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.

Answer: Option
B

Explanation:

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.

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}

Answer: Option
A

Explanation:

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.

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

Answer: Option
C

Explanation:

According to the question,

⇒ 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.

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

Answer: Option
B

Explanation:

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° = ?

Answer: Option
C

Explanation:

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,

Again from ΔADP,

Hence, option C is correct.