosceles triangle means any two sides are same.

Take $\Delta \text{ABC}$ and draw the median when $\text{AB}=\text{AC}$.

$\text{AL}$ is the median and altitude of the given triangle. # Question: 1

Find the value of the unknown exterior angle $x$ in the following diagrams:

(i) (ii) (iii) (iv) (v) (vi) ## Solution

Since, Exterior angle $=$ Sum of interior opposite angles, therefore

(i)             $x=50°+70°=120°$

(ii)          $x=65°+45°=110°$

(iii)      $x=30°+40°=70°$

(iv)       $x=60°+60°=120°$

(v)          $x=50°+50°=100°$

(vi)       $x=60°+30°=90°$

# Question: 2

Find the value of the unknown interior angle $x$ in the following figures:

(i) (ii) (iii) (iv) (v) (vi) ## Solution

(i)             $x+50°=115°$

(ii)          $70°+x=100°$

(iii)      $x+90°=125°$

(iv)       $60°+x=120°$

(v)          $30°+x=80°$

(vi)       $x+35°=75°$

# Question: 1

Find the value of the unknown $x$ in the following diagrams:

(i) (ii) (iii) (iv) (v) (vi) ## Solution

(i)             In $\Delta \text{ABC,}$

$\angle \text{BAC}+\angle \text{ACB}+\angle \text{ABC}=\text{180}°$

[By angle sum property of a triangle]

(ii)          In $\Delta \text{PQR,}$

$\angle \text{RPQ}+\angle \text{PQR}+\angle \text{PRQ}=\text{180}°$

[By angle sum property of a triangle]

(iii)      In $\Delta \text{XYZ,}$

$\angle \text{ZXY}+\angle \text{XYZ}+\angle \text{YZX}=180°$

[By angle sum property of a triangle]

(iv)       In the given isosceles triangle,

$x+x+50°=180°$

[By angle sum property of a triangle]

(v)          In the given equilateral triangle,

$x+x+x=180°$

[By angle sum property of a triangle]

(vi)       In the given right angled triangle,

[By angle sum property of a triangle]

# Question: 2

Find the values of the unknowns x and y in the following diagrams:

(i) (ii) (iii) (iv) (v) (vi) ## Solution

(i)             $50°+x=120°$ [ Exterior angle property of a $\Delta$ ]

Now, $50°+x+y=180°$ [Angle sum property of a $\Delta$ ]

(ii)          $y=80°$ …….(i)    [ Vertically opposite angles]

Now, $50°+x+y=180°$ [Angle sum property of a $\Delta$ ]

[From equation (i) ]

(iii)      $50°+60°=x$

[Exterior angle property of of a $\Delta$ ]

$⇒x=110°$

Now, $50°+\text{6}0°+y=180°$  [Angle sum property of a $\Delta$ ]

$⇒y=180°-110°$

(iv)       $\text{x}=60°$ …….(i) [Vertically opposite angle]

Now, $30°+x+y=180°$ [Angle sum property of a $\Delta$ ]

[From equation (i) ]

(v)          $y=90°$ …….(i) [Vertically opposite angle]

Now, $\text{y}+x+x=180°$ [Angle sum property of a $\Delta$ ]

$90°+\text{2}x=180°$ [From equation (i)]

(vi)       $x=y$ …….(i) [Vertically opposite angle]

Now, $x+x+y=180°$ [Angle sum property of a $\Delta$ ]

[From equation (i)]

# Question: 1

Is it possible to have a triangle with the following sides?

(i)

(ii)

(iii)

## Solution

Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side.

(i)

$2+3>5$ No

$2+5>3$ Yes

$3+5>2$ Yes

This triangle is not possible.

(ii)

$3+6>7$ Yes

$6+7>3$ Yes

$3+7>6$ Yes

This triangle is possible.

(iii)

$6+3>2$ Yes

$6+2>3$ Yes

$2+3>6$ No

This triangle is not possible

# Question: 2

Take any point $\text{O}$ in the interior of a triangle $\text{PQR}$. Is: (i)             $\text{OP}+\text{OQ}>\text{PQ}?$

(ii)          $\text{OQ}+\text{OR}>\text{QR}?$

(iii)      $\text{OR}+\text{OP}>\text{RP}?$

## Solution

Join $\text{OR}$, $\text{OQ}$ and $\text{OP}$. (i)             Is $\text{OP}+\text{OQ}>\text{PQ}?$

Yes, $\text{POQ}$ form a triangle.

(ii)          Is $\text{OQ}+\text{OR}>\text{QR}?$

Yes, $\text{RQO}$ form a triangle.

(iii)      Is $\text{OR}+\text{OP}>\text{RP}?$

Yes, $\text{ROP}$ form a triangle.

# Question: 3

$\text{AM}$ is a median of a triangle $\text{ABC}$.
Is $\text{AB}+\text{BC}+\text{CA}>\text{2AM}?$ (Consider the sides of triangles $\Delta \text{ABC}$ and $\Delta \text{AMC}$.) ## Solution

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore,  In $\Delta \text{ABM}$, $\text{AB}+\text{BM}>\text{AM}$         (i)

In $\Delta \text{AMC}$, $\text{AC}+\text{MC}>\text{AM}$         (ii)

$\text{AB}+\text{BM}+\text{AC}+\text{MC}>\text{AM}+\text{AM}$

Hence, it is true.

# Question: 4

$\text{ABCD}$ is quadrilateral. Is $\text{AB}+\text{BC}+\text{CD}+\text{DA}>\text{AC}+\text{BD}?$ ## Solution

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore,  In $\Delta \text{ABC}$, $\text{AB}+\text{BC}>\text{AC}$           (i)

In $\Delta \text{ADC}$, $\text{AC}+\text{DC}>\text{AC}$           (ii)

In $\Delta \text{DCB}$, $\text{DC}+\text{CB}>\text{DB}$           (iii)

In $\Delta \text{ADB}$, $\text{AD}+\text{AB}>\text{DB}$          (iv)

Adding eq. (i), (ii), (iii) and (iv), we get

$\text{AB}+\text{BC}+\text{AD}+\text{DC}+\text{DC}+\text{CB}+\text{AD}+\text{AB}>\text{AC}+\text{AC}+\text{DB}+\text{DB}$

Hence, it is true.

# Question: 5

$\text{ABCD}$ is quadrilateral. Is $\text{AB}+\text{BC}+\text{CD}+\text{DA}<2\left(\text{AC}+\text{BD)}?$

## Solution

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. Therefore,  In $\Delta \text{AOB}$, $\text{AB}<\text{OA}+\text{OB}$          (i)

In $\Delta \text{BOC}$, $\text{BC}<\text{OB}+\text{OC}$           (ii)

In $\Delta \text{COD}$, $\text{CD}<\text{OC}+\text{OD}$          (iii)

In $\Delta \text{AOD}$, $\text{DA}<\text{OD}+\text{OA}$         (iv)

Adding equations (i), (ii), (iii) and (iv), we get

$\text{AB}+\text{BC}+\text{CD}+\text{DA}<\text{OA}+\text{OB}+\text{OB}+\text{OC}+\text{OC}+\text{OD}+\text{OD}+\text{OA}$

Hence, it is proved.

# Question: 6

The lengths of two sides of a triangle are  and . Between what two measures should the length of the third side fall?

## Solution

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

It is given that two sides of triangle are  and .

Therefore, the third side should be less than .

And also the third side cannot be less than the difference of the two sides.

Therefore, the third side has to be more than .

Hence, the third side could be the length more than  and less than .

# Question: 1

$\text{PQR}$ is a triangle right angled at $\text{P}$. If $\text{PQ}=\text{10}$ cm and $\text{PR}=24$ cm, find $\text{QR}$.

## Solution

Given:  ,

Let $\text{QR}$ be .

In right angled triangle $\text{QPR}$, ${\left(\text{Hypotenuse}\right)}^{2}={\left(\text{Base}\right)}^{2}+{\left(\text{Perpendicular}\right)}^{2}$

[By  Pythagoras  theorem]

Thus, the length of $\text{QR}$ is .

# Question: 2

## Solution

Given:  $\text{AB}=25$ cm, $\text{AC}=\text{7}$ cm

Let $\text{BC}$ be $x$ cm.

In right angled triangle $\text{ABC}$, ${\left(\text{Hypotenuse}\right)}^{2}={\left(\text{Base}\right)}^{2}+{\left(\text{Perpendicular}\right)}^{2}$

[By  Pythagoras  theorem]

Thus, the length of $\text{BC}$ is .

# Question: 3

long ladder reached a window  high from the ground on placing it against a wall at a distance $a$. Find the distance of the foot of  the ladder from the wall. ## Solution

Let $\text{AC}$ be the ladder and $\text{A}$ be the window.

Given:  , ,

In right-angled triangle $\text{ACB}$,

${\left(\text{Hypotenuse}\right)}^{2}={\left(\text{Base}\right)}^{2}+{\left(\text{Perpendicular}\right)}^{2}$

[By  Pythagoras  theorem]

.

Thus, the distance of the foot of the ladder from the wall is $9$ m. # Question: 4

## Which of the following can be the sides of a right triangle?

(i)

(ii)

(iii)

## Solution

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

${\left(\text{Hypotenuse}\right)}^{2}$ $={\left(\text{Base}\right)}^{2}$ $+{\left(\text{Perpendicular}\right)}^{2}$

(i) In $\Delta \text{ABC,}$         ${\text{(AC)}}^{2}={\text{(AB)}}^{2}+{\text{(BC)}}^{2}$

Since,

Therefore, the given sides are of the right-angled triangle.

Right angle lies on the opposite to the greater side $6.5\text{\hspace{0.17em}}\text{cm}$. i.e., at $\text{B}$

(ii)

In the given triangle, ${\left(5\right)}^{2}={\left(2\right)}^{2}+{\left(2\right)}^{2}$

$\text{R}\text{.H}\text{.S}\text{.}={\left(2\right)}^{2}+{\left(2\right)}^{2}=4+4=8$

Since,   $\text{L}\text{.H}\text{.S}\text{.}\ne \text{R}\text{.H}\text{.S}\text{.}$

Therefore, the given sides are not of the right angled triangle.

(iii)

In $\Delta \text{PQR,}$        ${\text{(PR)}}^{2}={\text{(PQ)}}^{2}+{\text{(RQ)}}^{2}$ Since, $\text{L}\text{.H}\text{.S}\text{.}=\text{R}\text{.H}\text{.S}$.

Therefore, the given sides are of the right-angled triangle.

Right angle lies on the opposite to the greater side . i.e., at $\text{Q}$

# Question: 5

## Solution

Let $\text{A'CB}$ represents the tree before it got broken at the point $\text{C}$ and let the top $\text{A'}$ touches the ground at $\text{A}$ after it broke. Then $\Delta \text{ABC}$ is a right-angled triangle, right angled at B. and

Using Pythagoras theorem in $\Delta \text{ABC}$,

${\text{(AC)}}^{2}={\text{(AB)}}^{2}+{\text{(BC)}}^{2}$

Hence, the total height of the tree $=$ .

# Question: 6

Angles $\text{Q}$ and $\text{R}$ of a $\Delta \text{PQR}$ are $25°$ and $65°$. Write which of the following is true: (i)             ${\text{PQ}}^{2}+{\text{QR}}^{2}{\text{=RP}}^{2}$

(ii)          ${\text{PQ}}^{2}+{\text{RP}}^{2}{\text{=QR}}^{2}$

(iii)      ${\text{RP}}^{2}+{\text{QR}}^{2}{\text{=PQ}}^{2}$

## Solution

In $\Delta \text{PQR}$

$\angle \text{PQR+}\angle \text{QRP+}\angle \text{RPQ=}180°$

[By Angle sum property of a $\Delta$ ]

Thus, $\Delta \text{PQR}$ is a right-angled triangle, right-angled at $\text{P}$.

$\therefore {\left(\text{Hypotenuse}\right)}^{2}={\left(\text{Base}\right)}^{2}+{\left(\text{Perpendicular}\right)}^{2}$

[By  Pythagoras  theorem]

Hence, Option (ii) is correct.

# Question: 7

## Solution

Given diagonal , length Let breadth $\text{(QR)}$ be $x$ cm.

Now, in right angled triangle $\text{PQR}$,

${\text{(PR)}}^{2}={\text{(RQ)}}^{2}+{\text{(PQ)}}^{2}$       [By Pythagoras theorem]

Therefore, the breadth of the rectangle is .

Perimeter of rectangle $=2\text{(length}+\text{breadth)}$

$=2\left(9+40\right)$

Hence, the perimeter of the rectangle is .

# Question: 8

The diagonals of a rhombus measure  and . Find its perimeter.

## Solution

Given: Diagonals  and . Since the diagonals of the rhombus bisect at right angle to each other. Therefore, $\text{OD}=\frac{\text{DB}}{2}=\frac{16}{2}=8$ cm

And

Now, In right-angled triangle $\text{DOC}$,

${\text{(DC)}}^{2}={\text{(OD)}}^{2}+{\text{(OC)}}^{2}$    [By Pythagoras theorem]

Perimeter of rhombus

Thus, the perimeter of rhombus is .