# CRE 1 - Basics | Geometry - Triangles

**CRE 1 - Basics | Geometry - Triangles**

In a right angled triangle, one of the acute angles is four times the other. What is the measure of the smallest angle of the triangle?

- A.
15°

- B.
18°

- C.
20°

- D.
22°

Answer: Option B

**Explanation** :

In a right triangle, one angle is 90° and the sum of the other two angles is 90°.

These acute angles are divided in the ratio of 1 : 5.

∴ We get x + 4x = 90°

∴ x = 18°

Hence, we get the two acute angles as 18° and 72°.

∴ The smallest angle of the triangle is 18°.

Hence, option (b).

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**CRE 1 - Basics | Geometry - Triangles**

What is the length of the altitude of an equilateral triangle of side 4 cm?

- A.
4 cm

- B.
6 cm

- C.
2√3 cm

- D.
2 cm

Answer: Option C

**Explanation** :

Altitude of an equilateral triangle is given by √3/2 × a.

where a is the side of an equilateral triangle.

∵ a = 4 cm

∴ Altitude = √3/2 × 4 = 2√3 cm.

Hence, option (c).

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**CRE 1 - Basics | Geometry - Triangles**

If a triangle has two sides having lengths 8 units and 12 units, which of the following can be the length of the third side?

- A.
2 units

- B.
4 units

- C.
6 units

- D.
All of these

Answer: Option C

**Explanation** :

In a triangle the sum of any two sides should be more than the third side and the difference of any two sides should be less than the third side.

Only option (c) satisfies both the conditions.

Hence, the length of the third side can be 6 units.

Hence, option (c).

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**CRE 1 - Basics | Geometry - Triangles**

What is the area (in cm^{2}) of a triangle with sides 6 cm, 8 cm and 10 cm?

- A.
34

- B.
24

- C.
12

- D.
None of these

Answer: Option B

**Explanation** :

s = (a + b + c)/2

where s is the semi-perimeter and a, b and c are the sides of the triangle.

s = (6 + 8 + 10)/2 = 12 cm

Area = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$

= $\sqrt{12\left(12-6\right)\left(12-8\right)\left(12-10\right)}=\sqrt{12\times 6\times 4\times 2}$

= 24

**Alternately,**

6, 8 and 10 are pythagorean triplets i.e. 10^{2} = 8^{2} + 6^{2}

⇒ The given triangle is a right triangle

⇒ Area = ½ × 6 × 8 = 24

Hence, option (b).

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**CRE 1 - Basics | Geometry - Triangles**

A triangle has three internal angles in the ratio of 3 : 4 : 5. What is the value of the greatest external angle?

- A.
57°

- B.
90°

- C.
135°

- D.
Can't be determined

Answer: Option C

**Explanation** :

Sum of the interior angles of a triangle is 180°.

These angles are in the ratio 3 : 4 : 5.

∴ 3x + 4x + 5x = 180°

∴ x = 15°

∴ We get the three angles as 45°, 60° and 75° respectively.

The greatest external angle should be supplementary to the smallest internal angle which measures 45°.

∴ Measure of the greatest external angle = 180 − 45° = 135°

Hence, option (c).

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**CRE 1 - Basics | Geometry - Triangles**

In ∆ABC, D is a point on side BC such that BC = 5 × BD and A(∆ABC) is 100 sq. units. What is the area of ∆ABD in sq. units?

- A.
20

- B.
25

- C.
100

- D.
12.5

Answer: Option A

**Explanation** :

BC is the base of ∆ABC and BD is base of ∆ABD.

If we drop a perpendicular from A on the line BC, this line is the height for both, ∆ABC and ∆ABD

∴ The ratio of the areas of the two triangles is equal to the ratio of their bases.

∴ A(∆ABD) = 1/5 × A(∆ABC)

∴ A(∆ABD) = 1/5 × 100 = 20 sq. units

Hence, option (a).

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**CRE 1 - Basics | Geometry - Triangles**

How many triangles with integral valued sides can be formed having perimeter 14 units?

- A.
6

- B.
5

- C.
4

- D.
3

Answer: Option C

**Explanation** :

In a triangle with sides a, b and c, sum of any two sides is greater than the third side, i.e. a + b > c.

∴ For any three sides given, a triangle can be constructed only if sum of any two sides is greater than the third side.

It is given that, a + b + c = 14

Also, it is given that a, b, c are integers.

**Case 1**: The largest side is 6.

The possible combination of sides is (6, 4, 4), (6, 5, 3) and (6, 6, 2)

**Case 2**: The largest side is 5.

The possible combination of sides is (5, 5, 4)

∴ The possible sets are (4, 4, 6), (5, 5, 4), (6, 5, 3) and (6, 6, 2).

∴ Only 4 such triangles are possible.

Hence, option (c).

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**CRE 1 - Basics | Geometry - Triangles**

In a triangle one of the angles is twice and the other is thrice the third angle. What is the ratio of the side opposite to the smallest angle to the side opposite to the largest angle?

- A.
1/2

- B.
2/1

- C.
√3/1

- D.
1/√3

Answer: Option A

**Explanation** :

The three angles of the triangle are in the ratio 1 : 2 : 3.

Hence, the given triangle can only be a 30°-60°-90° triangle.

Here, the side opposite the largest angle is the hypotenuse and the side opposite the smallest angle is the side opposite the angle with measure 30°.

Hence, by the 30°-60°-90° theorem

Side opposite to 30° angle is ½ of hypotenuse

∴ Ratio of the length of the largest side to the length of the smallest side is the ratio of the hypotenuse to the side opposite the angle with measure 30°

∴ (Side opposite to 30°)/Hypotenuse = 1/2

Hence, option (a).

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**CRE 1 - Basics | Geometry - Triangles**

The largest angle (in degree) of a triangle whose sides are in the ratio 1 : 1 : √2 is

Answer: 90

**Explanation** :

Since two sides of the triangle are in the ratio 1 : 1, these two sides of the triangle should be equal.

∴ The given triangle is isosceles in nature.

One of the sides is √2 times the other two sides

∴ By the 45°-45°-90° theorem, the given triangle is an Isosceles Right angled triangle.

∴ The largest angle is 90°

Hence, 90.

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**CRE 1 - Basics | Geometry - Triangles**

In ∆ABC, AC = BC, what is the area in square cm enclosed by ∆ABC if external angle C is 150°?

- A.
50

- B.
10

- C.
15

- D.
20

- E.
25

Answer: Option E

**Explanation** :

Given external angle C = 150°

⇒ internal angle C = 180° - 150° = 30°

∴ m ∠ACB = 30°

AC = BC = 10 cm

Two sides and the angle included by them are known.

∴ Area of the triangle = 1/2 × (AC) × (BC) × sin θ

∴ Area of ∆ABC = 1/2 × (10) × (10) × sin 30° = 25 cm^{2}. [sin 30° = 1/2]

Hence, option (e).

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**CRE 1 - Basics | Geometry - Triangles**

In a triangle ABC, AB = 13, BC = 14 and AC = 15 cm. AD is the perpendicular drawn from A to BC. Find the length of AD.

- A.
9

- B.
8

- C.
10

- D.
12

- E.
None of these

Answer: Option D

**Explanation** :

We can draw the following diagram from the information given below.

In △ABD: AB^{2} = BD^{2} + AD^{2}

⇒ AD^{2} = 169 - x^{2} ...(1)

In △ACD: AC^{2} = CD^{2} + AD^{2}

⇒ AD^{2} = 225 - (14 - x)^{2} ...(2)

(1) = (2)

⇒ 169 - x^{2 }= 225 - (14 - x)^{2}

⇒ 28x = 140

⇒ x = 5

⇒ AD^{2} = 169 - 25 = 144 [from (1)]

⇒ AD = 12

Hence, option (d).

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**CRE 1 - Basics | Geometry - Triangles**

The interior and its adjacent exterior angle of a triangle are in the ratio 2 : 3, What is the sum of the other two angles of the triangle?

- A.
108°

- B.
72°

- C.
100°

- D.
None of these

Answer: Option A

**Explanation** :

Let the interior angle be 2x and exterior angle be 3x.

The interior and the corresponding exterior angle of a triangle are supplementary.

⇒ 2x + 3x = 180°

⇒ x = 36°

Now, exterior angle is same as the sum of two opposite interior angles.

∴ Sum of the other 2 interior angles of the triangle = 3 × 36° = 108°

Hence, option (a).

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**CRE 1 - Basics | Geometry - Triangles**

In the figure given below, ∠ABC = 30°, ∠ADC = 10° and ∠BCD = 20°. Find external ∠BAD.

- A.
70°

- B.
40°

- C.
50°

- D.
60°

Answer: Option D

**Explanation** :

Let us draw a line joining points C and A and extend it till E.

In ∆ABC, ∠EAB is the exterior angle, hence

∠EAB = ∠ABC + ∠BCA ...(1)

In ∆ADC, ∠EAD is the exterior angle, hence

∠EAD = ∠ADC + ∠DCA ...(2)

(1) + (2)

∠EAB + ∠EAD = ∠ABC + ∠BCA + ∠ADC + ∠DCA

⇒ ∠BAD = ∠ABC + ∠ADC + ∠BCD

⇒ ∠BAD = 30° + 10° + 20° = 60°

Hence, option (d).

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**CRE 1 - Basics | Geometry - Triangles**

In the figure given below, AD : DE = 3 : 2 and BE : EC = 3 : 4. Find the ratio of area of triangle ADB to area of triangle CDE?

- A.
1 : 1

- B.
9 : 8

- C.
1 : 2

- D.
2 : 1

- E.
Cannot be determined

Answer: Option B

**Explanation** :

∆BDE and ∆CDE lie between same 2 parallel lines, hence the ratio of their areas is same as the ratio of their bases.

⇒ $\frac{\mathrm{area}(\u2206\mathrm{BDE})}{\mathrm{area}(\u2206\mathrm{CDE})}$ = $\frac{3}{4}$ ...(1)

Similarly, $\frac{\mathrm{area}(\u2206\mathrm{ADB})}{\mathrm{area}(\u2206\mathrm{BDE})}$ = $\frac{3}{2}$ ...(2)

(1) × (2)

⇒ $\frac{\mathrm{area}(\u2206\mathrm{ADB})}{\mathrm{area}(\u2206\mathrm{CDE})}$ = $\frac{9}{8}$

Hence, option (b).

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**CRE 1 - Basics | Geometry - Triangles**

In the figure below, AD is the angle bisector of angle A. Which of the following is true?

- A.
DC > AD < BD

- B.
DC < AD < BD

- C.
DC > AD > BD

- D.
AD > BD > CD

- E.
Cannot be determined

Answer: Option C

**Explanation** :

AD is the angle bisector of ∠A = 100°.

∴ ∠BAD = ∠CAD = 50°

We know side opposite to larger angle is larger than side opposite to smaller angle.

In triangle ABD,

∠ABD > ∠BAD

⇒ AD > BD ...(1)

In triangle ADC,

∠DAC > ∠ACD

⇒ DC > AD ...(2)

From (1) and (2)

⇒ DC > AD > BD

Hence, option (c).

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**CRE 1 - Basics | Geometry - Triangles**

The sum of three altitudes of a triangle is

- A.
equal to the sum of three sides

- B.
less than the sum of three sides

- C.
greater than the sum of three sides

- D.
twice the sum of sides

Answer: Option B

**Explanation** :

In a triangle the sum of three altitudes of a triangle is less than the sum of sides.

Hence, option (b).

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**CRE 1 - Basics | Geometry - Triangles**

In a triangle measure of two sides is 6 cm and 4 cm and the common angle of these two sides is 60°. Find the third side (in cm)?

- A.
2√5

- B.
3

- C.
2√3

- D.
2√7

Answer: Option D

**Explanation** :

We can apply the cosine rule to find the third side.

AC^{2} = AB^{2} + BC^{2} - 2 × AB × BC × CosB

Here, AB = 6, BC = 4 and ∠B = 60°

⇒ AC^{2} = 6^{2} + 4^{2} - 2 × 6 × 4 × Cos60°

⇒ AC^{2} = 36 + 16 - 2 × 6 × 4 × 1/2 [Cos60° = 1/2]

⇒ AC^{2} = 36 + 16 - 6 × 4

⇒ AC^{2} = 28

⇒ AC = 2√7

Hence, option (c).

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**CRE 1 - Basics | Geometry - Triangles**

In a triangle ABC, AB = 12, BC = 5 and AC = 13 cm. Find the length of altitude drawn on side AC (in cm).

- A.
5

- B.
$\sqrt{\frac{60}{13}}$

- C.
2√3

- D.
√5

Answer: Option B

**Explanation** :

5, 12 and 13 form a pythagorean tiplet, hence the given triangle is a right triangle with AC as the hypotenuse

In a right triangle, the length of altitude drawn on hypotenuse is $\sqrt{\frac{\mathrm{product}\mathrm{of}\mathrm{smaller}\mathrm{sides}}{\mathrm{hypotenuse}}}$.

∴ altitude = $\sqrt{\frac{5\times 12}{13}}$ = $\sqrt{\frac{60}{13}}$

Hence, option (b).

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**CRE 1 - Basics | Geometry - Triangles**

What is the area (in cm^{2}) of an isosceles triangle ABC with sides AB = AC = 10 cm, and BC = 12 cm?

- A.
36

- B.
48

- C.
24

- D.
72

Answer: Option B

**Explanation** :

Since AB = AC, it is an isosceles triangle. The perpendicular drawn from A to BC will bisect side BC.

In right triangle ADB,

AB = 10 and BD = 6, hence AD = 8 [6, 8 and 10 is a pythagorean triplet]

∴ Area of triangle = ½ × BC × AD = ½ × 12 × 8 = 48.

Hence, option (b).

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**CRE 1 - Basics | Geometry - Triangles**

Two triangles A and B have same perimeter. A is an equilateral triangle whereas B is an isosceles triangle. Which of the following is true regarding areas of A and B.

- A.
A > B

- B.
A = B

- C.
A < B

- D.
Cannot be determined

Answer: Option A

**Explanation** :

For same perimeter area of an equilateral triangle is maximum.

Hence, option (a).

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**CRE 1 - Basics | Geometry - Triangles**

Two triangles A and B have the same area. A is an equilateral triangle whereas B is an isosceles triangle. Which of the following is true about their perimeters?

- A.
A > B

- B.
A = B

- C.
A < B

- D.
Cannot be determined.

Answer: Option C

**Explanation** :

For same area, perimeter of an equilateral triangle is minimum.

Hence, option (c).

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