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Directions: Study the following questions carefully and choose the right answer:
Important for :
CGL Tier 2
CGLTier 1
SSC 10+2
1
If the hypotenuse of a right triangle is 41 cm and the sum of the other two sides is 49 cm, find the difference between the other sides.
A
30 cm
B
31 cm
C
32 cm
D
29 cm
» Explain it
Correct Option:
B
Traditional Method :
Let the other two sides be x and y.
Given, x + y = 49 cm
and, 41
^{2}
= x
^{2}
+ y
^{2}
[By pythagoras theorem]
(x + y)
^{2}
= x
^{2}
+ y
^{2}
+ 2xy
⇒ 49
^{2}
= 41
^{2}
+ 2xy
⇒ 2401 = 1681 + 2xy
⇒ 2xy = 2401 – 1681 = 720
(x – y)
^{2}
= x
^{2}
+ y
^{2}
– 2xy
⇒ (x – y)
^{2}
= 41
^{2}
– 720 = 1681 – 720 = 961
⇒ x – y = 31 cm
______________________________________________
Intutive Method :
Following the Pythagorean triples, if the hypotenuse is 41 units, the only possible combination of the other two sides must be 40 units and 9 units.
And therefore, the difference between the other two sides will be 40 - 9 = 31 cm.
Hence, option B is correct.
1
A
B
C
D
» Explain it
सही विकल्प :
B
B
2
A point D is taken from the side BC of a right-angled triangle ABC, where AB is hypotenuse. Then
A
AB
^{2}
+ CD
^{2}
= BC
^{2}
+ AD
^{2}
B
CD
^{2}
+ BD
^{2}
= 2AD
^{2}
C
CD
^{2}
+ BD
^{2}
= 2AD
^{2}
D
AB
^{2}
= AD
^{2}
+ BD
^{2}
» Explain it
Correct Option:
A
In ΔABC,
AB
^{2}
= AC
^{2}
+ BC
^{2}
[By pythagoras theorem]
⇒ AC
^{2}
= AB
^{2}
– BC
^{2}
...(i)
In ΔACD,
AD
^{2}
= AC
^{2}
+ CD
^{2}
[By pythagoras theorem]
⇒ AD
^{2}
= AB
^{2}
– BC
^{2}
+ CD
^{2}
[From eq. (i)]
⇒ AB
^{2}
+ CD
^{2}
= BC
^{2}
+ AD
^{2}
Hence, option A is correct.
2
A
B
C
D
» Explain it
सही विकल्प :
A
A
3
In a right-angled triangle, the product of two sides is equal to half of the square of the third side i.e., hypotenuse. One of the acute angle must be
A
60°
B
30°
C
45°
D
15°
» Explain it
Correct Option:
C
According to the question,
AB × BC =
AC
^{2}
2
⇒ AC
^{2}
= 2 × AB × BC
⇒ AB
^{2}
+ BC
^{2}
= 2 × AB × BC [By pythagoras theorem, AC
^{2}
= AB
^{2}
+ BC
^{2}
]
⇒ AB
^{2}
+ BC
^{2}
– 2 × AB × BC = 0
⇒ (AB – BC)
^{2}
= 0
⇒ AB = BC
∴ ∠C = ∠A
In ΔABC,
We know that the sum of the angles of a triangle is 180°.
∠A + ∠B + ∠C = 180°
⇒ ∠A + 90° + ∠A = 180° [∵ ΔABC is an right-angled triangle]
⇒ 2∠A = 180° – 90° = 90°
⇒ ∠A = 45° and ∠C = 45° [∵ ∠A = ∠C]
Hence, option C is correct.
3
A
B
C
D
» Explain it
सही विकल्प :
C
C
4
ABC is an isosceles triangle such that AB = AC and AD is the median to the base BC with ∠ABC = 35°. Then ∠BAD is
A
35°
B
55°
C
70°
D
110°
» Explain it
Correct Option:
B
AD is the median to the base BC.
∴ ∠ADB = 90°
Given, ∠ABC = 35°
∴ ∠ABD = 35°
In ΔABD,
We know that the sum of the angles of a triangle is 180°.
∠ABD + ∠ADB + ∠BAD = 180°
35° + 90° + ∠BAD = 180°
∠BAD = 180° – 35° – 90° = 55°
Hence, option B is correct.
4
A
B
C
D
» Explain it
सही विकल्प :
B
B
5
An isosceles triangle ABC is right-angled at B. D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the side AB and AC respectively of ΔABC. If AP = a cm, AQ = b cm and ∠BAD = 15°, sin 75° = ?
A
2b
3
a
B
a
2b
C
3
a
2b
D
2a
3
b
» Explain it
Correct Option:
C
Given, ΔABC is an isosceles trialge and ∠B is right-angled.
∴ ∠A = ∠C
and ∠B = 90°
We know that the sum of the angles of a triangle is 180°.
∴ ∠A + ∠B + ∠C = 180°
⇒ ∠A + 90° + ∠A = 180° [∵ ∠B = 90° & ∠A = ∠C]
⇒ 2∠A = 180° – 90° = 90°
⇒ ∠A = 45° and ∠C = 45° [∵ ∠A = ∠C]
From ΔADP,
∠APD + ∠PAD + ∠ADP = 180°
[
∠BAD = 15° (given)
]
∴ ∠PAD = 15°
⇒ 90° + 15° + ∠ADP = 180° [PD ⊥ AB ∴ ∠APD = 90°]
⇒ ∠ADP = 180° – 90° – 15° = 75°
Now, ∠A = ∠BAD + ∠DAC
⇒ 45° = 15° + ∠DAC
⇒ ∠DAC = 45° – 15° = 30°
∴ ∠DAQ = 30°
From ΔADQ,
∠AQD + ∠DAQ + ∠ADQ = 180°
⇒ 90° + 30° + ∠ADQ = 180° [DQ ⊥ AC ∴ ∠AQD = 90°]
⇒ ∠ADQ = 180° – 90° – 30° = 60°
Again from ΔADQ,
sin 60° =
AQ
AD
⇒
3
=
b
2
AD
⇒ AD =
2b
3
Again from ΔADP,
sin 75° =
AP
=
a
=
a
3
AD
2b/
3
2b
Hence, option C is correct.
5
A
B
C
D
» Explain it
सही विकल्प :
C
C
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