EXERCISE 6.5
Introduction - Ex. 6.5, Triangle and its Properties - NCERT Class 7th Maths Solutions
Q 1, Ex 6.5 - The Triangle and its Properties - Chapter 6 - Maths Class 7th - NCERT
PQR is a triangle right angled at P . If PQ = 10 cm and PR = 24 cm , find QR
Sol :
By applying Pythagoras theorem in △PQR,
(PQ)2 + (PR)2 = (RQ)2
(10)2 + (24)2 = (RQ)2
100 + 576 = (QR)2
676 = (QR)2
QR = 26 cm
Q 2, Ex 6.5 - The Triangle and its Properties - Chapter 6 - Maths Class 7th - NCERT
QUESTION 2
ABC is a triangle right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
By appling Pythagoras theorem in △ABC ,
(AC)2 + (BC)2 = (AB)2
(BC)2 = (AB)2 - (AC)2
(BC)2 = (25)2 - (7)2
(BC)2 = 625 - 49
(BC)2 = 576
BC = 24 cm
Q 3, Ex 6.5 - The Triangle and its Properties - Chapter 6 - Maths Class 7th - NCERT
QUESTION 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 .
Sol :
By applying Pythagoras theorem,
(15)2 = (12)2 + a
225 = 144 + a2
a2 = 225 - 144
a2 = 81 m
a = 9 m
Therefore , the distance of the foot of the ladder from the wall is 9 cm
Q 4, Ex 6.5 - The Triangle and its Properties - Chapter 6 - Maths Class 7th - NCERT
QUESTION 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
In the case of the right-angled triangles , identify the right angles .
Sol :
(i)
2.5 cm, 6.5 cm, 6 cm
(2.5)2 = 6.25
(6.5)2 = 42.25
(6)2 = 36
It can be observed that ,
36 + 6.25 = 42.25
(6)2 + (2.5)2 = (6.5)2
The square of the length of one side is the sum of the squares of the lengths of the remaining two sides . Hence , these are the sides of a right- angled triangle . Right angle will be in front of the side of 6.5 cm measure .
(ii) 2 cm , 2 cm , 5 cm
(2)2 = 4
(2)2 = 4
(5)2 = 25
Here , (2)2+ (2)2 ≠ (5)2
The square of the length of one side is not equal to the sum of the squares of the remaining two sides . Hence , these sides are not of a right-angled triangle .
(iii) 1.5 cm , 2 cm , 2.5 cm
(1.5)2 + (2)2 = (2.5)2
The square of the length of one side is the sum of the squares of the lengths of the remaining two sides . Hence , these are the sides of a right-angled triangle .
Q 5, Ex 6.5 - The Triangle and its Properties - Chapter 6 - Maths Class 7th - NCERT
QUESTION 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.
Sol :
In the given figure , BC represents the unbroken part of the tree . Point C represents the point where the tree broke and CA represents the broken part of the tree . Triangle ABC , thus formed , is right-angled at B .
Applying Pythagoras theorem in △ABC ,
AC2 = BC2 + AB2
AC2 = (5 m)2 + (12 m)2
AC2 = 25 m2+ 144 m2
AC = 13 m
Thus , original height of the tree = AC + CB = 13 m + 5 m = 18 m
Q 6, Ex 6.5 - The Triangle and its Properties - Chapter 6 - Maths Class 7th - NCERT
QUESTION 6
Angles Q and R of a △PQR are 25° and 65°.
Write which of the following is true :
(i) PQ2 + QR2= RP2
(ii) PQ2 + RP2= QR2
(iii) RP2 + QR2 = PQ2
Sol :
The sum of the measures of all interior angles of a triangle is 180°
∠PQR + ∠PRQ + ∠QPR = 180°
25° + 65° + ∠QPR = 180°
90° + ∠QPR = 180°
∠QPR = 180° - 90°
∠QPR = 90°
Therefore , △PQR is right-angled at point P
Hence , (PR)2+ (PQ)2 = (QR)2
Thus , (ii) is true
Q 7, Ex 6.5 - The Triangle and its Properties - Chapter 6 - Maths Class 7th - NCERT
QUESTION 7
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Sol :
In a rectangle , all interior angles of 90° measure . Therefore , Pythagoras theorem can be applied here .
(41)2 = (40)2 + x2
1681 = 1600 + x2
x2 = 1681 - 1600
x2 = 81
x = 9 cm
Perimeter = 2(Length + Breadth)
= 2(x + 40)
= 2(9 + 40)
= 98 cm
Q 8, Ex 6.5 - The Triangle and its Properties - Chapter 6 - Maths Class 7th - NCERT
QUESTION 8
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Sol:
Let ABCD be a rhombus (all sides are of equal length) and its diagonals, AC and BD, are
intersecting each other at point 0. Diagonals in a rhombus bisect each other at 90°. It can be
observed that
$AO=\dfrac{AC}{2}=\dfrac{16}{2}=8 cm$
$BO=\dfrac{BD}{2}=\dfrac{30}{2}=15 cm $
By applying Pythagoras theorem in △AOB ,
OA2 + OB2 = AB2
82 + 152 = AB2
64 + 225 = AB2
289 = AB2
AB = 17
Therefore , the length of the sides of rhombus is 17 cm
Perimeter of rhombus = 4 × side of the rhombus
= 4 × 17
= 68 cm
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