In this chapter, we provide RD Sharma Class 10 Ex 4.7 Solutions Chapter 4 Triangles for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 4.7 Solutions Chapter 4 Triangles pdf, Now you will get step by step solution to each question.
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RD Sharma Solutions for Class 10 Chapter 4 Triangles Variables Ex 4.7 Download PDF
Chapter 4: Triangles Exercise – 4.7
If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.
Sides of triangle
AB = 3 cm
BC = 4 cm
AC = 6 cm
∴ AB2 = 32 = 9
BC2 = 42 = 16
AC2 = 62 = 36
Since, AB2 + BC2 ≠ AC2
Then, by converse of Pythagoras theorem, triangle is not a right triangle.
The sides of certain triangles are given below. Determine which of them right triangles are.
(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
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 converse of Pythagoras theorem, given triangle is a right triangle.
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, given triangle is not a right triangle.
a = 1.6 cm, b = 3.8 cm and C = 4 cm
∴ a2 = 64, b2 = 100 and c2 = 36
Since, a2 + c2 = 64 + 36 = 100 = b2
Then, by converse of Pythagoras theorem, given triangle is a light triangle.
A man goes 15 metres due west and then 8 metres due north. How far is lie from the starting point?
Let the starting point of the man be O and final point be A.
∴ In ∆ABO, by Pythagoras theorem AO2 = AB2 + BO2
⟹ AO2 = 82 + 152
⟹ AO2 = 64 + 225 = 289
⟹ AO = √289 = 17m
∴ He is 17m far from the starting point.
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.
In ∆ABC, by Pythagoras theorem
AB2 + BC2 = AC2
⟹ 152 + BC2 = 172
⟹ 225 + BC2 = 172
⟹ BC2 = 289 – 225
⟹ BC2 = 64
⟹ BC = 8 m
∴ Distance of the foot of the ladder from building = 8 m
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.
Let CD and AB be the poles of height 11 and 6 in.
Therefore CP = 11 – 6 = 5 m
From the figure we may observe that AP = 12m
In triangle APC, by applying Pythagoras theorem
Ap2 + PC2 = AC2
122 + 52 = AC2
AC2 = 144 + 25 = 169
AC = 13
Therefore distance between their tops = 13 m.
In an isosceles triangle ABC. AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.
AB = AC = 25 cm and BC = 14
In ∆ABD and ∆ACD
∠ADB = ∠ADC [Each = 90°]
AB = AC [Each = 25 cm]
AD = AD [Common]
Then, ∆ABD ≅ ∆ACD [By RHS condition]
∴ BD = CD = 7 cm [By c.p.c.t]
In ∆ADB, by Pythagoras theorem
AD2 + BD2 = AB2
⟹ AD2 + 72 = 252
⟹ AD2 = 625 – 49 = 576
⟹ AD = √576 = 24 cm
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?
Let length of ladder be AD = BE = l m
In ∆ACD, by Pythagoras theorem
AD2 = AC2 + CD2
⟹ l2 = 82 + 62 … (i)
In ∆BCE, by pythagoras theorem
BE2 = BC2 + CE2
⟹ l2 = BC2 + 82 … (ii)
Compare (i) and (ii)
BC2 + 82 = 82 + 62
⟹ BC2 + 62
⟹ BC = 6 m
Two poles of height 9 in and 14 m stand on a plane ground. If the distance between their feet is 12 in, find the distance between their tops.
AC = 14 m. DC = 12m and ED = BC = 9 m
Construction: Draw EB ⊥ AC
∴ AB = AC – BC = 14 — 9 = 5 m
And, EB = DC = 12m
In ∆ABE, by Pythagoras theorem,
AE2 = AB2 + BE2
⟹ AE2 = 52 + 122
⟹ AE2 = 25 + 144 = 169
⟹ AE= √169 = 13 m
∴ Distance between their tops = 13 m
Using Pythagoras theorem determine the length of AD in terms of b and c shown in Fig. 4.219
In ∆BAC, by Pythagoras theorem
BC2 = AB2 + AC2
⟹ BC2 = c2 + b2
In ∆ABD and ∆CBA
∠B = ∠B [Common]
∠ADB = ∠BAC [Each 90°]
Then, ∆ABD ͏~ ∆CBA [By AA similarity]
[Corresponding parts of similar ∆ are proportional]
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.
Let, AB = 5cm, BC = 12 cm and AC = 13 cm. Then, AC2 = AB2 + BC2. This proves that ∆ABC is a fight triangle. right angles at B. Let BD be the length of perpendicular from B on AC.
= 30 cm2
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