NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.5: The Triangle and its Properties
Ex 6.5 Class 7 Maths Question 1.
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Solution:
In right-angled triangle PQR, we have
By Pythagoras property.
\( QR^2 = PQ^2 + PR^2 \)
=\( (10)^2 + (24)^2\)
= 100 + 576 = 676
∴ QR = \(\sqrt {676} \) = 26 cm
The, the required length of QR = 26 cm.
Ex 6.5 Class 7 Maths Question 2.
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Solution:
In right-angled ∆ABC, we have
\(BC^2 + (7)^2 = (25)^2 \)(By Pythagoras property)
⇒\(BC^2\) + 49 = 625
⇒ \(BC^2\) = 625 – 49
⇒ \(BC^2\) = 576
∴ BC = \(\sqrt {576}\) = 24 cm
Thus, the required length of BC = 24 cm.
Ex 6.5 Class 7 Maths 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.
Solution:
Here, the ladder forms a right-angled triangle.
∴ \(a^2 + (12)^2 = (15)^2\) (By Pythagoras property)
⇒ \(a^2\)+ 144 = 225
⇒ \(a^2\) = 225 – 144
⇒ \(a^2\) = 81
∴ a = \( \sqrt {81}\) = 9 m
Thus, the distance of the foot from the ladder = 9m
Ex 6.5 Class 7 Maths 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
Solution:
(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 the 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 = 8 cm
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
Since 6.25 cm = 6.25 cm = 6.25 cm
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.
Ex 6.5 Class 7 Maths 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.
Solution:
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 ∆CAD,
\(AD^2 + AC^2 = CD^2\) (By Pythagoras property)
⇒\( (12)^2 + (5)^2 = CD^2\)
⇒ 144 + 25 = \(CD^2\)
⇒ 169 = \(CD^2\)
∴ 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.
Ex 6.5 Class 7 Maths Question 6.
Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.
(i) \( PQ^2 + QR^2 = RP^2\)
(ii)\(PQ^2 + RP^2 = QR^2\)
(iii) \(RP^2 + QR^2 = PQ^2 \)
Solution:
We know that
∠P + ∠Q + ∠R = 180° (Angle sum property)
∠P + 25° + 65° = 180°
∠P + 90° = 180°
∠P = 180° – 90° – 90°
∆PQR is a right triangle, right angled at P
(i) Not True
∴\( PQ^2 + QR^2 ≠ RP^2 \) (By Pythagoras property)
(ii) True
∴ \(PQ^2 + RP^2 = QP^2 \) (By Pythagoras property)
(iii) Not True
∴ \( RP^2 + QR^2 ≠ PQ^2 \) (By Pythagoras property)
Ex 6.5 Class 7 Maths Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Solution:
Given: Length AB = 40 cm
Diagonal AC = 41 cm
In right triangle ABC, we have
\( AB^2 + BC^2= AC^2 \) (By Pythagoras property)
⇒\( (40)^2 + BC^2 = (41)^2 \)
⇒ 1600 + \(BC^2 \) = 1681
⇒ \(BC^2 \) = 1681 – 1600
⇒ \(BC^2 \) = 81
∴ BC = \( sqrt {81} \) = 9 cm
∴ AB = DC = 40 cm and BC = AD = 9 cm (Property of rectangle)
∴ The required perimeter
= AB + BC + CD + DA
= (40 + 9 + 40 + 9) cm
= 98 cm
Ex 6.5 Class 7 Maths Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution:
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 ∆OAB,
\( AB^2 = OA^2 + OB^2 \) (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.