Test Level 3: Geometry - CAT MCQ

Area of ΔMUV = (1/3) x Area of ΔLMN = 40 sq. units
Similarly, area of ΔLWX = (1/4) x 120 sq. units = 30 sq. units
Let the area of ABCD be x sq. units.
Then, area of the shaded region = 30 sq. units – x + 40 sq. units – x + x = 70 sq. units – x = Area of the unshaded region
∴ 70 sq. units – x = (½) × 120 sq. units = 60 sq.units
∴ x = 10 sq. units

Draw BD || AC to cut OA at D.

In triangle ABC,
3² + AC² = 5²
AC = 4 m

Let radius = r.
OD² + DB² = OB²
DB = AC

DB² = r² - (r - a)²  
4² = 2ar - a²    (where a = 3)
16 + 9 = 2 x 3 x r
or, r = (25/6) m

PQ = PS = 2
⇒ OQ = 2 [∵ ΔOQP ≅ ΔOSP]
Let OA = r
OP = (OC - PC) = r - 2
Also, (OP)² = 2(OQ)²
(r - 2)² = 2(2)²
⇒ r² + 4 - 4r = 8
⇒ r² - 4r - 4 = 0
⇒ r = 2(1 +√2) = 2(√2 + 1)

Hence option c is correct

Let the radius of the circle be 'R' and ∠ODC = ∠ADE = θ.
If OM is drawn perpendicular to DC,
DM = R cos θ
OM = R sin θ
Length of rectangle ABCD,
AB = CD = 2DM = 2R cos θ
AD = BC = 2OM = 2R sin θ
Area of rectangle = AB(BC) = 2R cos θ x 2R sin θ = 2R² sin2θ
Area of circle = πR²
According to the question,
πR² : 2R² sin 2θ = π : √3

⇒ sin 2θ = √2 = sin 60°
θ = 30°
In triangle ADE, tan θ = AE/AD
AE : AD = tan 30° = 1 : √3

Let the radius of the circle be `r` cm.
∴ We get (r - 20)² + (r - 10)² = r²
∴ r² - 40r + 400 + r² - 20r + 100 = r²
∴ r² - 60r + 500 = 0
∴ r = 10 cm or r = 50 cm
But, r cannot be 10 cm.
∴ r = 50 cm

As shown in the figure, area of equilateral triangle 

(a represents the length of side of an equilateral triangle)
a = 24 cm
In triangle RDP,
∠RDP = 30°
cos 30° = RD/PD = 24/PD
PD = 16√3 cm
Area of quadrilateral RDSP

192√3 sq. cm²

As shown in the figure, Area of quadrilateral PQRS = Area of triangle PQR + Area of triangle PRS … (1)
Both these triangles are right-angled triangles.

Area of triangle PQR = (1/2) x 4 x 3 = 6 sq. units

Area of triangle PRS = (1/2) x 12 x 5 = 30 sq. units
Area of quadrilateral PQRS = 6 + 30 = 36 sq. units

Circumcentre will lie on one of the sides if the triangle is a right-angled triangle.
So, two of the vertices of the triangle will lie on a diameter of the circle.
So, the points must be equidistant to maximise the number of such triangles.
Let AN be one such chosen diameter and hence, one side of a triangle.
It is to be noted that there will be 13 such diameters possible.
So, third vertex of the triangle can be chosen from any of the other 24 points left (other 24 points leaving A and N).
Possible total number of such triangles = 13 × 24 = 312.

Angles subtended by equal chords are equal.
∠AFB = 25° = ∠BFC
In the cyclic quadrilateral ABCF,
∠AFC = 25° + 25° = 50°
∠ABC = 180° - 50° = 130° (because the sum of opposite angles of a cyclic quadrilateral is 180°)