Basic Problems MCQ Quiz - Objective Question with Answer for Basic Problems - Download Free PDF

Given:

∠C = 90° 

CD ⊥ AB

∠A = 65°

Concept used:

The sum of the angles of a triangle is 180°.

Calculation:

In ΔABC,

∠BAC + ∠CBA  + ∠ACB  = 180°

⇒ 65° + 90° + ∠CBA  = 180°

⇒ ∠CBA  = 25°

Important Points

We can arrive at the solution based on the perpendicular as well. But it is more time-consuming. Since ∠A and ∠C are given, it is wise to directly apply them in the concept and get the solution.

Given:

In triangle ABC,

∠A = 60^º

AB = 6 cm

AC = 4 cm

Formula Used:

Using the Cosine Rule:

BC² = AB² + AC² - 2 × AB × AC × cos(∠A)

Calculation:

Substitute the values:

AB = 6, AC = 4, cos(60^º) = 1/2

⇒ BC² = 6² + 4² - 2 × 6 × 4 × (1/2)

⇒ BC² = 36 + 16 - 24

⇒ BC² = 28

⇒ BC = √28

⇒ BC = √(4 × 7)

⇒ BC = 2√7

⇒ BC = (2/√7) cm

The correct length of BC is (2/√7) cm.

In triangle $ABC$, if the bisectors of $\mathrm{\angle }ABC$ and $\mathrm{\angle }ACB$ meet at point $O$ and $\mathrm{\angle }A={60}^{\circ }$, then the angle $\mathrm{\angle }BOC$ is given by:

$$\mathrm{\angle }BOC={90}^{\circ }+\frac{\mathrm{\angle }A}{2}$$

Substituting $\mathrm{\angle }A={60}^{\circ }$:

$$\mathrm{\angle }BOC={90}^{\circ }+\frac{{60}^{\circ }}{2}={120}^{\circ }$$

Thus, the measure of $\mathrm{\angle }BOC$ is:

$$\boxed{{\displaystyle {120}^{\circ }}}$$

Given:

In △ABC, DE || AC

BD = 8 cm, AD = 7 cm

Formula used:

By Basic Proportionality Theorem (Thales' theorem):

If DE || AC, then

Area of two similar triangles is proportional to the square of corresponding sides.

Ratio of areas = (BD / AB)²

Calculation:

AB = AD + BD = 7 + 8 = 15 cm

Ratio of areas of △BDE to △ABC = (8/15)² = 64/225

Area of trapezium ADEC = Area of △ABC - Area of △BDE

Required ratio:

Area(△BDE) : Area(ADEC)

= 64 : (225 - 64)

= 64 : 161

∴ The ratio of the area of △BDE to the trapezium ADEC is 64 : 161.

Given:

In ΔABC, ∠A = 90º

BC = 10 cm

AC = 8 cm

BP = 9 cm

Formula Used:

Using Pythagoras theorem in ΔABC:

$A{B}^2+A{C}^2=B{C}^2$

Using Pythagoras theorem in ΔABP:

$A{B}^2+A{P}^2=B{P}^2$

Calculation:

Using Pythagoras theorem in ΔABC:

AB² + 8² = 10²

⇒ AB² + 64 = 100

⇒ AB² = 36

⇒ AB = 6 cm

Using Pythagoras theorem in ΔABP:

6² + AP² = 9²

⇒ 36 + AP² = 81

⇒ AP² = 45

⇒ AP = √45

⇒ AP = 3√5 cm

The correct answer is option 4.

Given:

In ΔABC, AB = 8cm

∠A bisected internally to intersect BC at D.

AB = 8 cm, BD = 6 cm and DC = 7.5 cm

Concept used:

In a triangle, the angle bisector of an angle is divides the opposite side to the angle in the ratio of the remaining two sides.

$\frac{AB}{AC}=\frac{BD}{DC}$

Calculation:

AB = 8 cm, ∠A is bisected internally to intersect BC at D,

⇒ AB/AC = BD/CD

⇒ 8/AC = 6/7.5

Given that,

The two sides of a triangle are of length 3 cm and 8 cm and the length of its third side is x cm.

As we know,

Sum of two sides of triangle is always greater than third side of the triangle.

∴ Sum of two sides of triangle > third side of the triangle.

⇒ 3 + 8 > Third side

⇒ 11 > x

Also,

Another case will be,

⇒ x + 3 > 8

⇒ x > 8 - 3

⇒ x > 5

∴ 5 < x < 11

Given:

Three sides of the triangle are (k – 4), k and (k + 4).

Formula:

As we know,

Pythagoras Theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

Calculation:

According to the question

(k + 4)² = (k – 4)² + k²

⇒ k² + 16 + 8k = k² + 16 – 8k + k²

⇒ k² = 16k

⇒ k = 16

Concept:

Similar triangles:

Similar triangle are triangles that have the same shape, but their sizes may vary. 

Properties:

• Both have the same shape but sizes may be different
• Each pair of corresponding angles are equal
• The ratio of corresponding sides is the same

Calculation:

In ΔADE and ΔABC

∠A is common

∠D = ∠B and ∠E = ∠C

∴ ΔADE ∼ ΔABC

According to similar triangle property

$\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$

$\begin{array}{l}\frac{4}{4+3}=\frac{DE}{BC}\\ \frac{DE}{BC}=\frac{4}{7}\end{array}$

Similarly in ΔDOE & ΔBOC

∠DEO = ∠OBC (Corresponding angles are equal)

∠DOE = ∠BOC (vertically opposite angles)

∴ ΔDEO ∼ ΔOBC

$\begin{array}{l}\frac{DE}{BC}=\frac{DO}{OC}\\ \frac{DO}{OC}=\frac{4}{7}\end{array}$

Hence, $\frac{DO}{DC}=\frac{4}{4+7}=\frac{4}{11}$

Given:

Three sides of the triangle are 8 cm, 6 cm and 5 cm respectively.

Concept:

P² + B² = H² (For right angles triangle)

P² + B² > H² (For acute triangle)

P² + B² < H² (For obtuse triangle)

Calculation:

According to the question

8² > 6² + 5²

⇒ 64 > 36 + 25

⇒ 64 > 61

Hence, triangle is obtuse angled triangle.

P2 + B2 = H2 (For right angles triangle)

P2 + B2 > H2 (For acute triangle)

P2 + B2 < H2 (For obtuse triangle)

where P, B, and H are the three sides of any triangle, H being the largest side.

So, while applying the given values in the conditions for each type of triangle we need to careful not to put the wrong value in the wrong place.

⇒ Base of an isosceles triangle is BC = 10 cm

⇒ Altitude of isosceles triangle is AD = 10 × [6/5] = 12 cm

As we know, D is a point of BC

∴ BD = DC = [1/2] × BC = 5 cm

In right angled triangle BDA

⇒ (BA)² = (BD)² + (AD)²

⇒ (BA)² = 5² + 12²

⇒ BA = 13 cm = AC

⇒ Area of the ΔABC = [1/2] × BC × AD = [1/2] × 10 × 12 = 60 cm²

⇒ Semi perimeter of ΔABC = (13 + 13 + 10)/2 = 36/2 = 18 cm

As we know,

⇒ In-radius of ΔABC = (Area of Δ)/S = 60/18 = 10/3 cm

Given:

Two sides of a triangle = 12.8 m & 9.6 m 

Formula used:

Area of a triangle = 1/2 × Base × Height

Calculation:

Let the height of the corresponding side 12.8 m = h 

1/2 × 9.6 × 12 = 1/2 × 12.8 × h

⇒ (9.6 × 12)/12.8 = h 

⇒ h = 9 m 

∴ The height of the triangle corresponding to side with length 12.8 m long = 9 m 

Given:

ratio of sides of a triangle = 5 : 4 : 3

perimeter of the triangle = 84 cm

Formula used:

Perimeter of triangle = Sum of sides

Calculations:

Let the sides of triangle be 5x, 4x and 3x so that they are in ratio 5 : 4 : 3.

∴ 5x + 4x + 3x = 84

⇒ 12x = 84

⇒ x = 7 cm

So, the sides of triangles are 35, 28 and 21 meters.

∴ The length of the largest side is 35 m.

Given:

Length of side a = 13 inches

Length of side b = 15 inches

Length of side c = 14 inches

Formula Used:

Heron's formula: Area of a triangle = $\surd (s(s-a)(s-b)(s-c))$, where s is the semi-perimeter of the triangle.

Solution:

We can use Heron's formula to find the area of the triangle.

First, we need to calculate the semi-perimeter of the triangle:

s = (a + b + c)/2

⇒ (13 + 15 + 14)/2

⇒ 21

Next, we can use Heron's formula to find the area of the triangle:

Area = $\surd (s(s-a)(s-b)(s-c))$

⇒ $\surd (21(21-13)(21-15)(21-14))$

⇒  √(21(8)(6)(7))

⇒ √(2⁴ x 3² x 7²)

⇒ 84

Therefore, the area of the triangle is 84 square inches.

Calculation:

Consider the following figure :

MN || EF

⇒ ∆DMN is similar to ∆DEF

⇒ DM/DE= DN/DF = MN/EF

Given,

MN/EF = 2 ∶ 5 and DE = 60

⇒ 2/5 = DM/60

⇒ DM = 2 × 12 = 24 cm

∴ ME = DE - DM = 60 - 24 = 36 cm

∴ Option 4 is the correct answer.