Triangles — Competency Focused Questions

In the given figure, the bisectors of ∠B and ∠C intersect each other at O and ∠BAC = 50°. The measure of ∠BOC is :

1. 100°

2. 115°

3. 130°

4. 140°

Answer

In △ABC,

By angle sum property of triangle,

∠A + ∠B + ∠C = 180°

⇒ 50° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° - 50°

⇒ ∠B + ∠C = 130° .......(1)

From figure,

As, OB is bisector of angle B.

∠B = ∠ABO + ∠OBC = ∠OBC + ∠OBC = 2∠OBC

⇒ ∠OBC = $\dfrac{∠B}{2}$

As, OC is bisector of angle C.

∠C = ∠ACO + ∠OCB = ∠OCB + ∠OCB = 2∠OCB

⇒ ∠OCB = $\dfrac{∠C}{2}$

In △OBC,

By angle sum property of triangle,

⇒ ∠OBC + ∠BOC + ∠OCB = 180°

⇒ $\dfrac{∠B}{2}$ + ∠BOC + $\dfrac{∠C}{2}$ = 180°

⇒ ∠BOC + $\dfrac{∠B + ∠C}{2}$ = 180°

⇒ ∠BOC + $\dfrac{130°}{2}$ = 180° [Substituting from eq.(1)]

⇒ ∠BOC + 65° = 180°

⇒ ∠BOC = 180° - 65°

⇒ ∠BOC = 115°.

Hence, option 2 is the correct option.

In the given figure, △ABD ≅ △ACD. If ∠DAC = 30° and ∠BDC = 110°, then the measure of ∠DBA is :

1. 30°

2. 50°

3. 70°

4. 25°

Answer

Given,

△ABD ≅ △ACD

Since, corresponding parts of congruent triangles are equal.

⇒ ∠DBA = ∠ACD = y (let)

⇒ ∠ADB = ∠ADC = x (let)

From figure,

⇒ ∠ADB + ∠ADC + ∠BDC = 360°

⇒ x + x + 110° = 360°

⇒ 2x = 360° - 110°

⇒ 2x = 250°

⇒ x = $\dfrac{250°}{2}$

⇒ x = 125°

⇒ ∠ADC = 125°

In △ADC,

By angle sum property of triangle,

⇒ ∠ADC + ∠ACD + ∠CAD = 180°

⇒ 125° + y + 30° = 180°

⇒ 155° + y = 180°

⇒ y = 180° - 155°

⇒ y = 25°

⇒ ∠DBA = y = 25°.

Hence, option 4 is the correct option.

ABC is a triangle in which AC = BC and ∠BAC = 50°. Side BC is produced to D such that BC = CD. ∠BAD is equal to :

1. 45°

2. 50°

3. 90°

4. 100°

Answer

Given,

AC = BC

∠BAC = ∠ABC = 50°

In △ABC,

By angle sum property of triangle,

∠BAC + ∠ABC + ∠ACB = 180°

⇒ 50° + 50° + ∠ACB = 180°

⇒ 100° + ∠ACB = 180°

⇒ ∠ACB = 180° - 100°

⇒ ∠ACB = 80°

From figure,

∠ACD + ∠ACB = 180° (Linear pair)

⇒ ∠ACD + 80° = 180°

⇒ ∠ACD = 180° - 80°

⇒ ∠ACD = 100°

In △ACD,

AC = CD

∠CAD = ∠ADC = x (let)

By angle sum property of triangle,

⇒ ∠ADC + ∠CAD + ∠ACD = 180°

⇒ x + x + 100° = 180°

⇒ 2x = 180° - 100°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$

⇒ x = 40°.

⇒ ∠CAD = ∠ADC = 40°.

From figure,

∠BAD = ∠BAC + ∠CAD = 50° + 40° = 90°.

Hence, option 3 is the correct option.

ABD is a triangle such that ∠ADB = 20° and C is a point on BD such that AB = AC and CD = CA. The measure of ∠ABC :

1. 40°

2. 50°

3. 55°

4. 60°

Answer

In △ADC,

CD = CA

∠ADC = ∠CAD = 20° (Angles opposite to equal sides in a triangle are equal)

In △ACD,

By angle sum property of triangle,

⇒ ∠ACD + ∠ADC + ∠CAD = 180°

⇒ ∠ACD + 20° + 20° = 180°

⇒ ∠ACD + 40° = 180°

⇒ ∠ACD = 180° - 40°

⇒ ∠ACD = 140°

From figure,

∠ACB + ∠ACD = 180° (Linear pair)

⇒ ∠ACB + 140° = 180°

⇒ ∠ACB = 180° - 140°

⇒ ∠ACB = 40°

In △ABC,

AB = AC

∠ABC = ∠ACB = 40° (Angles opposite to equal sides in a triangle are equal)

Hence, option 1 is the correct option.

The lengths of the three sides of a triangle are 4 cm, 5 cm, and 7 cm. Which of the following cannot be the length of any one of the medians?

1. 2.5 cm

2. 3.8 cm

3. 5 cm

4. None of these

Answer

Suppose there is a triangle with sides of length a, b and c, then the median to side a is always less than the sum of other two sides.

Thus, in this case each of the following options can be the length of the median of triangle.

Hence, option 4 is the correct option.

In △ABC, ∠B = 35°, ∠C = 65° and the bisector AD of ∠BAC meets BC at D. Arrange the sides AD, BD and CD in ascending order of their lengths.

Answer

In △ADB,

⇒ ∠BAD + ∠ADB + ∠ABD = 180°

⇒ 40° + ∠ADB + 35° = 180°

⇒ ∠ADB + 75° = 180°

⇒ ∠ADB = 180° - 75°

⇒ ∠ADB = 105°.

We know that,

The shortest side of a triangle has the smallest angle opposite to it.

In triangle ABD,

Since,

⇒ ∠B < ∠A

⇒ AD < BD .......(1)

From figure,

∠ADB + ∠ADC = 180° (Linear pair)

⇒ ∠ADC + 105° = 180°

⇒ ∠ADC = 180° - 105°

⇒ ∠ADC = 75°

In triangle ACD,

Since,

⇒ ∠A < ∠C

⇒ CD < AD ........(2)

From eq.(1) and (2) we have:

⇒ CD < AD < BD

Hence, CD < AD < BD.

In the given figure, find the value of a + b + c + d + e + f.

Answer

In △AED,

⇒ a + b + ∠A = 180° ....(1)

In △DBF,

⇒ c + d + ∠B = 180° ....(2)

In △EFC,

⇒ e + f + ∠C = 180° ....(3)

In △ABC,

⇒ ∠A + ∠B + ∠C = 180° ....(4)

Adding eq.(1), (2) and (3), we get :

⇒ a + b + c + d + e + f + ∠A + ∠B + ∠C = 180° + 180° + 180°

⇒ a + b + c + d + e + f + 180° = 540°

⇒ a + b + c + d + e + f = 540° - 180°

⇒ a + b + c + d + e + f = 360°.

Hence, the value of a + b + c + d + e + f = 360°.