MCQ: Triangles - 3 - SSC CGL MCQ

Let the sides of the triangle be 3x, 4x and 6x units.
Clearly, (3x)² + (4x)² < (6x)²
∴ The triangle will be obtuse angled.
Hence, option C is correct.

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

Since, DE || BC

⇒ x (x –1) = (x +2)(x –2)
⇒ x² – x = x² – 4
⇒ x = 4
Hence, option B is correct.

The right bisectors of the sides of a triangle meet at a point. The point of intersection is called circum-centre. For an obtuse angled triangle, circum-centre lies outside the triangle.
Hence, option D is correct

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.

We know that "The sum of two sides of a triangle should be greater than the third side."
Following this we can get only two possible combinations using the given details as mentioned below.
(3, 5, 6) and (2, 5, 6)
Hence, option B is correct.

According to the question,

⇒ AC² = 2 x AB x BC
⇒ AB² + BC² = 2 x AB × BC [By Pythagoras theorem, AC² = AB² + BC²]
⇒ AB² + BC² – 2 x 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.

Let PS be altitude of ΔPQR.

As we know, PQ = 10 cm and PS = 8 cm
Now, by the Pythagorean theorem, we get
∴ In ΔPSQ, QS = 6 cm
∴ SR = QR – PS = (20 – 6) = 14 cm
In ΔPSR,
PR² = PS² + SR² = 82 + 142 = 260.
∴ PR = 260 cm
Hence, option A is correct.

Let, AB = 4x cm, BC = 5x cm, CA = 6x cm
Now, ΔOBA + ΔBOC + ΔAOC = ΔABC

⇒ 12 + 15 + 18 = 6h
⇒ 45 = 6h
⇒ h = 7.5 cm
Hence, option A is correct.

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.

∠CBD = ∠A + ∠C , ∠BCE = ∠B + ∠A.
[∵ An exterior angle of a triangle is equal to the sum of the opposite interior angles.]
∴ ∠CBD + ∠BCE = (∠A + ∠B + ∠C) + ∠A = 180° + A

Hence, option B is correct.

We know that, Exterior angle is sum of opposite interior angles

Similarly,
∠DCE = ∠DBC + ∠BDC
⇒ y = x + 50°
From equation (i)
∠A = 2(x + 50°) – 2x = 2x + 100° – 2x = 100°
Hence, option A is correct.

Given, ΔABC is an isosceles triangle 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,

Hence, option C is correct.

∠B = ∠ACB = 70°
∠BAC = 180° – (70° + 70°) = 40°
∠ADE = ∠CAD + ∠ACD = 40° + 110° = 150°
Hence, option A is correct.