NCERT Solutions for Class 7 Mathematics Chapter 6 - The Triangle and Its Properties

Answer:

Six elements i.e. three sides and three angles of $\angle$RBC are as follows :

Answer:

Side opposite to the vertex Q of $∆$PQR is PR.

Answer:

Angle opposite to side LM of $∆$LMN is $∆$MNL.

Answer:

Vertex opposite to the side RT of $∆$RST is S.

Answer:

(i) $∆$ABC has two equal sides AC and BC each of length 8 cm, also each angle is acute.
Hence, $∆$ABC is an isosceles acute angled triangle.
(ii) In $∆$PQR, no two sides equal in length also $\angle$QRP = 90$°$.
Hence, $∆$PQR is a right-angled scalene triangle.
(iii) $∆$LMN has two equal sides; MN and LN each of length 7 cm, also each angle is acute.
Hence, $∆$LMN is an isosceles acute angled triangle.
(iv) $∆$RST has all three equal sides; RS, ST and RT each of length 5.2 cm.
Hence, $∆$RST is an equilateral triangle.
(v) $∆$ABC has two equal sides, AB and BC each of length 3 cm. Also $\angle$ABC is obtuse.
Hence, $∆$ABC is an isosceles obtuse angled triangle.
(vi) $∆$PQR has two equal sides PQ and QR each of length 6 cm. Also $\angle$PQR = 90$°$
Hence, $∆$PQR is an isosceles right-angled triangle.

Answer:

Since there are three vertices and three sides of a triangle therefore a triangle has three medians. In the adjoining fig., D, E and F are the midpoints of line segments BC, AC and AB respectively. Therefore line segments; AD, BE and CF are three medians of $∆$ABC.

Answer:

Median does not lie wholly in the interior of the triangle. It lies in the triangular region.
Reason: In the adjoining fig., line segment AD is the median from A to D (the midpoint of BC) The locus (family) of all points on AD (excluding A and D) lie in the interior of $∆$ABC but points A and D lie on the boundary of $∆$ABC. Collectively we can say that median AD of triangle lie in the triangular region.

Answer:

Since there are three vertices and three sides of a triangle. Therefore a triangle has three altitudes.
In the adjoining fig., AD $\perp$ BC, BE $\perp$ AC
and CF $\perp$ AB. Therefore line segments; AD, BE and CF are three altitudes.

Answer:

Answer:

No; altitude always do not lie in the interior of a triangle. Reason: In the adjoining fig., ABC is an obtuse angled triangle; obtuse angled at B. The altitude AL drawn from A on produced CB lies in the exterior of triangle.

Answer:

Yes, in right triangle altitudes of the triangle are two of its sides.

Answer:

Yes, the altitude and median can be same only for equilateral triangle.
In an equilateral triangle; say $∆$ABC median AD and
altitude AL drawn from vertex A to the opposite side always coincide.
Similarly, altitude BM and median BE coincide.
Also, altitude CN and median CF coincides.

Answer:

Answer:

In the adjoining fig., we have $∆$ ABC; in which line segment BE is the median drawn from B to AC.

Answer:

In the adjoining fig., we have $∆$PQR; in which PQ and PR are the altitudes drawn from Q to PR and from R to PQ respectively. In such case we have only right-angled triangle PQR right-angled at P.

Answer:

In the adjoining fig., we have an obtuse angled triangle XYZ in which YL is an altitude drawn from Y to produced XZ.
Such altitude lies in the exterior of the triangle.

Answer:

In the adjoining fig., we have an isosceles triangle
PQR such that PQ = PR when we draw perpendicular
bisectors l, m and n of sides QR, PR and PQ respectively.
We find that out of three perpendicular bisectors only l passes through the vertex P; common to the equal sides PQ and PR.
Hence, only PS acts as median and altitude from
P to QR.
We conclude that in an isosceles triangle median and altitude drawn from the vertex (common to the equal sides) to the third side are always same.

Answer:

Let ABC is the required triangle whose exterior
angle is of measure 70$°$.

$\therefore$ By Exterior angle property
$\angle$A + $\angle$B = $\angle$ACD
or $\angle$A + 25$°$ = 70$°$
$\therefore$ $\angle$B = 25$°$ ...(Given)
$\angle$ACD = 70$°$ ...(Given)
or $\angle$A = 70$°$ – 25$°$
or $\angle$A = 45$°$
Now, since sum of three angles of a triangle is of measure 180$°$.
$\angle$A + $\angle$B + $\angle$C = 180$°$
45$°$ + 25$°$ + $\angle$C = 180$°$
$\angle$C = 180$°$ – 70$°$
$\angle$C = 110$°$
Hence, $\angle$A = 45$°$ and $\angle$C = 110$°$.

Answer:

The two interior opposite angles are of measure 60$°$ and 80$°$.
By Exterior angle property,
Exterior angle = Sum of interior opposite angles
= 60$°$ + 80$°$ = 140$°$

Answer:

As we know that:
Exterior angle = Sum of opposite interior angles
Therefore exterior angle = 50$°$ + 50$°$ = 100$°$
Hence, in the diagram; exterior angle of measure 50$°$ is irrelevant.

Answer:

Exterior angle = Sum of interior opp. angles
$\therefore$ x$°$ = 50$°$ + 70$°$
$\Rightarrow$ x$°$ = 120$°$
$\Rightarrow$ x = 120

Answer:

Exterior angle = Sum of interior opp. angles
$\therefore$ x$°$ = 45$°$ + 65$°$
$\Rightarrow$ x$°$ = 110$°$
$\Rightarrow$ x = 110

Answer:

Exterior angle = Sum of interior opp. angles
$\therefore$ x$°$ = 30$°$ + 40$°$
$\Rightarrow$ x$°$ = 70$°$
$\Rightarrow$ x = 70

Answer:

Exterior angle = Sum of interior opp. angles
$\therefore$ x$°$ = 60$°$ + 60$°$
$\Rightarrow$ x$°$ = 120$°$
$\Rightarrow$ x = 120

Answer:

Exterior angle = Sum of interior opp. angles
$\therefore$ x$°$ = 50$°$ + 50$°$
$\Rightarrow$ x$°$ = 100$°$
$\Rightarrow$ x = 100

Answer:

Exterior angle = Sum of interior opp. angles
$\therefore$ x$°$ = 30$°$ + 60$°$
$\Rightarrow$ x$°$ = 90$°$
$\Rightarrow$ x = 90

Answer:

Sum of interior opp. angles = Exterior angle
$\therefore$ x$°$ + 50$°$ = 115$°$
$\Rightarrow$ x$°$ = 115$°$ – 50$°$
$\Rightarrow$ x$°$ = 65$°$
$\Rightarrow$ x = 65

Answer:

Sum of Interior opp. angles = Exterior angle
$\therefore$ x$°$ + 70$°$ = 100$°$
$\Rightarrow$ x$°$ = 100$°$ – 70$°$
$\Rightarrow$ x$°$ = 30$°$
$\Rightarrow$ x = 30

Answer:

Sum of interior opp. angles = Exterior angle
$\therefore$ x$°$ + 90$°$ = 125$°$
$\Rightarrow$ x$°$ = 125$°$ – 90$°$
$\Rightarrow$ x$°$ = 35$°$
$\Rightarrow$ x = 35

Answer:

Sum of interior opp. angles = Exterior angle
$\therefore$ x$°$ + 60$°$ = 120$°$
$\Rightarrow$ x$°$ = 120$°$ – 60$°$
$\Rightarrow$ x$°$ = 60$°$
$\Rightarrow$ x = 60

Answer:

Sum of interior opp. angles = Exterior angle
$\therefore$ x$°$ + 30$°$ = 80$°$
$\Rightarrow$ x$°$ = 80$°$ – 30$°$
$\Rightarrow$ x$°$ = 50$°$
$\Rightarrow$ x = 50

Answer:

Sum of Interior opp. angles = Exterior angle
$\therefore$ x$°$ + 35$°$ = 75$°$
$\Rightarrow$ x$°$ = 75$°$ – 35$°$
$\Rightarrow$ x$°$ = 40$°$
$\Rightarrow$ x = 40

Answer:

By angle-sum property of a triangle
In $∆$ABC,
x$°$ + 50$°$ + 60$°$ = 180$°$
$\Rightarrow$ x$°$ = 180$°$ – 50$°$ – 60$°$
= 180$°$ – 110$°$
$\Rightarrow$ x$°$ = 70$°$
Hence, x = 70

Answer:

In $∆$PQR,
x$°$ + 30$°$ + 90$°$ = 180$°$
(By angle–sum property of a triangle)
$\Rightarrow$ x$°$ = 180$°$ – 30$°$ – 90$°$
= 180$°$ – 120$°$
$\Rightarrow$ x$°$ = 60$°$
Hence, x = 60

Answer:

In $∆$XYZ,
x$°$ + 30$°$ + 110$°$ = 180$°$
(By angle–sum property of a triangle)
$\Rightarrow$ x$°$ = 180$°$ – 30$°$ – 110$°$
= 180$°$ – 140$°$
$\Rightarrow$ x$°$ = 40$°$
Hence, x = 40

Answer:

In the given isosceles triangle:
x$°$ + x$°$ + 50$°$ = 180$°$
(By angle–sum property of a triangle)
$\Rightarrow$ 2x$°$ = 180$°$ – 50$°$
$\Rightarrow$ 2x$°$ = 130$°$
$\Rightarrow$ x$°$ = 130$°$/2
$\Rightarrow$ x$°$ = 65$°$
Hence, x = 65$°$

Answer:

In the given equilateral triangle:
x$°$ + x$°$ + x$°$ = 180$°$
(By angle–sum property of a triangle)
$\Rightarrow$ 3x$°$ = 180$°$
$\Rightarrow$ x$°$ = 180$°$/3
$\Rightarrow$ x$°$ = 60$°$
Hence, x = 60

Answer:

In the given right angled triangle:
x$°$ + 2x$°$ + 90$°$ = 180$°$
(By angle–sum property of a triangle)
$\Rightarrow$ 3x$°$ = 180$°$ – 90$°$
$\Rightarrow$ 3x$°$ = 90$°$
$\Rightarrow$ x$°$ = 90$°$/3
$\Rightarrow$ x$°$ = 30$°$
Hence, x = 30

Answer:

As we know that in a triangle;
Sum of interior opposite angles = Exterior angle
$\therefore$ x$°$ + 50$°$ = 120$°$
$\Rightarrow$ x$°$ = 120$°$ – 50$°$
$\Rightarrow$ x$°$ = 70$°$
Now in given triangle;
x$°$ + y$°$ + 50$°$ = 180$°$
(By angle-sum property of a triangle)
$\Rightarrow$ 70$°$ + y$°$ + 50$°$ = 180$°$
(putting x$°$ = 70$°$ as found above)
$\Rightarrow$ y$°$ = 180$°$ – 50$°$ – 70$°$
$\Rightarrow$ y$°$ = 180$°$ – 120$°$
$\Rightarrow$ y$°$ = 60$°$
Hence, x = 70$°$, y = 60$°$.

Answer:

y$°$ = 80$°$ (vertically opposite angles)
$\Rightarrow$ y = 80$°$
Now in the given triangle:
x$°$ + y$°$ + 50$°$ = 180$°$
(By angle-sum property of a triangle)
$\Rightarrow$ x$°$ + 80$°$ + 50$°$ = 180$°$
(putting y$°$ = 80$°$ as found above)
$\Rightarrow$ x$°$ = 180$°$ – 80$°$ – 50$°$
$\Rightarrow$ x$°$ = 180$°$ – 130$°$
$\Rightarrow$ x$°$ = 50$°$
Hence, x = 50$°$, y = 80$°$.

Answer:

As we know that in a triangle
Exterior angle = Sum of interior opposite angles
$\therefore$ x$°$ = 50$°$ + 60$°$
$\Rightarrow$ x$°$ = 110$°$
Now in given triangle:
y$°$ + 50$°$ + 60$°$ = 180$°$
(By angle-sum property of a triangle)
$\Rightarrow$ y$°$ = 180$°$ – 50$°$ – 60$°$
$\Rightarrow$ y$°$ = 180$°$ – 110$°$
$\Rightarrow$ y$°$ = 70$°$
Hence, x = 110$°$, y = 70$°$.

Answer:

x$°$ = 60$°$ [Vertically opposite angles]
$\Rightarrow$ x = 60
Now in given triangle:
x$°$ + y$°$ + 30$°$ = 180$°$
[By angle-sum property of a triangle]
$\Rightarrow$ 60$°$ + y$°$ + 30$°$ = 180$°$
[putting x$°$ = 60$°$ as found above]
$\Rightarrow$ y$°$ = 180$°$ – 60$°$ – 30$°$
$\Rightarrow$ y$°$ = 180$°$ – 90$°$
$\Rightarrow$ y$°$ = 90$°$
Hence, x = 60$°$, y = 90$°$.

Answer:

y$°$ = 90$°$ (Vertically opposite angles)
$\Rightarrow$ y = 90
Now in given triangle:
x$°$ + x$°$ + y$°$ = 180$°$
[By angle-sum property of a triangle]
$\Rightarrow$ 2x$°$ + 90$°$ = 180$°$
[Putting y$°$ = 90$°$ as found above]
$\Rightarrow$ 2x$°$ = 180$°$ – 90$°$
$\Rightarrow$ 2x$°$ = 90$°$
$\Rightarrow$ x$°$ = 90$°$/2
$\Rightarrow$ x$°$ = 45$°$
Hence, x = 45$°$, y = 90$°$.

Answer:

In each case:
y$°$ = x$°$
(Vertically opposite angles)
Now in given triangle:
y$°$ + y$°$ + y$°$ = 180$°$
[By angle-sum property of a triangle]
$\Rightarrow$ 3y$°$ = 180$°$
$\Rightarrow$ y$°$ = 180$°$/3 = 60$°$
Hence, x = 60$°$; y = 60$°$.

Answer:

Let third angle be x$°$
$\therefore$ x$°$ + 30$°$ + 80$°$ = 180$°$
[By angle-sum property of a triangle]
$\Rightarrow$ x$°$ = 180$°$ – 30$°$ – 80$°$
$\Rightarrow$ x$°$ = 180$°$– 110$°$
$\Rightarrow$ x$°$ = 70$°$
Hence, (x = 70)

Answer:

Let each of the two equal angles be x$°$. Therefore by angle-sum property of a triangle:

Answer:

Let three angles of a triangle be x$°$, 2x$°$ and x$°$.
Therefore,
x$°$ + 2x$°$ + x$°$ = 180$°$
[By angle-sum property of a triangle]
$\Rightarrow$ 4x$°$ = 180$°$
$\Rightarrow$ x$°$ = 180$°$/4
$\Rightarrow$ x$°$ = 45$°$
Now three angles of given triangle are: x$°$ = 45$°$,
2x$°$ = 2 × 45$°$ = 90$°$
and x$°$ = 45$°$.
We observe that one angle is 90$°$.
Each of the two other angles is of measure 45$°$.
Therefore sides opposite to two equal angles are equal.
Thus, we may classify the triangle in two different
ways as follows:
(i) Triangle is a right-angled triangle.
(ii) Triangle is an isosceles triangle.

Answer:

Given triangle is isosceles
$\therefore$ x$°$ = 40$°$
[$\therefore$ Angles opposite to two equal sides of an isosceles triangle are equal]
$\Rightarrow$ x = 40

Answer:

Let isosceles triangle be $∆$ABC in which
AB = AC
$\therefore$ m $\angle$A = 45$°$
[$\therefore$ Angles opposite to equal sides of an isosceles
triangle are equal]

Now in $∆$ABC:
x$°$ + $\angle$A + 45$°$ = 180$°$
(Angle–sum property of a triangle)
$\Rightarrow$ x$°$ + 45$°$ + 45$°$ = 180$°$
[$\therefore$ $\angle$A = 45$°$ as found above]
$\Rightarrow$ x$°$ = 180$°$– 45$°$ – 45$°$
$\Rightarrow$ x$°$= 90$°$
$\Rightarrow$ x = 90

Answer:

x$°$ = 50$°$
[$\therefore$ Angles opposite to equal sides of an isosceles triangle are equal]
$\Rightarrow$ x = 50

Answer:

Answer:

Answer:

Answer:

Let isosceles triangle be DABC such that
AB = AC
$\angle$ACB and $\angle$ACD

form a linear pair.
Therefore,
$\angle$ACB + 120$°$ = 180$°$
$\Rightarrow$ $\angle$C = 180$°$ – 120$°$
$\Rightarrow$ $\angle$C = 60$°$
Now, AB = AC implies that
x$°$ = $\angle$C
x$°$ = 60$°$
[$\therefore$ Angles opposite to equal sides of a triangle are equal]

Answer:

Let isosceles triangle be $∆$ABC in which
AB = AC
$\therefore$ $\angle$B = x$°$
[$\therefore$ Angles opposite to equal sides of a triangle are always equal]
Also, we have
$\angle$B + x$°$ = 110$°$
[$\therefore$ Exterior angle of a triangle is equal to the sum of its interior opposite angles]
$\Rightarrow$ x$°$ + x$°$ = 110$°$
$\Rightarrow$ 2x$°$ = 110$°$
$\Rightarrow$ x$°$ = 110$°$/2
$\Rightarrow$ x$°$ = 55$°$
$\Rightarrow$ x = 55

Answer:

Let isosceles triangle be DABC in which
AB = BC
In fig., $\angle$BAC = 30$°$
[Vertically opposite angles]

Now in $∆$ABC; we have
AB = BC
$\therefore$ x$°$ = $\angle$BAC
[$\therefore$ Angles opposite to equal sides of a triangle are equal]
$\Rightarrow$ x$°$ = 30$°$
[$\therefore$$\angle$BAC = 30$°$ as found above]
$\Rightarrow$ x = 30

Answer:

Answer:

Answer:

Answer:

Given sides are 2 cm, 3 cm and 5 cm
Now, 2 + 3 = 5 ...(1)
3 + 5 > 2
5 + 2 > 3
$\therefore$ of (1), it is not possible

Answer:

Given sides are 3 cm, 6 cm, 7 cm
Now, 3 + 6 > 7
6 + 7 > 3
7 + 3 > 6
$\therefore$ Sum of the lengths of any two sides would be greater than the length of the third side.

Answer:

Given sides are 6 cm, 3 cm, 2 cm
Now, 6 + 3 > 2
3 + 2 < 6
$\therefore$ of this, it is not possible.

Answer:

(i) Yes, OP + OQ > PQ because sum of the lengths of any two sides of $∆$POQ is always greater than the third side.
(ii) Yes, OQ + OR > QR because sum of the lengths of any two sides of $∆$ROQ is always greater than the third side.

(iii) Yes, OR + OP > RP because sum of the lengths of any two ides of $∆$ROP is always greater than the third.

Answer:

Yes; AB + BC + CA > 2 AM
Proof: In $∆$ABM; AB + BM > AM ...(i)
[$\therefore$ Sum of the lengths of any two sides of a triangle is greater than the third side]
In $∆$ACM; AC + MC > AM ...(ii)
[using same reason as above]
Adding (i) and (ii), we get:
(AB + BM) + (AC + MC) > AM + AM
or AB + (BM + MC) + AC > 2 AM
or AB + BC + AC > 2 AM
[$\therefore$ AM is the median of triangle ABC
$\therefore$ BM = MC. Also M is a point on BC
$\therefore$BC = BM + MC]

Answer:

Yes; AB + BC + CD + DA > AC + BD
Proof: In $∆$ABC; AB + BC > AC ...(i)
[$\therefore$ Sum of lengths of any two sides of a triangle is always greater than the third side]

In $∆$ADC; CD + DA > AC ...(ii)
(using same reason as above)
In $∆$ABD; AB + DA > BD ...(iii)
(using same reason as above)
In $∆$BCD; BC + CD > BD ...(iv)
(using same reason as above)
Adding (i), (ii), (iii) and (iv), we get:
(AB + BC) + (CD + DA) + (AB + DA) + (BC + CD) > AC + AC + BD + BD
or (AB + AB) + (BC + BC) + (CD + CD) +
(DA + DA) > 2AC + 2BD
or 2AB + 2BC + 2CD + 2DA > 2AC + 2BD
or 2 (AB + BC + CD + DA) > 2 (AC + BD)
or AB + BC + CD + DA > AC + BD.

Answer:

Yes; AB + BC + CD + DA < 2 (AC + BD)
Proof: In quadrilateral ABCD; let diagonals AC and BD interset each other at point O.
Now in $∆$AOB, AO + OB > AB .... (i)
[$\therefore$ sum of the lengths of any two sides of a triangle is greater than the third side]

Similarly in $∆$BOC;
OB + OC > BC ...(ii)
(using same reason as above)
In $∆$COD;
OC + OD > CD ...(iii)
(using same reason as above)
Also in $∆$AOD;
AO + OD > DA ...(iv)
(using same reason as above)
Adding (i), (ii), (iii) and (iv), we get:
(AO + OB) + (OB + OC) + (OC + OD) + (AO + OD) > AB + BC + CD + DA
$\Rightarrow$ 2 (AO + OB + OC + OD) > AB + BC + CD + DA
$\Rightarrow$ 2 [(AO + OC) + (OB + OD)] > AB + BC + CD + DA
$\Rightarrow$ 2 [AC + BD] > AB + BC + CD + DA
or AB + BC + CD + DA < 2 (AC + BD) Proved.

Answer:

If x can be the length of the third side, we should have:

The length cannot be negative. Therefore we reject it.
Thus, the numbers between 3 and 27 satisfy
these.
$\therefore$The length of the third side could be any
length between 3 cm and 27 cm.

Answer:

No; this is not the condition required to draw a triangle. Based on the angles, the only condition to draw a triangle is that the sum of three angles of a triangle should be 180$°$.
For example: If we consider three angles as 50$°$, 30$°$, and 100$°$. We have 50$°$ + 30$°$ + 100$°$ but triangle with these angles is possible to draw because 50$°$ + 30$°$ + 100$°$ = 180$°$; angle-sum property is satisfied.

Answer:

Answer:

Answer:

Let the ladder be AC where C is the foot of the ladder and A be the top of ladder at 12 m high from the ground. Wall is a rt. angled to the ground at point B.
Now in rt. angled triangle ABC; using Pythagoras property;

Answer:

Therefore the triangle with given sides is a rig ht-angled triangle.
In this right-angled triangle; the longest side with length 6.5 cm is the hypotenuse and angle opposite to this side is the right angle.

Answer:

Answer:

Answer:

Let the original height of tree be BD and tree is broken at point C; at a height of 5 m from the ground.

Let top of the tree D touches the ground at point A, 12 m away from the base of the tree.
Since it form a right-angled triangle;
$∆$ABC right-angled at point B.
Now using Pythagoras property,

Answer:

In $∆$PQR ;
$\angle$P + $\angle$Q + $\angle$R = 180$°$
[Angle-sum property of a triangle]
$\Rightarrow$ $\angle$P + 25$°$ + 65$°$ = 180$°$
$\Rightarrow$ $\angle$P = 180$°$ – 25$°$ – 65$°$
$\Rightarrow$ $\angle$P = 90$°$
It implies that $∆$PQR is right-angled triangle right angled at point P.
As we know that the side opposite to the right angle is the hypotenuse so side QR is the hypotenuse.
Using Pythagoras property in rt. angled triangle PQR,
we have PQ² + RP² = QR²
Hence, option (ii); PQ² + RP² = QR² is true.

Answer:

Perimeter of rectangle ABCD = 2 (length + breadth)
= 2 (AB + BC)
= 2 (40 + 9)
= 2 × 49
= 98 cm.

Answer:

Answer:

As we know that in any right-angled triangle; the hypotenuse is the longest side.

Answer:

In $∆$ABC right-angled at B the side opposite to $\angle$B is the hypotenuse AC.
As we know that hypotenuse is the longest side.
So, AC is the longest side.

Answer:

Hypotenuse is the longest side of a right triangle.

Answer:

Pythagoroas property gives the relation between the lengths of the sides of a right triangle. According to the Pythagoras property:

‘‘The square described on the hypotenuse has an area equal to the sum of the areas of the squares described on its other two legs. In the given fig., ABC is right-angled triangle right angled at C.

An Indian mathematician Baudhayan gave an equivalent form of this property and stated the theorem in its most general form.
According to the Baudhayan theorem:
‘‘The square descirbed on the diagonal of a rectangle has an area equal to the sum of the areas of the squares described on its two sides.
In the given fig. ABCD is a rectangle.