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# Test: Geometry- 1

## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL | Test: Geometry- 1

Description
This mock test of Test: Geometry- 1 for Quant helps you for every Quant entrance exam. This contains 15 Multiple Choice Questions for Quant Test: Geometry- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Geometry- 1 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Test: Geometry- 1 exercise for a better result in the exam. You can find other Test: Geometry- 1 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1

### In the given figure, MN and KL are parallel lines. ∠LKO = 70°, ∠KON = 100° Find ∠MNO.

Solution: Hence, ∠MNO = 30°

QUESTION: 2

### The supplement of an angle is five times the angle. What is the measurement of the angle?

Solution:

Let an angle be x.
So, acc. to question its supplement angle will be 5x.
Adding both the angles will give us sum of 180°.

► x + 5x = 180°
►  6x= 180°
►  x = 30°

QUESTION: 3

### What is the area of the triangle below? Solution:

The area of a triangle may be found by using the formula,

►  A=1/2bh

where represents the base and h represents the height.

Thus, the area may be written as

= 1/2(11)(6) = 33

Hence, the area of the triangle is 33 cm2.

QUESTION: 4

ABCD is an isosceles trapezium with lines AB parallel to CD.
If ∠DCB = 40°, ∠BAD equal Solution:

Isosceles trapezium is always cyclic. The sum of opposite angles of a cyclic quadrilateral is 180°.

∴ ∠BCD + ∠BAD = 180°
► 40° + ∠BAD = 180°

QUESTION: 5

In the given figure, AB and CD are two chords of a circle intersecting at O. If AO = 4 cm and OB = 6 cm and OC = 3 cm. Find OD. Solution:

► AO x OB = OC x OD

► 4 x 6 = 3 x OD

► OD = 8 CM

QUESTION: 6

In the above figure, O is the centre of the circle, and
∠AOB = 120°. Find ∠ACB. Solution: ► 1 / 2 ∠AOB = ∠ACB
► ∠ACB = 60°

• The angle subtended by an arc of a circle at its center is twice of the angle it subtends anywhere on the circle’s circumference.
• The proof of this theorem is quite simple, and uses the exterior angle theorem – an exterior angle of a triangle is equal to the sum of the opposite interior angles.
• If the two opposite interior angles happen to be equal, then the exterior angle will be twice of any of the opposite interior angles.
QUESTION: 7

PQRS is a cyclic quadrilateral. If PQR is an equilateral triangle, find ∠RSP.

Solution:

The sum of opposite angles of a cyclic quadrilateral is 180°.
PQR is an equilateral triangle
So, Measurement of all its angles is 60°.

► ∠PQR + ∠RSP = 180°
► 60° + ∠RSP = 180°
► ∠RSP = 120° QUESTION: 8

In a triangle ABC, the incentre is at O. If ∠BOC  = 100°, find ∠BAC.

Solution: In ∠BOC

► x + y = 80
► 2x + 2y = 160°

Also, 2x + 2y + 2z = 180°

► 160° + 2z = 180°
► ∠BAC = 2z = 20°

QUESTION: 9

In the above figure, QR = 4 cm and RS = 12 cm. TS = 8 cm and QU is extended to T. Find PQ. Solution: In ∠QST and ∠QRU (∠QST and ∠QRU are similar triangles)

► QR / QS = RU / TS
⇒ 4 / 16 = RU / 8
RU = 2

Also,

► 1 / PQ = 1 / 2 - 1 / 8
= (4 - 1) / 8
= 3 / 8

PQ = 8 / 3 cm

QUESTION: 10

How many different regular polygons can be formed with the interior angle exceeding the exterior angle and the sum of the interior angles not exceeding 180°?

Solution:

The sum of interior angles is the sum of the angles formed by the polygon like in square each interior angle is 90 hence the total sum is 90 * 4 = 360.

Here the option must be none of these as only a triangle (that too equilateral in this case) has a sum of interior angles equal to 180.

QUESTION: 11

The ratio of the sides of Δ ABC is 1 : 2 : 4. What is the ratio of the altitudes drawn onto these sides?

Solution:

A triangle is only possible when the sum of two sides is greater than the third side of the triangle.

► We have ratio of sides as 1 : 2 : 4
► So sides are 1x, 2x and 4x.
► But 1x + 2x < 4x

So, the triangle is not possible. Since triangle is not possible, altitudes are not possible.

QUESTION: 12

The three sides of a triangle measure 6 cm, 8 cm and 10 cm respectively. A rectangle equal in area to the triangle has a length of 8 cm. The perimeter of the rectangle is:

Solution:

Clearly, triangle is a right angled triangle Area of triangle:

► 1 / 2 x 6 x 8 = 24 cm2
► Now, 8 x b = 24
► b = 3

Required perimeter = 2(8 + 3) = 22 cm

QUESTION: 13

A rectangular enclosure 40 m x 36 m has a horse tethered to a corner with a rope of 14 m in length. What is the ratio of the respective areas it can graze, if it is outside the enclosure and if it is inside the enclosure?

Solution: ► Imagine a circle on the corner of the rectangle

► 3 quarters of the circle lie outside the rectangle and 1 quarter lies inside.

► Hence Required ratio = 3 : 1

QUESTION: 14

Read the passage below and solve the questions based on it.

The area of a square is equal to the area of a rectangle. Moreover, the perimeter of the square is also equal to the perimeter of the rectangle.

Q. The length of the rectangle is equal to the:

Solution:

► This is possible only when both length and breadth of rectangle are equal to the side of the square.

QUESTION: 15

A quadrilateral is inscribed in a circle. If an angle is inscribed in each of the segments outside the quadrilateral, then what is the sum of the four angles?

Solution: Required sum
= θ + 180° - θ + 180° - θ + 180° - θ + θ + 8 = 540°