Baudhayan theorem, also commonly known as the Pythagoras theorem, is the key concept in this exercise. Students will get a clear picture of the concept when solving the problems of the exercise. To guide students in the right direction, RD Sharma Solutions Class 10 is prepared by subject experts at BYJU’S. Additionally, the RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.7 PDF is provided below.
RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.7
Access RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.7
1. If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.
Solution:
We have,
Sides of the triangle as
AB = 3 cm
BC = 4 cm
AC = 6 cm
On finding their squares, we get
AB2 = 32 = 9
BC2 = 42 = 16
AC2 = 62 = 36
Since, AB2 + BC2 ≠ AC2
So, by the converse of Pythagoras theorem, the given sides cannot be the sides of a right triangle.
2. The sides of certain triangles are given below. Determine which of them are right triangles.
(i) a = 7 cm, b = 24 cm and c = 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm
(iii) a = 1.6 cm, b = 3.8 cm and c = 4 cm
(iv) a = 8 cm, b = 10 cm and c = 6 cm
Solutions:
(i) Given,
a = 7 cm, b = 24 cm and c = 25 cm
∴ a2 = 49, b2 = 576 and c2 = 625
Since, a2 + b2 = 49 + 576 = 625 = c2
Then, by the converse of Pythagoras’ theorem
The given sides are of a right triangle.
(ii) Given,
a = 9 cm, b = 16 cm and c = 18 cm
∴ a2 = 81, b2 = 256 and c2 = 324
Since, a2 + b2 = 81 + 256 = 337 ≠ c2
Then, by converse of Pythagoras theorem
The given sides cannot be of a right triangle.
(iii) Given,
a = 1.6 cm, b = 3.8 cm and C = 4 cm
∴ a2 = 2.56, b2 = 14.44 and c2 = 16
Since, a2 + b2 = 2.56 + 14.44 = 17 ≠ c2
Then, by the converse of Pythagoras theorem
The given sides cannot be of a right triangle.
(iv) Given,
a = 8 cm, b = 10 cm and C = 6 cm
∴ a2 = 64, b2 = 100 and c2 = 36
Since, a2 + c2 = 64 + 36 = 100 = b2
Then, by the converse of Pythagoras theorem
The given sides are of a right triangle
3. A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?
Solution:
Let the starting point of the man be O and the final point be A.
In ∆ABO,
by Pythagoras theorem AO2 = AB2 + BO2
⇒ AO2 = 82 + 152
⇒ AO2 = 64 + 225 = 289
⇒ AO = √289 = 17 m
∴ the man is 17m far from the starting point.
4. A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.
Solution:
In ∆ABC, by Pythagoras theorem
AB2 + BC2 = AC2
⇒ 152 + BC2 = 172
225 + BC2 = 172
BC2 = 289 – 225
BC2 = 64
∴ BC = 8 m
Therefore, the distance between the foot of the ladder and the building = 8 m
5. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Solution:
Let CD and AB be the poles of height 11m and 6m.
Then, it’s seen that CP = 11 – 6 = 5m.
From the figure, AP should be 12m (given)
In triangle APC, by applying Pythagoras theorem, we have
AP2 + PC2 = AC2
122 + 52 = AC2
AC2 = 144 + 25 = 169
∴ AC = 13 (by taking sq. root on both sides)
Thus, the distance between their tops = 13 m.
6. In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.
Solution:
Given,
∆ABC, AB = AC = 25 cm and BC = 14.
In ∆ABD and ∆ACD, we see that
∠ADB = ∠ADC [Each = 90°]
AB = AC [Given]
AD = AD [Common]
Then, ∆ABD ≅ ∆ACD [By RHS condition]
Thus, BD = CD = 7 cm [By corresponding parts of congruent triangles]
Finally,
In ∆ADB, by Pythagoras theorem
AD2 + BD2 = AB2
⇒ AD2 + 72 = 252
AD2 = 625 – 49 = 576
∴ AD = √576 = 24 cm
7. The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?
Solution:
Let’s assume the length of the ladder to be, AD = BE = x m
So, in ∆ACD, by Pythagoras theorem
We have,
AD2 = AC2 + CD2
⇒ x2 = 82 + 62 … (i)
Also, in ∆BCE, by Pythagoras theorem
BE2 = BC2 + CE2
⇒ x2 = BC2 + 82 … (ii)
Compare (i) and (ii)
BC2 + 82 = 82 + 62
⇒ BC2 + 62
⇒ BC = 6 m
Therefore, the tip of the ladder reaches to a height of 6 m.
8. Two poles of height 9 in and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Solution:
Comparing with the figure, it’s given that
AC = 14 m, DC = 12m and ED = BC = 9 m
Construction: Draw EB ⊥ AC
Now,
It’s seen that AB = AC – BC = (14 – 9) = 5 m
And, EB = DC = 12m [distance between their feet]
Thus,
In ∆ABE, by Pythagoras theorem, we have
AE2 = AB2 + BE2
AE2 = 52 + 122
AE2 = 25 + 144 = 169
⇒ AE = √169 = 13 m
Therefore, the distance between their tops = 13 m
9. Using Pythagoras theorem determine the length of AD in terms of b and c shown in Fig. 4.219
Solution:
We have,
In ∆BAC, by Pythagoras theorem, we have
BC2 = AB2 + AC2
⇒ BC2 = c2 + b2
⇒ BC = √(c2 + b2)
In ∆ABD and ∆CBA
∠B = ∠B [Common]
∠ADB = ∠BAC [Each 90°]
Then, ∆ABD ͏~ ∆CBA [By AA similarity]
Thus,
AB/ CB = AD/ CA [Corresponding parts of similar triangles are proportional]
c/ √(c2 + b2) = AD/ b
∴ AD = bc/ √(c2 + b2)
10. A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm.
Solution:
From fig. AB = 5cm, BC = 12 cm and AC = 13 cm.
Then, AC2 = AB2 + BC2.
⇒ (13)2 = (5)2 + (12)2 = 25 + 144 = 169 = 132
This proves that ∆ABC is a right triangle, right angled at B.
Let BD be the length of the perpendicular from B on AC.
So, the area of ∆ABC = (BC x BA)/ 2 [Taking BC as the altitude]
= (12 x 5)/ 2
= 30 cm2
Also, the area of ∆ABC = (AC x BD)/ 2 [Taking BD as the altitude]
= (13 x BD)/ 2
⇒ (13 x BD)/ 2 = 30
BD = 60/13 = 4.6 (to one decimal place)
11. ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of ∆ FBE = 108cm2, find the length of AC.
Solution:
Given,
ABCD is a square. And, F is the mid-point of AB.
BE is one-third of BC.
Area of ∆ FBE = 108 cm2
Required to find: length of AC
Let’s assume the sides of the square to be x.
⇒ AB = BC = CD = DA = x cm
And, AF = FB = x/2 cm
So, BE = x/3 cm
Now, the area of ∆ FBE = 1/2 x BE x FB
⇒ 108 = (1/2) x (x/3) x (x/2)
⇒ x2 = 108 x 2 x 3 x 2 = 1296
⇒ x = √(1296) [taking square roots of both sides]
∴ x = 36 cm
Further in ∆ ABC, by Pythagoras theorem, we have
AC2 = AB2 + BC2
⇒ AC2 = x2 + x2 = 2x2
⇒ AC2 = 2 x (36)2
⇒ AC = 36√2 = 36 x 1.414 = 50.904 cm
Therefore, the length of AC is 50.904 cm.
12. In an isosceles triangle ABC, if AB = AC = 13cm and the altitude from A on BC is 5cm, find BC.
Solution:
Given,
An isosceles triangle ABC, AB = AC = 13 cm, AD = 5 cm
Required to find: BC
In ∆ ADB, by using Pythagoras theorem, we have
AD2 + BD2 = 132
52 + BD2 = 169
BD2 = 169 – 25 = 144
⇒BD = √144 = 12 cm
Similarly, applying Pythagoras theorem is ∆ ADC we can have,
AC2 = AD2 + DC2
132 = 52 + DC2
⇒ DC = √144 = 12 cm
Thus, BC = BD + DC = 12 + 12 = 24 cm
13. In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
(i) AD = a√3 (ii) Area (∆ABC) = √3 a2
Solution:
(i) In ∆ABD and ∆ACD, we have
∠ADB = ∠ADC = 90o
AB = AC [Given]
AD = AD [Common]
So, ∆ABD ≅ ∆ACD [By RHS condition]
Hence, BD = CD = a [By C.P.C.T]
Now, in ∆ABD, by Pythagoras theorem
AD2 + BD2 = AB2
AD2 + a2 = 2a2
AD2 = 4a2 – a2 = 3a2
AD = a√3
(ii) Area (∆ABC) = 1/2 x BC x AD
= 1/2 x (2a) x (a√3)
= √3 a2
14. The lengths of the diagonals of a rhombus is 24 cm and 10 cm. Find each side of the rhombus.
Solution:
Let ABCD be a rhombus and AC and BD be the diagonals of ABCD.
So, AC = 24cm and BD = 10cm
We know that diagonals of a rhombus bisect each other at right angle. (Perpendicular to each other)
So,
AO = OC = 12 cm and BO = OD = 3 cm
In ∆AOB, by Pythagoras theorem, we have
AB2 = AO2 + BO2
= 122 + 52
= 144 + 25
= 169
⇒ AB = √169 = 13cm
Since the sides of the rhombus are all equal,
Therefore, AB = BC = CD = AD = 13cm.
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