MCQ: Geometry - 2 - SSC CGL MCQ

According to question, Given that
∠BAD = ∠CAD. AB = 4 cm, AC = 5.2 cm, BD = 3 cm
In ΔABC, AD is the bisector of ∠A

But BC = BD + CD = 3cm + 3.9 cm = 6.9 cm

Since DE || BC,

Let ABC and DEF be the two similar Δs having area 81 cm2 and 144 cm2 respectively:
Let BC = 27 cm
Then since ΔABC ∼ ΔDEF

∴ x = 36 cm.

Draw a figure as per given question,
In a Right triangle ADC, Use the (Pythagoras Theorem)

In a Right triangle BCE, Use the formula

∴ Width of the street (AC + BC) = AB = 12 + 9 = 21 m.

Lat AB = BC = AC = a
AB² = AD² + BD2 (Pythagoras Theorem)

In ΔACE
∠A + ∠C + ∠E = 180°
Similarly in ΔDFB
∠D + ∠F + ∠B = 180°
∴ (∠A + ∠C + ∠E) + (∠D + ∠F + ∠B) = 360°
∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°

As we know the formula,

∠BOC = 125°

∠DCK = ∠FDG = 55° (corr. ∠s)
∴ ∠ACE = ∠DCK = 55° (vert. opp. ∠s)
So, ∠AEC = 180° – (40° + 55°) = 85°
∴ ∠HAB = ∠AEC = 85° (corr. ∠s)
Hence, x = 85°.

∠AGH = ∠DHG (alt. int. angles)

Thus, lines GM and HL are intersected by a transversal GH at G and H respectively such that pair of alternate angles are equal, i.e.,
∴ ∠HGM = ∠GHL
∴ GM || HL
Similarly, GL || HM
So, GMHL is a ||gm.
Since AB || CD and EF is a transversal
∴ ∠BGH + ∠DHG = 180° [co-interior angles]

But ∠LGH + ∠LHG + ∠GLH = 180°
∴ 90° + ∠GLH = 180° ⇒ ∠GLH = 90°
Thus, in ||gm GMHL, we have ∠GLH = 90°
Hence, GMHL is a rectangle.

Let, the angle be A
⇒ Its complement angle = 90° – A
According to the question
(90 – A) = A + 60°
⇒ 90 - 60 = A + A
⇒ 2A = 30°
⇒ A = 15°

Since OP bisects ∠BOC,
∴ ∠BOC = 2∠POC
Again, OQ bisects ∠AOC,
∴ ∠AOC = 2∠QOC
Since ray OC stands on line AB, ∴,
∠AOC + ∠BOC = 180°
⇒ 2∠QOC + 2∠POC = 180°
⇒ 2∠QOC + ∠POC = 180°
⇒ ∠QOC + ∠POC = 90°
⇒ ∠POQ = 90°.
The above sum can also be restated as follows; The angle between the bisectors of a linear pair of angles is a right angle.

According to question,
∠B = ∠C = 55° , ∠D = 25°
We can say ,
AB = AC ( ∴ ∠B = ∠C = 55° )
In Triangle ABC,
∠A + ∠B + ∠C = 180°
⇒ ∠A + 55° + 55° = 180°
⇒ ∠A + 110° = 180°
⇒ ∠A = 180° - 110°
⇒ ∠A = 70° ..........................(1)
As per given figure,
∠ACD + ∠ACB = 180° ( ∠ACB = ∠C = 55°)
⇒ ∠ACD + 55° = 180°
⇒ ∠ACD = 180° - 55°
⇒ ∠ACD = 125° ....................... (2)
Now in Triangle ACD,
∠CAD + ∠ACD + ∠CDA = 180°
⇒ ∠CAD + 125° + 25° = 180°
⇒ ∠CAD + 150° = 180°
⇒ ∠CAD = 30° ...........................(3)
( In a Δ, greater angle has longer side opposite to it )
From the equation (1) , (2) and (3);
∠B < ∠A and ∠CAD > ∠D ;
∴ BC > CA and CA < CD

Given in the question, QR = 26 cm, PM = 6 cm, MR = 8 cm
According to question, in Δ PMR
(Pythagoras Theorem)

According to question, in Δ PQR
(Pythagoras Theorem)

∴ area of triangle ∆PQR = Base length x Height / 2
⇒ area of triangle ∆PQR = PR x PQ / 2
⇒ area of triangle ∆PQR = 10 x 24 / 2 = 10 x 12
⇒ area of triangle ∆PQR = 120

Clearly DE || BC (by converse of BPT) ∴ ΔADE ∼ ABC (∠A = ∠A and ∠ADE = ∠B)