Properties of Triangles MCQ Quiz - Objective Question with Answer for Properties of Triangles - Download Free PDF

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x \).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12\,x\,BC = \tfrac12\,x\,\sqrt{9 - 6x}\)

To maximize, differentiate w.r.t. \(x\) and set to zero:

\(\displaystyle \frac{d}{dx}\bigl(x\sqrt{9-6x}\bigr) = \sqrt{9-6x} \;-\;\frac{6x}{2\sqrt{9-6x}} = 0 \;\Longrightarrow\; x = 1 \)

At \(x = 1\), we get \(BC = \sqrt{9 - 6} = \sqrt{3}\), so

\(A_{\max} = \tfrac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}\)

∴ The maximum area is \(\frac{\sqrt{3}}{2}\) square unit.

Hence, the correct answer is Option  1.

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x\).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12\,x\,\sqrt{9 - 6x} \)

To maximize, set up

\(A^2 = \tfrac14\,x^2\,(9 - 6x)\)

\(\displaystyle \frac{d(A^2)}{dx} = \tfrac14\bigl(2x(9 - 6x) + x^2(-6)\bigr) = \frac{18x(1 - x)}{4} = 0 \)

Hence \(x = 1\) (discarding x=0, so

\(BC = \sqrt{9 - 6}= \sqrt{3},\quad AC = 3 - 1 = 2\)

Therefore,

\(\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} \implies A = \frac{\pi}{3}\)

∴ \(\angle A = \frac{\pi}{3}\).

Hence, the correct answer is Option 3.

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x \).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12\,x\,BC = \tfrac12\,x\,\sqrt{9 - 6x}\)

To maximize, differentiate w.r.t. \(x\) and set to zero:

\(\displaystyle \frac{d}{dx}\bigl(x\sqrt{9-6x}\bigr) = \sqrt{9-6x} \;-\;\frac{6x}{2\sqrt{9-6x}} = 0 \;\Longrightarrow\; x = 1 \)

At \(x = 1\), we get \(BC = \sqrt{9 - 6} = \sqrt{3}\), so

\(A_{\max} = \tfrac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}\)

∴ The maximum area is \(\frac{\sqrt{3}}{2}\) square unit.

Hence, the correct answer is Option  1.

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x\).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12\,x\,\sqrt{9 - 6x} \)

To maximize, set up

\(A^2 = \tfrac14\,x^2\,(9 - 6x)\)

\(\displaystyle \frac{d(A^2)}{dx} = \tfrac14\bigl(2x(9 - 6x) + x^2(-6)\bigr) = \frac{18x(1 - x)}{4} = 0 \)

Hence \(x = 1\) (discarding x=0, so

\(BC = \sqrt{9 - 6}= \sqrt{3},\quad AC = 3 - 1 = 2\)

Therefore,

\(\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} \implies A = \frac{\pi}{3}\)

∴ \(\angle A = \frac{\pi}{3}\).

Hence, the correct answer is Option 3.

Explanation:

We are given the triangle with angles:

\( \angle B = x = 30^\circ\)

\( \angle A = 3x = 90^\circ \)

\(\angle C = 180^\circ - 120^\circ = 60^\circ \)

Step 1: Check if the sum of the angles is 180°:

\( \angle A + \angle B + \angle C = 90^\circ + 30^\circ + 60^\circ = 180^\circ \)

This confirms that the angles satisfy the angle sum property of a triangle.

Statement I. The triangle is right-angled.

Since\( \angle A = 90^\circ \), the triangle is right-angled.

Statement III: III. The angles A, C and B of the triangle are in AP.

The angles \(30^\circ 60^\circ , \text and 90^\circ \) are in Arithmetic Progression because:

\( 60^\circ - 30^\circ = 30^\circ \quad \text{and} \quad 90^\circ - 60^\circ = 30^\circ \)

This confirms that the angles are in AP.

Statement II is not correct because there is no mention of a side being 3 times the other.

∴ The correct answer is Option (I) and (III) are correct.

Hence, the correct answer is Option 3.

Concept:

If three sides of Δ ABC are a, b and c units. Then

\(\cos A = \frac{{{b^2} + {c^2} - {a^2}\;}}{{2bc}}\)

\(\cos B = \frac{{{c^2} + {a^2} - {b^2}\;}}{{2ac}}\)

\(\cos C = \frac{{{a^2} + {b^2} - {c^2}\;}}{{2ab}}\)

Calculation:

Here, the sides of the Δ ABC are a = 12 units, b = 14 units and c = 16 units.

We know that \(\cos B = \frac{{{c^2} + {a^2} - {b^2}\;}}{{2ac}}\), by substituting the values of a, b and c in the formula, we get

⇒ \(\cos B = \frac{{{a^2} + {c^2} - {b^2}\;}}{{2ac}} = \frac{{{{12}^2} + {{16}^2} - {{14}^2}}}{{2 \times 12 \times 16}} \)

\(=\frac{204}{384}= \frac{{17}}{{32}}\)

Concept:

• Sine Rule:

In a triangle Δ ABC, where a is the side opposite to A, b is the side opposite to B, c is the side opposite to C and where R is the circum-radius:

\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)

Calculation:

Given: Angles of a triangle are in the ratio 1: 2: 3 and the circum-radius is 10 cm.

Let A: B: C = 1: 2: 3 or A = k, B = 2k and C = 3k where k is any real number and R = 10 cm.

 As we know that, A + B + C = 180°

⇒ A + B + C = k + 2k + 3k = 180°

⇒ k = 30°

⇒ A = 30°, B = 60° and C = 90°

As we know that, \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)

\( \Rightarrow \frac{a}{{\sin 30^\circ }} = \frac{b}{{\sin 60^\circ }} = \frac{c}{{\sin 90^\circ }} = 20\)

⇒ a = 20 × sin 30° = 10 cm, b = 20 × sin 60° = 10√3 cm and c = 20 × sin 90° = 20 cm.

Concept:

For a Δ ABC, with sides a, b and c. 

\(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\), \(\cos B = \frac{{{c^2} + {a^2} - {b^2}}}{{2ac}}\) and \(\cos C = \frac{{{b^2} + {a^2} - {c^2}}}{{2ab}}\)

Calculation:

Here, the sides of the Δ ABC are a = 3 units, b = 5 units and c = 3 units and we know that, \(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\).

⇒ \(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{{5^2} + {3^2} - {3^2}}}{{2 \times 5 \times 3}}\)

⇒ cos A = 5/6

Given:

D and E are points on the sides AB and AC

AD = CE

BE and CD intersect at F

Concept used:

Concept of congruency of the triangle,

The exterior angle is always equal to the sum of the interior opposite angle. 

Calculation:

ΔCBE ≅ ΔACD [SAS congruency]

So, the three angles of these two triangles are the same,

Let ∠EBC be θ from this ∠ACD is also θ

Now,

∠BEC = 180° - (60° + θ)

⇒ 120° - θ

Now, In ΔECF

Exterior angle ∠CFB = (120° - θ) + θ

⇒ 120°

∴ ∠CFB is 120°

Concept:

Angle sum property: The sum of angles of a triangle is 180°

Calculation:

Here, we have to find the measure of the two equal angles of a right isosceles triangle.

Let Δ ABC be a right isosceles triangle with ∠ B = 90° and AB = BC.

As we know, the angles opposite to equal sides are also equal.

⇒ ∠ACB = ∠BAC = x

Now by angle sum property, we have

⇒ x + x + 90° = 180°

⇒ x = 45°

Hence, the measure of the two equal angles of a right isosceles triangle is 45°.

Key Points

In a right isosceles triangle, one angle will be 90° and the other two sides will be equal.

The angles opposite to equal sides will also be equal.

Given:

 sin (C + D) = √3/2

sec (C - D) = 2/√3

Calculations:

If  sin (C + D) = √3/2 and sec (C - D) = 2/√3

Then,

⇒ C + D = 60°.............(1)

⇒ C - D = 30°..............(2)

Solving 1 & 2 .

C = 45°

D = 15°

∴ Option 1 is the correct answer.

Concept:

Cosine Rule of Triangle:

The square of the length of any side of a given triangle is equal to the sum of the squares of the length of the other sides minus twice the product of the other two sides multiplied by the cosine of angle included between them.

Consider, a, b, and c are lengths of the side of a triangle ABC as shown, then;

\(cos(x)=\frac{b^2+c^2-a^2}{2bc}\)

\(cos(y)=\frac{a^2+c^2-b^2}{2ac}\)

\(cos(z)=\frac{a^2+b^2-c^2}{2ab}\)

Calculation:

Let m = n = 1 unit

Then,

 \(\rm \sqrt{m^2+n^2+mn} \ \\\Rightarrow\rm \sqrt{1^2+1^2+1}= \sqrt 3\) 

Using cosine rule;

\(\rm \cos θ = {1^2+ 1^2 - {\sqrt 3}^2\over2\times 1 \times 1}\)

⇒ cos θ = -1/2

∴ θ = 120° 

Now, the sum of the acute angles of the triangle = 180° - 120° = 60° 

Concept:

If a, b and c are the sides of the Δ ABC such that, a + b + c = 2S then

•  \(\rm \tan \frac{A}{2} = \sqrt {\frac{{\left( {S - b} \right)\left( {S - c} \right)}}{{S(S -a)}}} \)
• \(\rm \tan \frac{B}{2} = \sqrt {\frac{{\left( {S - c} \right)\left( {S - a} \right)}}{{S(S -b)}}} \)
• \(\rm \tan \frac{C}{2} = \sqrt {\frac{{\left( {S - a} \right)\left( {S - b} \right)}}{{S(S - c)}}} \)

Calculation:

Given: For ΔABC we have a = 13, b = 14 and c = 15

Here, we have to find the value of tan (C/2)

As we know that, if a, b and c are the sides of the Δ ABC then 2S = a + b + c

⇒ 2S = 13 + 14 + 15 = 42

⇒ S = 21

As we know that, \(\rm \tan \frac{C}{2} = \sqrt {\frac{{\left( {S - a} \right)\left( {S - b} \right)}}{{S(S - c)}}} \)

⇒ \(\rm \tan \frac{C}{2} = \sqrt {\frac{{\left( {21 - 13} \right) \times \left( {21 - 14} \right)}}{{21 \times (21-15)}}} \)

= \(\rm \sqrt {\frac {8 \times 7}{21 \times 6}} = \frac 2 3\)

Hence, option 2 is the correct answer.

CONCEPT:

If a, b and c are the sides of the Δ ABC such that, a + b + c = 2S then \(\sin \frac{A}{2} = \sqrt {\frac{{\left( {S - b} \right)\left( {S - C} \right)}}{{bc}}} \)

CALCULATION:

Given: For ΔABC we have a = 18, b = 24 and c = 30

Here, we have to find the value of  sin (A/2)

As we know that, if a, b and c are the sides of the Δ ABC then 2S = a + b + c

⇒ 2S = 18 + 24 + 30 = 72

⇒ S = 36

As we know that, \(\sin \frac{A}{2} = \sqrt {\frac{{\left( {S - b} \right)\left( {S - C} \right)}}{{bc}}} \)

\(\Rightarrow \sin \frac{A}{2} = \sqrt {\frac{{\left( {36 - 24} \right) \times \left( {36 - 30} \right)}}{{24 \times 30}}} = \frac{1}{{\sqrt {10} }}\)

Hence, option A is the correct answer.

Concept:

• sin (A + B) = sin A cos B + cos A sin B  ___(1)
• The sum of the three angles of a triangle is 180° 
• sin(180 - θ) = sin θ  ___(2)

Calculation:

In triangle ABC, sum of angles = A + B + C = 180° 

⇒ B + C = 180 - A   ___(3)

Given, sec A (sin B cos C + cos B sin C)

⇒ sec A sin (B + C)     ∵ {Using (1)}

⇒ sec A. sin (180 - A)   ∵ {Using (2)}

⇒ sec A. sin A   ∵ {Using (3)}

⇒  \({\sin A \over \cos A }\)

⇒ tan A

∴ The correct answer is option (3).