Aptitude GeometryPage 2

9.	In figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55o and ∠BAC = 45o, find ∠BCD. (a)45o (b)60o (c)80o (d)90o

Answer!

Answer is: C∠CAD = ∠DBC = 55^o (Angles in the same segment)
∠DAB = ∠CAD + ∠BAC = 55^o + 45^o = 100^o
But ∠DAB + ∠BCD = 180^o (Opposite angles of a cyclic quadrilateral)
∠BCD = 180^o - 100^o = 80^o

10.	In figure, ∠ABC = 69o, ∠ACB = 31o, find ∠BDC (a)45o (b)60o (c)80o (d)100o

Answer!

Answer is: CIn ∠ABC,
∠ABC + ∠ACB + ∠BAC = 180^o
69^o + 31^o + ∠BAC = 180^o
∠BAC = 180^o - 100^o
∠BAC = 80^o
But ∠BAC = ∠BDC (Angles in the same segment of a circle are equal)
Hence, ∠BDC = 80^o

11.	In the figure, find the value of X?. (a)45o (b)60o (c)60o (d)120o

Answer!

Answer is: DIn △ABC,
25^o + 35^o + ∠ACB = 180^o
∠ACB = 120^o
Now, ∠ACB + ∠ACD = 180^o (Linear pair)
Or 120^o + ∠ACD = 180^o
Or ∠ACD = 60^o = ∠ECD
Again in the △CDE, CE is produced to A.
Hence, ∠AED = ∠ECD + ∠EDC
X = 60^o + 60^o = 120^o

12.	In the given figure given below, E is the mid-point of AB and F is the midpoint of AD. if the area of FAEC is 13, what is the area of ABCD ? (a)19.5 (b)26 (c)39 (d)None of these

Answer!

Answer is: BAs F is the mid-point of AD, CF is the median of the triangle ACD to the side AD.
Hence area of the triangle FCD = area of the triangle ACF.
Similarly area of triangle BCE = area of triangle ACE.
∴ Area of ABCD = Area of (CDF + CFA + ACE + BCE)
= 2 Area (CFA + ACE) = 2 x 13 = 26 sq. units.

13.	In the given figure, AB is chord of the circle with centre O, BT is tangent to the circle. The values of X and Y are (a)52o, 52o (b)58o, 52o (c)58o, 58o (d)60o, 64o

Answer!

Answer is: CGiven AB is a circle and BT is a tangent, ∠BAO = 32^o
Here, ∠OBT = 90^o [∴ Tangent is ┴ to the radius at the point of contact]
OA = OB [Radii of the same circle]
∴ ∠OBA = ∠OAB = 32^o [Angles opposite to equal side are equal]
∴ ∠OBT = ∠OBA + ∠ABT = 90^o or 32^o + X = 90^o.
∠X = 90^o - 32^o = 58^o
Also, ∠AOB = 180^o - ∠OAB - ∠OBA
= 180^{o o} 32^o - 32^o = 116^o
Now Y = (1 / 2) AOB
[Angle formed at the center of a circle is double the angle formed in the remaining part of the circle]
Y = (1 / 2) x 116^o = 58^o

14.	OA is perpendicular to the chord PQ of a circle with center O. If QR is a diameter, AQ = 4cm, OQ = 5 cm, then PR is equal to (a)6 cm (b)4 cm (c)8 cm (d)10 cm

Answer!

Answer is: AAO = √(OQ² - AQ²)
AO = √(5² - 4²)
AO = √9 = 3
Now, from similar △s QAO and △QOR
OR = 2OA = 2 x 3 = 6 cm.

15.	In a quadrilateral ABCD, the bisectors of ∠A and ∠B meet at O. If ∠C = 70 and ∠D = 130, then measure of ∠AOB is (a)40o (b)60o (c)80o (d)100o

Answer!

Answer is: D
A + B + C + D = 360^o
A + B = 360 - (130 + 70) = 160^o
(A / 2) + (B / 2) = 80^o
In △AOB,
(A / 2) + (B / 2) + O = 180^o
O = 180^o - 80^o = 100^o

16.	In a circle with center O, AB is a chord, and AP is a tangent to the circle. If ∠AOB = 140o, then the measure of ∠PAB is (a)40o (b)60o (c)70o (d)100o

Answer!

Answer is: C
In △AOB,
∠A + ∠B + ∠O = 180^o
∠A + ∠B = 180^o - 140^o
∠A = ∠B = 20^o {AO = BO}
∠PAO = 90^o
∠PAB + ∠BAO = 90^o
∠PAB = 90^o - 20^o = 70^o

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