NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Q.1 In $\mathrm{?}$PQR, D is the mid-point of $\overline{QR}$ ,
(i) $\overline{PM}$ is..........
(ii) $\overline{PD}$ is...........
(iii) Is QM = MR?

Answer :

(i) $\overline{PM}$ is altitude.
(ii) $\overline{PD}$ is median.
(iii) No, QM $?$ MR.

Q.2 Draw rough sketches for the following:
(a) In $\mathrm{?}$ABC, BE is a median.
(ib) In $\mathrm{?}$PQR, PQ and PR are altitudes of the triangle.
(c) In $\mathrm{?}$XYZ, YL is an altitude in the exterior of the triangle.

Answer :

Q.3 Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.

Answer :

Draw a Line segment AD $?BC$. It is an altitude for this triangle. Here we observe that length of QS and SR is also same. So PS is also a median of this triangle.

Q.1 Find the value of the unknown exterior angle x in the following diagrams:

Answer :

(i) $\mathrm{?}$x = 50° + 70° = 120° (Exterior angle is equal to sum of its interior opposite angles)
(ii) $\mathrm{?}$x = 65°+ 45° = 110° (Exterior angle is equal to sum of its interior opposite angles)
(iii) $\mathrm{?}$x = 30° + 40° = 70° (Exterior angle is equal to sum of its interior opposite angles)
(iv) $\mathrm{?}$x = 60° + 60° = 120° (Exterior angle is equal to sum of its interior opposite angles)
(v) $\mathrm{?}$x = 50° + 50° =100° (Exterior angle is equal to sum of its interior opposite angles)
(vi) $\mathrm{?}$x = 30° + 60° = 90° (Exterior angle is equal to sum of its interior opposite angle)

Q.2 Find the value of the unknown interior angle x in the following figures:

Answer :

(i) $\mathrm{?}$x + 50° = 115° (Exterior angle of a triangle)
$?$ $\mathrm{?}$x = 115°- 50° = 65°
(ii) $\mathrm{?}$x + 70° = 110° (Exterior angle of a triangle)
$?$ $\mathrm{?}$x = 110° – 70° = 40°
(iii) $\mathrm{?}$ x + 90° = 125° (Exterior angle of a right triangle)
$?$ $\mathrm{?}$x = 125° – 90° = 35°
(iv) $\mathrm{?}$x + 60° = 120° (Exterior angle of a triangle)
$?$ $\mathrm{?}$x = 120° – 60° = 60°
(v)$\mathrm{?}$ X + 30° = 80° (Exterior angle of a triangle)
$?$ $\mathrm{?}$x = 80° – 30° = 50°
(vi) $\mathrm{?}$ x + 35° = 75° (Exterior angle of a triangle)
$?$ $\mathrm{?}$ x = 75° – 35° = 40°

Q.1 Find the value of the unknown x in the following diagrams:

Answer :

(i) we know that sum of all angle of a triangle = 180
$\mathrm{?}$ x + 50° + 60° = 180°
Or, $\mathrm{?}$ x + 110° = 180°
$?$ $\mathrm{?}$x = 180° – 110° = 70°

(ii) we know that sum of all angle of a triangle = 180
$\mathrm{?}$x + 90° + 30 = 180° [$\mathrm{?}$ is right angled triangle]
Or, $\mathrm{?}$x + 120° = 180°
Or, $\mathrm{?}$x – 180° – 120° = 60°
(iii) we know that sum of all angle of a triangle = 180
$\mathrm{?}$x + 30° + 110° =180°
Or, $\mathrm{?}$x + 140° = 180°
$?$ $\mathrm{?}$x = 180° – 140° = 40°

(iv ) we know that sum of all angle of a triangle = 180
$\mathrm{?}$x + $\mathrm{?}$x + 50° = 180°
Or, 2x + 50° = 180°
Or, 2x = 180° – 50°
Or, 2x = 130°
$?$ x=130/2=65°

(v) we know that sum of all angle of a triangle = 180
$\mathrm{?}$x + $\mathrm{?}$x +$\mathrm{?}$x =180°
Or, 3 $\mathrm{?}$x = 180°
$?$ $\mathrm{?}$x=180°/3=60°

(vi) we know that sum of all angle of a triangle = 180
x + 2 x + 90° = 180° ($\mathrm{?}$ is right angled triangle)
Or, 3x + 90° = 180°
Or, 3x = 180° – 90°
Or, 3x = 90°
$?$ x=90/3=30°

Q. 2 Find the values of the unknowns x and y in the following diagrams:

Answer :

(i) $\mathrm{?}$x + 50° = 120° (Exterior angle of a triangle)
$?$ $\mathrm{?}$x = 120°- 50° = 70°
$\mathrm{?}$x + $\mathrm{?}$y + 50° = 180° (Angle sum property of a triangle)
70° + $\mathrm{?}$y + 50° = 180°
$\mathrm{?}$y + 120° = 180°
$\mathrm{?}$y = 180° – 120°
$?$ $\mathrm{?}$y = 60°
Thus $\mathrm{?}$x = 70 and $\mathrm{?}$y – 60°

(ii) $\mathrm{?}$y = 80° (Vertically opposite angles are same)
$\mathrm{?}$x + $\mathrm{?}$y + 50° = 180° (Angle sum property of a triangle)
$?$ $\mathrm{?}$x + 80° + 50° = 180°
$?$ $\mathrm{?}$x + 130° = 180°
$?$$\mathrm{?}$x = 180° – 130° = 50°
Thus, $\mathrm{?}$x = 50° and $\mathrm{?}$y = 80°

(iii) $\mathrm{?}$y + 50° + 60° = 180° (Angle sum property of a triangle)
$\mathrm{?}$y + 110° = 180°
$?$$\mathrm{?}$y = 180°- 110° = 70°
$\mathrm{?}$x + $\mathrm{?}$y = 180° (Linear pairs)
Or, $\mathrm{?}$x + 70° = 180°
$?$ $\mathrm{?}$x = 180° – 70° = 110°
Thus, $\mathrm{?}$x = 110° and y = 70°

(iv) $\mathrm{?}$x = 60° (Vertically opposite angles)
$\mathrm{?}$x + $\mathrm{?}$y + 30° = 180° (Angle sum Or, 60° + $\mathrm{?}$y + 30° = 180°
Or, $\mathrm{?}$y + 90° = 180°
Or, $\mathrm{?}$y = 180° – 90° = 90°
Thus, $\mathrm{?}$x = 60° and $\mathrm{?}$y = 90°

(v) $\mathrm{?}$y = 90° (Vertically opposite angles)
$\mathrm{?}$x + $\mathrm{?}$x + $\mathrm{?}$y = 180° (Angle sum property of a triangle)
Or, 2 $\mathrm{?}$x + 90° = 180°
Or, 2$\mathrm{?}$x = 180° – 90°
Or, 2$\mathrm{?}$x = 90°
$?$ $\mathrm{?}$x=90°/2=45°
Thus, $\mathrm{?}$x = 45° and $\mathrm{?}$y = 90

(vi) From the given figure, we have
$\mathrm{?}$y = $\mathrm{?}$x
$\mathrm{?}$1 = $\mathrm{?}$x
$\mathrm{?}$2 = $\mathrm{?}$x
Adding both sides, we have Adding both sides, we have Adding both sides, we have $\mathrm{?}$y + $\mathrm{?}$1 + $\mathrm{?}$2 = 3$\mathrm{?}$x
Or, 180° = 3$\mathrm{?}$x [Angle sum property of a triangle]
$?$ $\mathrm{?}$x=180°/3=60°
$\mathrm{?}$x = 60°, $\mathrm{?}$y = 60°

Q.1 Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm

Answer :

We know that for a triangle, the sum of any two sides must be greater than the third side.
(i) Given sides are 2 cm, 3 cm, 5 cm
(2 + 3) = 5
5 = 5
Hence, the triangle is not possible.
(i) Given sides are 3 cm, 6 cm, 7 cm
(3 + 6) = 9 > 7
(6 + 7) = 13 > 3
(7 + 3) = 10 > 6
Hence, the triangle is possible.
(iii) Given sides are 6 cm, 3 cm, 2 cm
(3 + 2) = 5 < 6
Hence, the triangle is not possible.

Q.2 Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?

Answer :

(i) Yes, $\mathrm{?}$OPQ has sides OP, OQ and PQ.
So, OP + OQ > PQ
(ii) Yes, $\mathrm{?}$OQR has sides OR, OQ and QR.
So, OQ + OR > QR
(iii) Yes, $\mathrm{?}$ORP has sides OR, OP and PR.
So, OR + OP > RP

Q.3 AM is a median of a $\mathrm{?}$ ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles $\mathrm{?}$ABM and $\mathrm{?}$AMC.)

Answer :

We know that,
The sum of the length of any two sides is always greater than the third side.
Now In $\mathrm{?}$ABM,
Here, AB + BM > AM … [equation i]
Then, In $\mathrm{?}$ACM
Here, AC + CM > AM … [equation ii]
By adding equation [i] and [ii] we get,
AB + BM + AC + CM > AM + AM
From the figure we have, BC = BM + CM
AB + BC + AC > 2 AM
Hence, the given expression is true.

Q.4 ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?

Answer :

In $\mathrm{?}$ABC, we have
AB + BC > AC …(i) [Sum of any two sides is greater than the third side]
In $\mathrm{?}$BDC, we have
BC + CD > BD …(ii)
In $\mathrm{?}$ADC, we have CD + DA > AC .... (iii)
In $\mathrm{?}$DAB, we have DA + AB > BD …(iv)
Adding eq. (i), (ii), (iii) and (iv), we get
2AB + 2BC + 2CD + 2DA > 2AC + 2BD
Or, AB + BC + CD + DA > AC + BD [Dividing both sides by 2]
Hence, proved

Q.5 ABCD is a quadrilateral.
Is AB + BC + CD + DA < 2(AC + BD)?

Answer :

$\mathrm{?}$AOB, we have AB < AO + BO …(i) [Any side of a triangle is less than the sum of other two sides]
In $\mathrm{?}$BOC, we have
BC < BO + CO …(ii)
In $\mathrm{?}$COD, we have
CD < CO + DO …(iii)
In $\mathrm{?}$AOD, we have
DA < DO + AO …(iv)
Adding eq. (i), (ii), (iii) and (iv), we have
AB + BC + CD + DA < 2AO + 2BO + 2CO + 2DO
Or, AB + BC + CD + DA < 2(AO + BO + CO + DO)
Or, AB + BC + CD + DA < 2 [(AO + CO) + (BO + DO)]
Or, AB + BC + CD + DA < 2(AC + BD)
Thus, AB + BC + CD + DA < 2(AC + BD)
Hence, proved.

Q.6 The length of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Answer :

Sum of two sides
= 12 cm + 15 cm = 27 cm
Difference of the two sides
= 15 cm – 12 cm = 3 cm
$?$ The measure of third side should fall between 3 cm and 27 cm.

Q.1 PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Answer :

In right angled triangle PQR, we have
$Q{R}^2=P{Q}^2+P{R}^2$ [From Pythagoras property]
$Q{R}^2={10}^2+{24}^2$
$Q{R}^2=100+576=676$
$?QR=\sqrt{676}=26cm$ The, the required length of QR = 26 cm.

Q.2 ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Answer :

In right angled?ABC, we have
$B{C}^2+7^2={25}^2$(By Pythagoras property)
Or, $B{C}^2+49=625$
Or, $B{C}^2=625–49$
Or, $B{C}^2=576$
$?BC=\sqrt{576}=24cm$
Thus, the required length of BC = 24 cm.

Q.3 A 15 m long ladder reached a window 12 m 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.

Answer :

Here, the ladder forms a right angled triangle.
$?$ $a^2+{12}^2={15}^2$(By Pythagoras property)
Or, $a^2+144=225$
Or, $a^2=225–144$
Or, $a^2=81$
$?$ a = $\sqrt{81}=9$ m
Thus, the distance of the foot from the ladder = 9m

Q.4 Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm

Answer :

(i) Given sides are 2.5 cm, 6.5 cm, 6 cm.
Square of the longer side = $(6.5{)}^2=42.25$ cm.
Sum of the square of other two sides = $(2.5{)}^2+(6{)}^2=6.25+36$ = 42.25 cm.
Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.
$?$The given sides form a right triangle.

(ii) Given sides are 2 cm, 2 cm, 5 cm .
Square of the longer side =$(5{)}^2=25$cm
Sum of the square of other two sides =$(2{)}^2+(2{)}^2=4+4=8cm$ Since 25 cm $?$8 cm
$?$The given sides do not form a right triangle.

(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm
Square of the longer side = $(2.5{)}^2=6.25$cm
Sum of the square of other two sides = $(1.5{)}^2+(2{)}^2=2.25+4=6.25cm=6.25cm=6.25cm$
Since the square of longer side in a triangle is equal to the sum of square of other two sides.
$?$The given sides form a right triangle.

Q.5 A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree . Find the original height of the tree.

Answer :

Let AB be the original height of the tree and broken at C touching the ground at D such that
AC = 5 m
and AD = 12 m
In right triangle $\mathrm{?}$CAD,
$A{D}^2+A{C}^2=C{D}^2$ (By Pythagoras property)
Or, $(12{)}^2+(5{)}^2=C{D}^2$
Or, 144 + 25 =$C{D}^2$ Or, 169 = $C{D}^2$
$?$CD =$\sqrt{169}$ = 13 m
But CD = BC
AC + CB = AB
5 m + 13 m = AB
$?$ AB = 18 m .
Thus, the original height of the tree = 18 m.

Q.6 Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.
(i) $P{Q}^2+Q{R}^2=R{P}^2$
(ii)$P{Q}^2+R{P}^2=Q{R}^2$
(iii)$R{P}^2+Q{R}^2=P{Q}^2$

Answer :

We know that
$\mathrm{?}P+\mathrm{?}Q+\mathrm{?}R=180°$(Angle sum property)
$\mathrm{?}$P + 25° + 65° = 180°
$\mathrm{?}$P + 90° = 180°
$\mathrm{?}$P = 180° – 90° – 90°
$\mathrm{?}$PQR is a right triangle, right angled at P
(i) Not True
$?P{Q}^2+Q{R}^2?R{P}^2$ (By Pythagoras property)
(ii) True
$?$ $P{Q}^2+R{P}^2=Q{P}^2$ (By Pythagoras property)
(iii) Not True
$?R{P}^2+Q{R}^2?P{Q}^2$(By Pythagoras property)

Q.7 Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Answer :

Length of rectangle = 40cm
Diagonal = 41 cm
$Breat{h}^2={41}^2?{40}^2$
Or, $breat{h}^2=1681–1600$
Or, breath = $\sqrt{81}$ cm
Breath = 9 cm
Perimeter of rectangle = 2×(40+9) = 2×49 = 98 cm

Q.8 The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer :

Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm
Since, the diagonals of a rhombus bisect each other at 90°.
$?$ OA = OC = 8 cm and OB = OD = 15 cm
In right $\mathrm{?}$OAB,
$A{B}^2=O{A}^2+O{B}^2$ (By Pythagoras property)
= $(8{)}^2+(15{)}^2=64+225$ = 289
$?$AB =$\sqrt{289}$= 17 cm
Since AB = BC = CD = DA (Property of rhombus)
$?$Required perimeter of rhombus = 4 × side = 4 × 17 = 68 cm.