Test Level 2: Geometry - 2 - CAT MCQ

As shown in the figure, the required acute angle is $\displaystyle \angle$AEB.

∠ACB = 1/2(∠AOB) = (1/2)(80°) = 40°

[∵ the angle subtended by an arc or a chord of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.]

Similarly:
∠DBC = (1/2)(∠DOC) = (1/2)(70°) = 35°
∴ ∠AEB = ∠ECB + ∠EBC =∠ACB +∠DBC [∵ ∠AEB is the exterior angle of ΔBEC.]
= 40° + 35° = 75°

We know that the angle made by a chord at the centre is twice that made by it at any point on the circle.

It is also given that AB is perpendicular to OC. Let us suppose they intersect at M.

Now, in triangle AMC:
∠AMC + ∠ACM + ∠CAM = 180°
i.e.
90° + 35° + ∠ACM = 180°
∠ACM = 180° - 125° = 55°
∠ACM = ∠ACO
So, ∠OCA = 55°

The smallest radius means the smallest circumference. The circumference is 2πr. For six equidistant points with distance between them to be an integer, this integer has to be the smallest, i.e. 1. Therefore, the circumference is 6 units. 
So, 2πr = 6 or r = 3/π.

In ΔACB, ∠BAC = θ = 30°

As AB is a diameter, CB² = CD.CA.

Thus, CB² = 2CD.CB, which implies CB = 2.CD =2√3

Again, in ΔACB, ∠BAC = θ = 30°, so tan 30° = CB/AB

We obtain, AB = √3 x CB = √3 x 2√3 = 6

Hence, radius of the circle = 6/2 = 3 units

Option 1 is the answer.

Now, from the property of angle bisector theorem, an angle bisector divides opposite sides in the same ratio as the side containing the angle.

Perimeter of ΔABC = AB + BC + CA = 24 + 36 + 30 = 90 units

We know that ABCDE is a regular pentagon.

Therefore, all its sides are equal and each angle measures 108°.

Now, in ΔBCD, ∠BCD = 108°

∴ ∠BDC = ∠DBC = 36°

Similarly, in ΔABC, ∠BAC = $%5Cangle$ACB = 36°

∴ ∠AOB = θ = ∠DBC + ∠ACB = 36° + 36° = 72° {∵ ∠AOB is the exterior angle of ΔOBC}

In cyclic quadrilateral ABCD, ∠CAD + ∠BAC + ∠BCA + ∠ACD = 180° and these angles form an AP with the first term equal to 15°.
15 + 15 + d + 15 + 2d + 15 + 3d = 180
6d = 120
d = 20
Hence, the common difference is 20° and ∠BCA = 15 + 2d = 55°.

Given above is the figure based on the data provided in the question.
In triangle OAP,
OP² = 3² + 4²
OP = 5 cm
Also, OP = OM + MP
So, MP = (5 - 3) cm = 2 cm

By symmetry, the centre of the largest circle is the centre of the smaller circle in the middle. By making the constructions as shown and with appropriate representation of the lengths, we get r² = 30² + 10² = 1000. Thus, A = πr² = π(1000) = 1000π cm^2.

After joining all the opposite vertices (let them meet at point X), we find that right triangle AOF ≌ Right triangle AOX.
⇒ Area of ΔAOX = Area of ΔAOF

According to the given information, AP/PB =4/3

PQ || AC and QD || CP.
Triangles ABC and PBQ are similar (by AAA).
Therefore,
CQ/QB = 4/3

Similarly, triangles PBC and DBQ are similar and 

If AP = 4x and PB = 3x, then DB = 9x/7 and PD = 12x/7

So,

∠BDA = 90°, ∠AHD = 36°, AB = 2AD
∠DBA = 30°
So, DCA = 30° (because angles by the same segment at the circumference are equal)
Now, we know that triangle ABD is a right triangle.
So, ∠DAB = 60° and∠DBA = 30°
∠TDA =∠DBA (Angles in alternate segment)

Let the dogs be tied at P, Q, R and S, where P, Q and R are in one plane, and S is in a different plane.
Note: PS = QS = RS = PQ = QR = RS

ΔPQS, ΔPSR and ΔQSR are congruent and equilateral.

According to the given condition, AB = 2CD.
⇒ KP = CD = PF = AB/2

Also, AB = 4KL
⇒ KL = PG = AB/4

Now, in the right triangle FPG,
tan (∠FGO) = PF/PG = 2

So, none of the options is satisfying the above equation.

According to the given information: ∠CAT = 100° and ∠BAC = 80°

∠ABC = 50° = ∠ACT ( Angles are in the alternate segment)
∠BCA = 180° - (80° + 50°) = 180° - 130° = 50°
Hence, ∠BOA = 2 x∠BCA = 100°

An acute angle of the parallelogram is 60° (given).
Therefore, ∠ADC and ∠ABC = 60°
Thus, ∠BAD and ∠BCD = 120°
EFGH forms a rectangle in the above given figure.
Consider ΔDEC, DC = 8, EC = 4 and ED = 4√3 (By 30°, 60°, 90° theorem)
Considering ΔBFC, BC = 6, FC = 3 and FB = 3√3 (By 30°, 60°, 90° theorem)
Considering ΔAHD, AD = 6; therefore, AH = 3 and DH = 3√3
Now, side of the rectangle, EH = DE – DH = 4√3 – 3√3 = √3 and EF = EC – FC = 4 – 3 = 1
Area of rectangle EFGH = √3 square units

Join AP & AQ.
Now, let Area of ΔABC = 2n
Area of ΔABD = Area of ΔADC = 1/2 Area of ΔABC = n

As PD = BC = DQ;
Area of ΔAPD = Area of ΔABC = Area of ΔADQ = 2n
Area of ΔPED = Area of ΔPAE = 1/2 Area of ΔPAD = n

Similarly, Area of ΔEDQ = Area of ΔQEA = 1/2 Area of ΔQDA = n

Now,
Area of ΔLPB = Area of ΔLBD = a; Area of ΔLDE = Area of ΔLEA = b;
Area of ΔMDC = Area of ΔMQC = d; Area of ΔAEM = Area of ΔEDM = c
Area of ΔPDE = 2a + b = n
Area of ΔABD = 2b + a = n

Solving above two equations; we get, a = b = (n/3)

Similarly; we get c = d = (n/3)

Area of pentagon = a + b + c + d = 

In ΔEDG and in ΔFEC:

1.8y = 0.9x + 0.9y
0.9y = 0.9x
x = y

Now, in ΔABE and ΔFCE:

12x = 3.78 + 3.6x
8.4x = 3.78
x = 0.45 m

We can see that OP = OR

Suppose ST = y and PT = r

then SU = ST = y and PO = RO = RU = PT = r

So, RS = r + y and OS = 2r - y

In right angle ΔSOR

OS² + OR² = RS²

⇒ (2r - y)² + r² = (r + y)²

⇒ 4r² + y² - 4ry + r² = r² + y² + 2ry

⇒ 4r² - 4ry = 2ry

Area of the bigger semi-circular field

Total area of the grazing surface for horses =

Then the non-grazing part of the field 

2πr = 3a = 4b (where r is the radius of the circle, a is the side of the triangle and b is the side of the square).
Now, take any values of r, a and b, which satisfy the given equation.
Let r = 21 cm,
Therefore, 2πr = 3a

b = 33 cm

Therefore, area of circle =πr² = (22/7) x 441 = 1386 cm²

Area of triangle

Area of square = b² = 33² = 1089 cm²
In each case, one will find that: Area of circle > Area of square > Area of triangle
Or c > s > t