Chapter-6 The Triangle and Its Properties

Solution: -

(i) 

 is altitude.

Point p is the vertex of the triangle and as shown in the figure we can say that one end of the line PM is the vertex of the triangle and the other end on the line opposite to the vertex.

(ii) 

PD is median

When a line joins the vertex of the triangle with the mid-point of the opposite line, this line is known as the median. A median divide the triangle into the half portion,

(iii) 

No

From the figure we can say that, D is the mid-point of the line QR and the point m is on thee left-side on the point D.

As, the M is not the mid-point of the line QR, QM=MR is not possible.

Solution: -

(a)

When a line joins the vertex of the triangle with the mid-point of the opposite line, this line is known as the median. A median divide the triangle into the half portion,

Here, in ΔABC BE is the median. So, AE=EC.

(b) 

When a line has it’s one point on the vertex of the triangle and the other point on the line opposite to the vertex then, the line is known as altitude.

Here, RP and QP are the line having one end on the vertex of the triangle and the other point on the line opposite to the vertex.

So, RP and QP are altitude.

(c)

When a line has it’s one point on the vertex of the triangle and the other point on the line opposite to the vertex then, the line is known as altitude.

Here, from the figure we can see that in some cases we have to extent the opposite side so that an altitude can be drawn. 

In the above figure, we can see line XZ is extended so that the altitude can be draw.

Solution: -

The triangle which has two side of the same length and the angle made by them with the base is same are known as Isosceles triangles. 

Here, in the above figure, line PS ⊥ BC which means PS is the altitude of the triangle.

Now, S is the mid-point of the line QR and QS=SR. So, the line PS is also the median of the triangle.

So, PS is altitude and median both of the triangle.

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to Exterior angle of the triangle. 

x=50o +70o

x=120o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

x=65o+45o

x=110o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

x=30o+40o

x=70o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

x=60o+60o

x=120o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

x=50o+50o

x=100o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

x=30o+60o

x=90o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

x+50o=115o

By taking 50o from left-hand sided to right-hand side, it becomes –50o

x =115o–50o

x=65o

The value of the unknown interior angle x is 65o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

= 70o + x = 100o

By taking 70o from left-hand sided to right-hand side, it becomes –70o

x=100o–70o

x=30o

The value of the unknown interior angle x is 30o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

By looking to the figure, we can say that it is a right-angle triangle. So, the other opposite interior angle is 90o.

x+90o=125o

By taking 90o from left-hand sided to right-hand side, it becomes –90o

x=125o–90o

x= 35o

The value of the unknown interior angle x is 35o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

x+60o=120o

By taking 60o from left-hand sided to right-hand side, it becomes –60o

x=120o–60o

x=60o

The value of the unknown interior angle x is 60o

Solution: -

x+30o=80o

By taking 30o from left-hand sided to right-hand side, it becomes –30o

x=80o–30o

x=50o

The value of the unknown interior angle x is 50o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

x+35o=75o

By taking 35o from left-hand sided to right-hand side, it becomes –35o

x=75o–35o

x=40o

The value of the unknown interior angle x is 40o

Solution: -

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

∠BAC+∠ABC+∠BCA=180o

x+50o+60o=180o

x+110o=180o

By taking 110o from left-hand sided to right-hand side, it becomes –110o

x=180o–110o

x=70o

The value of the unknown interior angle x is 70o

Solution: -

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o and one of the given angle is right-angle so ∠QPR is 90o

So, from the figure we can write that

∠QPR+∠PQR+∠PRQ=180o

90o+30o+x=180o

120o+x=180o

By taking 120o from left-hand sided to right-hand side, it becomes –120o

x=180o–120o

x=60o

The value of the unknown interior angle x is 60o

Solution: -

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

∠XYZ+∠YXZ+∠XZY=180o

110o+30o+x=180o

140o+x=180o

By taking 140o from left-hand sided to right-hand side, it becomes –140o

x=180o–140o

x=40o

The value of the unknown interior angle x is 40o

Solution: -

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

50o+x+x=180o

50o+2x=180o

By taking 50o from left-hand sided to right-hand side, it becomes –50o and simultaneously dividing both the side of the equation by 2.

2x=180o–50o

2x=130o

x=130o/2

x=65o

The value of the unknown interior angle x is 65o

Solution: -

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

x+x+x=180o

3x=180o

x=180o/3

x=60o

The value of the unknown interior angle x is 60o

Here, the specified triangle is an equiangular triangle.

Solution: -

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o and one of the given angle is right-angle so ∠QPR is 90o

So, from the figure we can write that

90o+2x+x =180o

90o+3x=180o

By taking 90o from left-hand sided to right-hand side, it becomes –90o and simultaneously dividing both the side of the equation by 3.

3x=180o–90o

3x=90o

x=90o/3

x=30o

So, 2x = 2 × 30o = 60o

The value of the unknown interior angle x and 2x is 30o and 60o

Solution: -

Fron the geometry of the triangle, we can say that, sum of the opposite interior angles is equal to triangle’s Exterior angle. 

50o+x=120o

By taking 50o from left-hand sided to right-hand side, it becomes –50o

x=120o–50o

x=70o 

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

50o+x+y=180o

50o+70o+y=180o

120o+y=180o

By taking 120o from left-hand sided to right-hand side, it becomes –120o

y=180o–120o 

y=60o

The value of the unknown interior angle x and y is 70o and 60o

Solution: -

We know from the geometry of the figure that vertically opposite angles are same in measurement.

y=80o

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

50o+80o+x =180o

130o +x=180o

By taking 130o from left-hand sided to right-hand side, it becomes –130o

x=180o –130o 

x=50o

The value of the unknown interior angle x and y is 50o and 80o

Solution: -

We know from the geometry of the figure that the sum of the angles on the some line should be equal to 180o.

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

50o+60o+y =180o

110o+y=180o

By taking 110o from left-hand sided to right-hand side, it becomes –110o

y=180o–110o

y=70o

Now, we know from the geometry of the figure that the sum of the angles on the some line should be equal to 180o.

x+y=180o

x+70o=180o

By taking 70o from left-hand sided to right-hand side, it becomes –70o

x=180o–70

x=110o

The value of the unknown interior angle y and exterior angle x is 70o and 110o

Solution: -

We know from the geometry of the figure that vertically opposite angles are same in measurement.

x=60o

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

30o+x+y=180o

30o+60o+y=180o

90o +y=180o

By taking 90o from left-hand sided to right-hand side, it becomes –90o

y=180o –90o 

y=90o

The value of the unknown interior angle y and x is 90o and 60o

Solution: -

We know from the geometry of the figure that vertically opposite angles are same in measurement.

y=90o

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

x+x+y=180o

2x+90o=180o

By taking 90o from left-hand sided to right-hand side, it becomes –90o and simultaneously dividing both the side of the equation by 2.

2x=180o –90o 

2x=90o

x=90o/2

x=45o

The value of the unknown interior angle y and x is 90o and 45o

Solution: -

We know from the geometry of the figure that vertically opposite angles are same in measurement.

x=y

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

So, from the figure we can write that

x+x+x=180o

3x=180o

x=180o/3

x=60o

The value of the unknown interior angle y and exterior angle x is 60o and 60o

Solution: -

Here, from the geometry of the triangle, we can say that drawing a triangle is only possible when the sum of any two side is greater than the remaining third side.

(2+3)=5

5=5

Since, the sum is not greater than the third side we are not able to make a triangle with the sides 2cm,3cm and 5cm.

(ii) 

Here, from the geometry of the triangle, we can say that drawing a triangle is only possible when the sum of any two side is greater than the remaining third side.

(3+6)=9>7

(6+7)=13>9

(7+3)=10>6

Since, the sum is greater than the third side we are able to make a triangle with the sides 3cm, 6cm and 7cm.

(iii) 

Here, from the geometry of the triangle, we can say that drawing a triangle is only possible when the sum of any two side is greater than the remaining third side.

(3+2)=5<6

Since, the sum is greater than the third side we are able to make a triangle with the sides 6cm, 3cm and 2cm.

Solution: -

Here, if we are going to take any point inside the triangle and join that point with the vertex of the triangle, then the present triangle will divide into three triangles ΔOPQ, ΔOQR and ΔORP

Here, from the geometry of the triangle, we can say that drawing a triangle is only possible when the sum of any two side is greater than the remaining third side.

(i) Yes, in ΔOPQ, their three sides are OP, OQ and PQ.

Now, OP + OQ > PQ

(ii) Yes, in ΔOQR their three sides are OR, OQ and QR.

Now, OQ + OR > QR

(iii) Yes, in ΔORP their three sides are OR, OP and PR.

Now, OR + OP > RP

Solution: -

Here, from the geometry of the triangle, we can say that drawing a triangle is only possible when the sum of any two side is greater than the remaining third side.

Here, taking first triangle as ΔABM,

We know that, AB + BM > AM … [equation i]

Now, taking second as triangle ΔACM,

We, know that, AC + CM > AM … [equation ii]

Now, by solving equations [i] and [ii], 

AB + BM + AC + CM > AM + AM

It is given in the question that m in the median so, BC = BM + CM

AB + BC + AC > 2 AM

Therefore, the given condition is true.

Solution: -

Here, from the geometry of the triangle, we can say that drawing a triangle is only possible when the sum of any two side is greater than the remaining third side.

Here, taking first triangle as ΔABC,

We know that, AB + BC > CA … [equation i]

Here, taking second triangle as ΔBCD,

We know that, BC + CD > DB … [equation ii]

Here, taking third triangle as ΔCDA

We know that, CD + DA > AC … [equation iii]

Here, taking fourth triangle as ΔDAB

We know that, DA + AB > DB … [equation iv]

Now, by adding all equations [i], [ii], [iii] and [iv], we obtained,

AB + BC + BC + CD + CD + DA + DA + AB > CA + DB + AC + DB

2AB + 2BC + 2CD + 2DA > 2CA + 2DB

Dividing both the side if the equation by 2.

2(AB + BC + CA + DA) > 2(CA + DB)

AB + BC + CA + DA > CA + DB

Therefore, the given condition is true.

Solution: -

Here, if we try to solve the given equation by above method then it will be very difficult. So, let’s assume that the diagonals are intersecting each other at a point P.

Here, from the geometry of the triangle, we can say that drawing a triangle is only possible when the sum of any two side is greater than the remaining third side.

Here, taking first triangle as ΔPAB,

We know that, PA + PB < AB … [equation i]

Here, taking second triangle as ΔPBC

We know that, PB + PC < BC … [equation ii]

Here, taking third triangle as ΔPCD

We know that, PC + PD < CD … [equation iii]

Here, taking fourth triangle as ΔPDA

We know that, PD + PA < DA … [equation iv]

Now, by adding all equations [i], [ii], [iii] and [iv], we obtained,

PA + PB + PB + PC + PC + PD + PD + PA < AB + BC + CD + DA

2PA + 2PB + 2PC + 2PD < AB + BC + CD + DA

2PA + 2PC + 2PB + 2PD < AB + BC + CD + DA

2(PA + PC) + 2(PB + PD) < AB + BC + CD + DA

Now, P is the point of intersection of two diagonals, soC = PA + PC and BD = PB + PD

Now,

2AC + 2BD < AB + BC + CD + DA

2(AC + BD) < AB + BC + CD + DA

Therefore, the given condition is true.

Solution: -

Here, from the geometry of the triangle, we can say that drawing a triangle is only possible when the sum of any two side is greater than the remaining third side.

So, let’s assume that the third side is greater than the given side.

So, the third side should be greater than the sum of the given two sides

12+15=27cm

There is also one condition that the length of the third should also not less than the difference between the given two sides.

So, 15–12=3cm

Therefore, the length of the third side should be more than 3 and less than 27.

Solution: -

Let’s draw the diagram for the given value of PQ and PR with the right-angle at P.

For the right-angle triangle, we can use Pythagoras’ Theorem to find the value of unknowns.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

From the above figure we can say that, RQ is the hypotenuse

QR2=PQ2+PR2

QR2=102+242

QR2=100+576

QR2=676

QR=√676

QR=26 cm

So, the length of hypotenuse QR is 26cm.

Solution: -

Let’s draw the diagram for the given value of AB and AC with the right-angle at C.

For the right-angle triangle, we can use Pythagoras’ Theorem to find the value of unknowns.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

From the above figure we can say that, AB is the hypotenuse

AB2=AC2+BC2

252=72+BC2

625=49+BC2

Now, by transferring 49 from right-hand side to left-hand side, it becomes –49

BC2=625–49

BC2=576

BC=√576

BC=24 cm

So, the length of the BC is 24cm

Solution: -

For the right-angle triangle, we can use Pythagoras’ Theorem to find the value of unknowns.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

From the above figure we can say that, ladder is the hypotenuse

152=122+a2

225=144+a2

Now, by transferring 144 from left-hand side to right-hand side, it becomes –144

a2=225–144

a2=81

a=√81

a=9 m

So, the distance of the ladder from the wall is 9m.

Solution: -

(i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm

For the given value we can say that the largest one should be the hypotenuse side, i.e., b=6.5cm.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

So, by using Pythagoras’ theorem,

b2=a2+c2

6.52=2.52+62

42.25=6.25+36

42.25=42.25

Here, both the side of the equation having same values and we know that the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The angle opposite to the hypotenuse sided is the right-angle.

Therefore, the triangle given to us is a right-angled triangle and the right angle is the one opposite to the hypotenuse side, 6.5cm.

(ii) Let a = 2 cm, b = 2 cm, c = 5 cm

For the given value we can say that the largest one should be the hypotenuse side, i.e., c=5cm.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

So, by using Pythagoras’ theorem,

c2=a2+b2

52=22+22

25=4+4

25≠8

Here, both the side of the equation doesn’t have same values.

Therefore, the triangle given to us is not a right-angled triangle.

(iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm

For the given value we can say that the largest one should be the hypotenuse side, i.e., c=2.5cm.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

So, by using Pythagoras’ theorem,

b2=a2+c2

2.52=1.52+22

6.25=2.25+4

6.25=6.25

Here, both the side of the equation having same values and we know that the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The angle opposite to the hypotenuse sided is the right-angle.

Therefore, the triangle given to us is a right-angled triangle and the right angle is the one opposite to the hypotenuse side, 2.5cm.

Solution: -

Let’s us assume that the tree is making right-angle with the ground.

After getting broken from the height 5m it touches the ground 12m from the root of the tree.

Due to this a right-angle triangle will be formed.

So, let’s call this triangle as ABC, B be the point from where the tree is broken and C is the point where tree-top touch the ground. So, AC = 12m,

For the given value we can say that the largest one should be the hypotenuse side, BC.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

BC2=AB2+AC2

BC2=52+122

BC2=25+144

BC2=169

BC=√169

BC=13m

So, the height of the tree is AB + BC

=5+13

=18 m

Solution: -

In the question it is given that ∠Q=25o, ∠R=65o

So, first we have to find ∠P.

As we know from the geometry of the triangle, sum of all the three interior angles should be equal to 180o.

∠PQR+∠QRP+∠RPQ=180o

25o+65o+∠RPQ=180o

90o+∠RPQ=180o

∠RPQ=180o–90o

∠RPQ=90o

Here, we obtained one of the angle as right-angle. So, the side opposite to the right-angle is the hypotenuse side. Here QR is the hypotenuse.

Therefore, QR2=PQ2+PR2

So, option (ii) is correct.

Solution: -

Let’s suppose that ABCD be the rectangular plot.

So, AB=40cm and AC=41cm

Now, we have to find, BC.

For the given value we can say that the largest one should be the hypotenuse side, AC.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

AC2=AB2+BC2

412=402+BC2

BC2=412–402

BC2=1681–1600

BC2=81

BC=√81

BC=9cm

So, to find the perimeter of the rectangle plot

 P=2(length+breadth)

Where, length=40cm,breadth=9cm

Now,

P=2(40+9)

P=2×49

P=98cm

Solution: -

Let’s suppose that PQRS is a rhombus, in a rhombus all sides are equal in length.

Now, the diagonal PR and SQ are crossing each other at point O. Diagonals in the rhombus bisect each other at 90o.

Hence, we can write PO=(PR/2)

PO=16/2

PO=8cm

And, SO=(SQ/2)

SO=30/2

SO=15cm

Now, we have to find, PS.

For the given value we can say that the largest one should be the hypotenuse side, PS.

In Pythagoras’ theorem, the sum of the square of hypotenuse side is equal to the sum of the square of the other two side for a right-angle triangle. The side opposite to the right-angle triangle is known as the hypotenuse side.

PS2=PO2+SO2

PS2=82+152

PS2=64+225

PS2=289

PS=√289

PS=17cm

So, the length is 17cm

Now, the perimeter of the rhombus, P = 4 × side of the rhombus

P=4×17

P=68cm

Therefore, the perimeter of the rhombus is 68cm.