# CRE 1 - Basics | Geometry - Quadrilaterals & Polygons

**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

Sum of all the interior angles of a polygon is 3060°. Find the number of sides of the polygon?

Answer: 19

**Explanation** :

We know sum of all interior angles of a polygon = (n - 2) × 180°

∴ (n - 2) × 180° = 3060°

⇒ n = 19

Hence, 19.

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

The perimeter of a square, a regular octagon and a hexagon are equal. If their areas are denoted by S, O and H respectively, then which of the following is true.

- A.
S > O > H

- B.
O > S > H

- C.
H > S > O

- D.
O > H > S

Answer: Option D

**Explanation** :

When perimeter is same, then the polygon with higher number of sides has greater area.

Hence, option (d).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

How many regular polygons with number of sides 'n' exist such that all the angles (in degrees) of the polygon are integers?

Answer: 22

**Explanation** :

We know sum of all interior angles of a polygon = (n - 2) × 180°

∴ Each interior angle of a regular polygon = $\frac{\left(\mathrm{n}-2\right)\times 180\xb0}{n}=180\xb0-\frac{360\xb0}{\mathrm{n}}$.

Now for interior angles to be integer, $\frac{360\xb0}{\mathrm{n}}$ should be an integer i.e., n should be a factor of 360 and greater than 2.

Total factors of 360 = 24

Regular Polygons satisfying the conditions specified above are 22 (n > 2).

Hence, 22.

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

What is the fourth angle of a cyclic quadrilateral in which three angles are 50°, 130° and 120°?

- A.
120°

- B.
60°

- C.
30°

- D.
20°

Answer: Option B

**Explanation** :

For a cyclic quadrilateral, the sum of the interior angles is always 360°.

Assume the fourth angle to be ‘x’.

∴ x + 50° + 130° + 120° = 360°

∴ x + 300° = 360°

∴ x = 60°

Hence, option (b).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

The diagonals of which of the following quadrilaterals do not intersect each other at right angles: Kite, Rhombus, Rectangle and Square?

- A.
Square

- B.
Rectangle

- C.
Rhombus

- D.
Kite

Answer: Option B

**Explanation** :

As shown in the figure above, Squares, Rhombuses and Kites are quadrilaterals in which the diagonals intersect each other at right angles.

But for a Rectangle, the diagonals don’t intersect each other at right angles. (except when the rectangle is a square.)

Hence, option (b).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

What is the area of a square (in cm^{2}) in which the sides are double that of a square with area 30 cm^{2}?

- A.
60

- B.
120

- C.
90

- D.
None of these

Answer: Option B

**Explanation** :

∵ We know area of a square ∝ (side)^{2}.

Hence, if side is double, the area will become quadruple.

⇒ The new area will be 4 × 30 = 120 cm^{2}.

Hence, option (b).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

What is the perimeter (in cm) and area (in cm^{2}) of the parallelogram shown in the figure? (All the values are given in cm)

- A.
18, 20

- B.
20, 32

- C.
26, 32

- D.
26, 16

Answer: Option C

**Explanation** :

∵ A parallelogram has its opposite sides equal in length, we have,

∴ AB = CD = 8 cm and AD = BC = 5 cm

Now, the perimeter of a parallelogram, P = AB + BC + CD + DA= 2 × (AB + BC)= 2 × (8 + 5) = 26 cm.

and Area of a parallelogram, A = Base × Height = CD × Height = 8 × 4 = 32 cm^{2}

Hence, option (c).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

What is the length of the diagonal of a rectangle with an area 48 cm^{2} and one of its sides as 6 cm?

- A.
10 cm

- B.
9 cm

- C.
12 cm

- D.
8 cm

Answer: Option A

**Explanation** :

Opposite sides of a rectangle are equal in length.

∴ AD = BC = 6 cm and AB = CD

Area of a rectangle, A = Length × Breadth

∴ 48 = 6 × CD

∴ CD = 8 cm

∴ AB = CD = 8 cm

A rectangle has all interior angles as right angles i.e. 90°.

In ∆BCD, m∠BAD = 90°

Using Pythagoras theorem, we have,

AB^{2} + AD^{2} = BD^{2}

∴ BD^{2} = 6^{2} + 8^{2}

∴ BD = 10 cm

Hence, option (a).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

Two diagonals of a quadrilateral meet at an angle of 30°. If the area of the quadrilateral is 50 cm^{2} and length of one of the diagonals is 5 cm, what is the length of its other diagonal?

- A.
20 cm

- B.
40 cm

- C.
60 cm

- D.
30 cm

- E.
None of these

Answer: Option B

**Explanation** :

Area of a quadrilateral when two diagonals and an included angle is known

= 1/2 × d_{1} × d_{2} × Sin θ

∴ 50 = 1/2 × 5 × d_{2} × Sin 30°

⇒ 50 = 1/2 × 5 × d_{2} × 1/2

⇒ d_{2} = 40 cm

Hence, option (b).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

What is the area (in cm^{2}) of a cyclic quadrilateral with sides 62 cm, 51 cm, 38 cm, and 23 cm?

Answer: 1680

**Explanation** :

The given quadrilateral is a cyclic quadrilateral.

Area of a cyclic quadrilateral = $\surd \left(\right(s-a\left)\right(s-b\left)\right(s-c\left)\right(s-d))$

where s is the semi perimeter of the cyclic quadrilateral and is given as s = (a + b + c + d)/2

s = (62+51+38+23)/2 = 87 cm

Area of □ABCD = $\surd \left(\right(87-62\left)\right(87-51\left)\right(87-38\left)\right(87-23))$

= $\surd (25\times 36\times 49\times 64)$ = 5 × 6 × 7 × 8 = 1680 cm2.

Hence, option (a).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

Which of the following has the maximum area for the same perimeter?

- A.
Rectangle

- B.
Parallelogram

- C.
Square

- D.
Rhombus

Answer: Option C

**Explanation** :

A square is the most symmetrical quadrilateral and therefore has the maximum area for a given perimeter of a quadrilateral.

Hence, option (d).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

Which of the following has the maximum perimeter for the same area?

- A.
Circle

- B.
Hexagon

- C.
Triangle

- D.
Rhombus

Answer: Option C

**Explanation** :

For polygons with same area, the polygon with higher number of sides has lesser perimeter.

∴ Triangle has the least number of sides from the options given, ∴ For same area, a triangle will have the maximum perimeter.

Hence, option (d).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

Which of the following has the maximum area for the same perimeter?

- A.
Quadrilateral

- B.
Pentagon

- C.
Hexagon

- D.
Circle

Answer: Option D

**Explanation** :

For same perimeter, polygon with higher number of sides will have greater area.

Hence, option (d).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

Which of the following has the maximum perimeter for the same area?

- A.
Quadrilateral

- B.
Pentagon

- C.
Hexagon

- D.
Circle

Answer: Option A

**Explanation** :

For same area, polygon with higher lesser number of sides will have greater perimeter.

Hence, option (a).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

Area of the given figure is 60 cm^{2}. Length of the diagonal BD is equal to

- A.
16 cm

- B.
12 cm

- C.
8 cm

- D.
10 cm

- E.
None of these

Answer: Option B

**Explanation** :

If the length of a diagonal and that of the two perpendiculars drawn from that diagonal to the other two vertices are known, the area of the quadrilateral is given as:

Area of a quadrilateral = 1/2 × (h_{1} + h_{2}) × d where d is the length of the diagonal and h_{1} and h_{2} are the lengths of the two perpendiculars.

∴ 60 = 1/2 × (4 + 6) × BD

∴ BD = 12 cm.

Hence, option (b).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

Find the area (in cm^{2}) of an octagon (which is not a regular polygon) whose perimeter is 40 cm and which circumscribes a circle of radius 6 cm.

Answer: 120

**Explanation** :

Area of the polygon = inradius × semi-perimeter = 6 × 40/2 = 120 cm^{2}.

Hence, 120.

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

A square of side 's' units is cut down into a regular octagon. What is the side of the regular octagon?

- A.
$\frac{s}{\sqrt{2}}$

- B.
$\frac{s}{\sqrt{2}+1}$

- C.
$\frac{s}{\sqrt{2}+2}$

- D.
None of these

Answer: Option B

**Explanation** :

We will have to cut a right angled triangle from the 4 corners of the square to form an octagon.

Let 'a' be the side of the right angled triangle, such that the hypotenuse of the right angled triangle be the side of the octagon.

a + √2a + a = s

⇒ $a=\frac{s}{2+\sqrt{2}}$

⇒ $\sqrt{2}a=\frac{s}{\sqrt{2}+1}$

Hence, option (c).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

What is the area of a regular octagon of side 16 cm?

- A.
256(1 + √2)

- B.
128(4 + 2√2)

- C.
128(3 + 2√2)

- D.
64(10 + √2)

Answer: Option D

**Explanation** :

Area of the octagon = Area of the bigger square – 4 × (Area of the corner right triangle)

The side of the bigger square will be 16 + 16/√2 + 16/√2 = 16 × (1 + √2).

∴ Area of the octagon = (16(1 + √2))^{2 }- 4 × 1/2 × (8√2)^{2}

= 256(3 + 2√2) – 256

= 256(2 + 2√2)

Hence, option (d).

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

What is the area (in cm^{2}) of an isosceles trapezium in which the two parallel sides are 12 cm and 6 cm and the two non-parallel sides are 5 cm each?

Answer: 36

**Explanation** :

In an isosceles trapezium ABCD. AD = BC = 5 and DE = FC = 3

In right triangle BCF, BC = 5, FC = 3 hence BF = 4 (pythagorean triplet)

∴ Area of the trapezium = ½ × (6 + 12) × 4 = 36 cm^{2}.

Hence, 36.

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**CRE 1 - Basics | Geometry - Quadrilaterals & Polygons**

The area of a parallelogram (in cm^{2}) with base and height equal to 12 cm and 6 cm respectively is?

Answer: 72

**Explanation** :

Area of a parallelogram = Base × Height

∴ Area = 12 × 6 = 72 cm^{2}.

Hence, 72.

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