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Directions: Study the following questions carefully and choose the right answer:
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.
» Explain it
B
Traditional Method :

Let the other two sides be x and y.

Given, x + y = 49 cm

and,  412 = x2 + y2         [By pythagoras theorem]

(x + y)2 = x2 + y2 + 2xy

⇒  492 = 412 + 2xy

⇒  2401 = 1681 + 2xy

⇒  2xy = 2401 – 1681 = 720

(x – y)2 = x2 + y2 – 2xy

⇒  (x – y)2 = 412 – 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.
 
2
A point D is taken from the side BC of a right-angled triangle ABC, where AB is hypotenuse. Then
» Explain it
A
 
In ΔABC,

AB2 = AC2 + BC2         [By pythagoras theorem]

⇒  AC2 = AB2 – BC2       ...(i)

In ΔACD,

AD2 = AC2 + CD2         [By pythagoras theorem]

⇒  AD2 = AB2 – BC2 + CD2       [From eq. (i)]

⇒  AB2 + CD2 = BC2 + AD2

Hence, option A is correct.

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
» Explain it
C
 
According to the question,

AB × BC =  AC2
2

⇒  AC2 = 2 × AB × BC

⇒  AB2 + BC2 = 2 × AB × BC         [By pythagoras theorem, AC2 = AB2 + BC2]

⇒  AB2 + BC2 – 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.
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
» Explain it
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.
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° = ?
» Explain it
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.