RS AGGARWAL CLASS 9 CHAPTER 9 CONGRUENCE OF TRIANGLES AND INEQUALITIES IN A TRIANGLE MCQ

 MULTIPLE CHOICE QUESTIONS

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Question 1:

Which of the following is not a criterion for congruence of triangles?
(a) SSA
(b) SAS
(c) ASA
(d) SSS

Answer 1:

(a) SSA
SSA is not a criterion for congruence of triangles.

Question 2:

If AB = QR, BC = RP and CA = PQ, then which of the following holds?
(a) ∆ABC ≅ ∆PQR
(b) ∆CBA ≅ ∆PQR
(c) ∆CAB ≅ ∆PQR
(d) ∆BCA ≅ ∆PQR

Answer 2:

(c) $∆CAB\cong ∆PQR$
As,
AB = QR,   (given)
BC = RP,     (given)
CA = PQ     (given)


∴ $∆CAB\cong ∆PQR$

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Question 3:

If ∆ABC ≅ ∆PQR  then which of the following is not true?
(a) BC = PQ
(b) AC = PR
(c) BC = QR
(d) AB = PQ

Answer 3:

(a) BC = PQ

If $∆ABC\cong ∆PQR$, then
BC = QR
Hence, the correct answer is option (a).

Question 4:

In ∆ABC, AB = AC and ∠B = 50°. Then, ∠A = ?
(a) 40°
(b) 50°
(c) 80°
(d) 130°

Answer 4:

In $△ABC$, we have:
AB = AC 
∠B = 50°
Since ABC is an isosceles triangle, we have:
$\angle C=\angle B$
$\angle C=50°$
In triangle ABC, we have:
$$\begin{gathered} \angle A+\angle B+\angle C=180° \\ \Rightarrow \angle A+50+50=180° \\ \Rightarrow \angle A=180°-100° \\ \therefore \angle A=80° \end{gathered}$$
Hence, the correct answer is option (c).

Question 5:

In ∆ABC, BC = AB and ∠B = 80°. Then, ∠A = ?
(a) 50°
(b) 40°
(c) 100°
(d) 80°
 

Answer 5:

Given: In ∆ABC, BC = AB and ∠B = 80°. 

In ∆ABC,

As, $AB=BC$

$\Rightarrow \angle A=\angle C$

Let $\angle A=\angle C=x$

Using angle sum property of a triangle,

$$\begin{gathered} \angle A+\angle B+\angle C=180° \\ \Rightarrow x+80°+ x=180° \\ \Rightarrow 2x=180°-80° \\ \Rightarrow 2x=100° \end{gathered}$$

$$\begin{gathered} \Rightarrow x=\frac{100°}{2} \\ \Rightarrow x=50° \\ \Rightarrow \angle A=50° \end{gathered}$$

Hence, the correct option is (a).

Question 6:

In ∆ABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?
(a) 4 cm
(b) 5 cm
(c) 8 cm
(d) 2.5 cm

Answer 6:

Given: In ∆ABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm.

In ∆ABC,

As, ∠C = ∠A                   (Given)

Therefore, $BC=AB$       (Sides opposite to equal angles.)

$\Rightarrow BC=AB=4 \mathrm{cm}$

Hence, the correct option is (a).

Question 7:

Two sides of a triangle are of length 4 cm and 2.5 cm. The length of the third side of the triangle cannot be
(a) 6 cm
(b) 6.5 cm
(c) 5.5 cm
(d) 6.3 cn

Answer 7:

Since, 4 + 2.5 = 6.5

So, 6.5 cm cannot be the third side of the triangle, as the sum of two sides of a triangle is always greater than the third side.

Hence, the correct option is (b).

Question 8:

In ∆ABC, if ∠C > ∠B, then
(a) BC > AC
(b) AB > AC
(c) AB > AC
(d) BC > AC

Answer 8:

(b) AB > AC

In $∆ABC$, we have:
$\angle C>\angle B$
The side opposite to the greater angle is larger.
$\therefore AB>AC$

Question 9:

It is given that ∆ABC ≅ ∆ FDE in which AB = 5 cm, ∠B = 40°, ∠A = 80° and FD = 5 cm. Then, which of the following is true?
(a) ∠D = 60°
(b) ∠E = 60°
(c) ∠F = 60°
(d) ∠D = 80°

Answer 9:

(b)​ $\angle E=60°$

$∆ABC\cong ∆FDE$
AB = 5cm, $\angle B=40°,\angle A=80°$ and FD = 5cm

$$\begin{gathered} \mathrm{Then} \angle A+\angle B+\angle C=180° \\ \Rightarrow 80°+40°+\angle C=180° \\ \Rightarrow \angle C=60° \\ Also, \angle C=\angle E \end{gathered}$$
∴ $\angle E=60°$

Question 10:

In ∆ABC, ∠A = 40° and ∠B = 60°. Then the longest side of ∆ABC is
(a) BC
(b) AC
(c) AB
(d) cannot be determined

Answer 10:

(c) AB

In triangle ABC, we have:
$\angle A=40°,\angle B=60°$        ...(Given)
$$\begin{gathered} \mathrm{Here}, \angle A+\angle B+\angle C=180° \\ \Rightarrow 60°+40°+\angle C=180° \\ \Rightarrow \angle C=80° \end{gathered}$$
∴ The side opposite to$\angle C$, i.e., AB, is the longest side of triangle ABC.

Question 11:

In the given figure, AB > AC. Then which of the following is true?
(a) AB < AD
(b) AB = AD
(c) AB > AD
(d) Cannot be determined

Answer 11:

(c) AB > AD

$AB>AC$ is given.

∴ $\angle ACB>\angle ABC$

 $$\begin{gathered} \mathrm{Now}, \angle ADB>\angle ACD (\mathrm{exterior} \mathrm{angle}) \\ \Rightarrow \angle ADB>\angle ACB>\angle ABC \\ \Rightarrow \angle ADB>\angle ABD\Rightarrow AB>AD \end{gathered}$$

Question 12:

In the given figure, AB > AC. If BO and CO are the bisectors of ∠B and ∠C respectively, then
(a) OB = OC
(b) OB > OC
(c) OB < OC

Answer 12:

(b) OB > OC

AB >AC    (Given)
$\Rightarrow \angle C>\angle B$
$\Rightarrow \frac{1}{2}\angle C>\frac{1}{2}\angle B$
$\Rightarrow \angle OCB>\angle OBC$  (Given)
$\Rightarrow OB>OC$

Question 13:

In the given figure, AB = AC and OB = OC. Then, ∠ABO : ∠ACO = ?
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) None of these

Answer 13:

(a) 1:1

​In $∆OAB \mathrm{and} ∆OAC, \mathrm{we} \mathrm{have}:$
$$\begin{gathered} \mathrm{AB}=\mathrm{AC} \left(\mathrm{Given}\right) \\ \mathrm{OB}=\mathrm{OC} \left(\mathrm{Given}\right) \\ \mathrm{OA} =\mathrm{OA} (\mathrm{Common} \mathrm{side}) \end{gathered}$$
$\mathrm{Thus}, ∆\mathrm{OAB}\cong \mathrm{OAC} (\mathrm{SSS} \mathrm{criterion})$
i.e., $\angle ABO=\angle ACO$
∴ $\angle ABO : \angle ACO=1 : 1$

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Question 14:

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is
(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

Answer 14:

(b) isosceles


In $∆ABC, \mathrm{BL}\parallel \mathrm{AC}.$
$\mathrm{CM}\parallel \mathrm{AB} \mathrm{such} \mathrm{that} \mathrm{BL}=\mathrm{CM}.$

To prove: AB = AC

$$\begin{gathered} \mathrm{In} △\mathrm{ABL} \mathrm{and} △\mathrm{ACM}, \\ \mathrm{BL}=\mathrm{CM} \left(\mathrm{Given}\right) \\ \angle \mathrm{BAL}=\angle \mathrm{CAM} (\mathrm{Common} \mathrm{angle}) \\ \angle \mathrm{ALB}=\angle \mathrm{AMC} (\mathrm{Each} 90°) \\ △\mathrm{ABL}\cong △\mathrm{ACM} (\mathrm{AAS} \mathrm{criterion}) \\ \therefore \mathrm{AB}=\mathrm{AC} \left(\mathrm{CPCT}\right) \end{gathered}$$

Question 15:

In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have
(a) ∠A = ∠D
(b) ∠B = ∠E
(c) ∠C = ∠F
(d) none of these

Answer 15:

(b) $\angle B=\angle E$


In $△ABC \mathrm{and} △DEF, \mathrm{we} \mathrm{have}:$
AB = DE      (Given)
BC = EF       (Given)
In order that $△ABC\cong DEF$, we must have $\angle B=\angle E$.

Question 16:

In ∆ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABC ≅ ∆DEF, we must have
(a) AB = DF
(b) AC = DE
(c) BC = EF
(d) ∠A = ∠D

Answer 16:

In order that $△ABC\cong △DEF$, we must have BC = EF.
Hence, the correct answer is option (c).

Question 17:

In ∆ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles

Answer 17:

(a) isosceles but not congruent

$$\begin{gathered} AB =AC \\ \Rightarrow \angle C=\angle B \\ \Rightarrow \angle P=\angle Q [\because \angle C=\angle P \mathrm{and} \angle B=\angle Q] \end{gathered}$$
Thus, both the triangles are isosceles but not congruent.

Question 18:

Which is true?
(a) A triangle can have two right angles.
(b) A triangle can have two obtuse angles.
(c) A triangle can have two acute angles.
(d) An exterior angle of a triangle is less than either of the interior opposite angles.

Answer 18:

(c) A triangle can have two acute angles.
The sum of two acute angles is always less than 180^o, thus satisfying the angle sum property of a triangle.
Therefore, a triangle can have two acute angles.

Question 19:

Fill in the blanks with < or >.
(a) (Sum of any two sides of a triangle) ...... (the third side)
(b) (Difference of any two sides of a triangle) ...... (the third side)
(c) (Sum of three altitudes of a triangle) ...... (sum of its three side)
(d) (Sum of any two sides of a triangle) ...... (twice the median to the 3rd side)
(e) (Perimenter of a triangle) ...... (sum of its three medians)

Answer 19:

a) Sum of any two sides of a triangle > the third side

b) Difference of any two sides of a triangle < the third side

c) Sum of three altitudes of a triangle < sum of its three side

d) Sum of any two sides of a triangle > twice the median to the 3rd side

e) Perimeter of a triangle > sum of its three medians

Question 20:

Fill in the blanks.
(a) Each angle of an equilateral triangle measures ...... .
(b) Medians of an equilateral triangle are ...... .
(c) In a right triangle the hypotenuse is the ...... side.
(d) Drawing a ∆ABC with AB = 3 cm, BC = 4 cm and CA = 7 cm is ...... .

Answer 20:

a) Each angle of an equilateral triangle measures $60°$.

b) Medians of an equilateral triangle are equal.

c) In a right triangle, the hypotenuse is the longest side.

d) Drawing a $△ABC$ with AB = 3cm, BC = 4cm and CA = 7cm is not possible.