Identify the smallest and the greatest side of ΔΡQR.
Answers
ANSWER:
(i)
In the right angled triangle LMN, ∠M = 90∘. Hence, side LN is the hypotenuse.
According to Pythagoras' theorem,
l(LN)2 = l(LM)2 + l(MN)2
⇒(x)2 = (7)2 + (24)2
⇒x2 = 49 + 576
⇒x2 = 625
⇒x2 = (25)2
⇒x = 25
∴ the value of x is 25.
(ii)
- In the right angled triangle PQR, ∠Q = 90∘. Hence, side PR is the hypotenuse.
According to Pythagoras' theorem,
l(PR)2 = l(QR)2 + l(PQ)2
⇒(41)2 = (x)2 + (9)2
⇒1681 = x2 + 81
⇒x2 = 1681 − 81
⇒x2 = 1600
⇒x2 = (40)2
⇒x = 40
∴ the value of x is 40.
(iii)
In the right angled triangle EDF, ∠D = 90∘. Hence, side EF is the hypotenuse.
According to Pythagoras' theorem,
l(EF)2 = l(ED)2 + l(DF)2
⇒(17)2 = (x)2 + (8)2
⇒289 = x2 + 64
⇒x2 = 289 − 64
⇒x2 = 225
⇒x2 = (15)2
⇒x = 15
∴ the value of x is 15.
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Question 2:
In the right-angled ∆PQR, ∠ P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of seg QR.
ANSWER:
In the right angled triangle PQR, ∠P = 90∘. Hence, side QR is the hypotenuse.
According to Pythagoras' theorem,
l(QR)2 = l(PQ)2 + l(PR)2
⇒l(QR)2 = (24)2 + (10)2
⇒l(QR)2 = 576 + 100
⇒l(QR)2 = 676
⇒l(QR)2 = (26)2
⇒l(QR) = 26
∴ Length of seg QR = 26 cm.
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Question 3:
In the right-angled ∆LMN, ∠ M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of seg MN.
ANSWER:
In the right angled triangle LMN, ∠M = 90∘. Hence, side LN is the hypotenuse.
According to Pythagoras' theorem,
l(LN)2 = l(MN)2 + l(LM)2
⇒(20)2 = l(MN)2 + (12)2
⇒400 = l(MN)2 + 144
⇒l(MN)2 = 400 − 144
⇒l(MN)2 = 256
⇒l(MN)2 = (16)2
⇒l(MN) = 16
∴ Length of seg MN = 16 cm.
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Question 4:
The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder ?
ANSWER:
Let LN be ladder of length 15 m that is resting against a wall. Let M be the base of the wall and L be the position of the window.
The window is 9 m above the ground. Now, MN is the distance between base of the wall and that of the ladder.
In the right angled triangle LMN, ∠M = 90∘. Hence, side LN is the hypotenuse.
According to Pythagoras' theorem,
l(LN)2 = l(MN)2 + l(LM)2
⇒(15)2 = l(MN)2 + (9)2
⇒225 = l(MN)2 + 81
⇒l(MN)2 = 225 − 81
⇒l(MN)2 = 144
⇒l(MN)2 = (12)2
⇒l(MN) = 12
∴ Length of seg MN = 16 m.
Hence, the distance between base of the wall and that of the ladder is 12 m.
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Question 1:
Find the Pythagorean triplets from among the following sets of numbers.
(i) 3, 4, 5
(ii) 2, 4, 5
(iii) 4, 5, 6
(iv) 2, 6, 7
(v) 9, 40, 41
(vi) 4, 7, 8
ANSWER:
It is known that, if in a triplet of natural numbers, the square of the biggest number is equal to sum of the squares of the other two numbers, then the three numbers form a Pythgorean triplet.
(i) The given set of numbers is (3, 4, 5).
The biggest number among the given set is 5.
52 = 25; 42 = 16; 32 = 9
Now, 16 + 9 = 25
∴ 42 + 32 = 52
Thus, (3, 4, 5) forms a Pythagorean triplet.
(ii) The given set of numbers is (2, 4, 5).
The biggest number among the given set is 5.
52 = 25; 42 = 16; 22 = 4
Now, 16 + 4 = 20 ≠ 25
∴ 42 + 22 ≠ 52
Thus, (2, 4, 5) does not form a Pythagorean triplet.
(iii) The given set of numbers is (4, 5, 6).
The biggest number among the given set is 6.
62 = 36; 52 = 25; 42 = 16
Now, 25 + 16 = 41 ≠ 36
∴ 52 + 42 ≠ 62
Thus, (4, 5, 6) does not form a Pythagorean triplet.
(iv) The given set of numbers is (2, 6, 7).
The biggest number among the given set is 7.
72 = 49; 62 = 36; 22 = 4
Now, 4 + 36 = 40 ≠ 49
∴ 22 + 62 ≠ 72
Thus, (2, 6, 7) does not form a Pythagorean triplet.
(v) The given set of numbers is (9, 40, 41).
The biggest number among the given set is 41.
92 = 81; 402 = 1600; 412 = 1681
Now, 81 + 1600 = 1681
∴ 92 + 402 = 412
Thus, (9, 40, 41) forms a Pythagorean triplet.
(vi) The given set of numbers is (4, 7, 8).
The biggest number among the given set is 8.
82 = 64; 72 = 49; 42 = 16
Now, 16 + 49 = 65 ≠ 64
∴ 42 + 72 ≠ 82
Thus, (4, 7, 8) does not form a Pythagorean triplet.
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Question 2:
The sides of some triangles are given below. Find out which ones are right-angled triangles?
(i) 8, 15, 17
(ii) 11, 12, 15
(iii) 11, 60, 61
(iv) 1.5, 1.6, 1.7
(v) 40, 20, 30
ANSWER:
It is known that, if in a triplet of natural numbers, the square of the biggest number is equal to sum of the squares of the other two numbers, then the three numbers form a Pythgorean triplet. If the lengths of the sides of a triangle form such a triplet, then the tria