This set of Class 9 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Inequalities in a Triangles”.

1. Which of the following figures does not represent a triangle?

a)

b)

c)

d)

View Answer

Explanation: Sum of any two sides of a triangle should be greater than the third side.

So, triangle with sides 3cm, 5cm, 9cm does not satisfy the condition. (3 + 5 = 8 < 9).

2. Which among the following statements are not correct?

a) ∠P is the largest angle

b) ∠R is the smallest angle

c) ∠P is the smallest angle

d) ∠P > ∠Q

View Answer

Explanation: In a triangle, angle opposite to the longest side is largest and angle opposite to the shortest side is the smallest. So, in triangle PQR, QR is the longest side. Hence ∠P is the largest angle. Similarly, QP is the shortest side. Hence ∠R is the smallest angle. Therefore, ∠P > ∠Q > ∠R.

3. Identify the correct relation if x > y.

a) AB < AC

b) AB > AC

c) AB < 2AC

d) AB = AC

View Answer

Explanation: Given that, x > y.

From Figure, ∠EBC + ∠CBA = 180° (Linear Pair)

And ∠CAD + ∠CAB = 180° (Linear Pair)

So, ∠EBC + ∠CBA = ∠CAD + ∠CAB

⇒ x + ∠CBA = y + ∠CAB

The above condition is valid only when ∠CBA < ∠CAB as x > y.

So, ∠CBA < ∠CAB

⇒ AC < AB (side opposite to larger angle is the longer side).

4. Identify the correct relation if x : y : z = 1 : 3 : 5.

a) BC > AB > AC

b) AC < BC < AB

c) BC > AC > AB

d) BC < AC < AB

View Answer

Explanation: x : y : z = 1 : 3 : 5

In ΔABC, ∠A + ∠B + ∠C = 180° (Angle sum property of triangle)

⇒ x + y + z = 180°

⇒ k + 3k + 5k = 180°

⇒ k = 20°

∠A = k = 20°, ∠B = 3k = 60°, ∠C = 5k = 100°

Hence, ∠A < ∠B < ∠C

So, BC < AC < AB as the side opposite to the largest angle is the longest and side opposite to smallest angle is the shortest.

5. Which among the following relation is correct if AD = AC?

a) AB < AD

b) AB > AD

c) AB = AD

d) AB = ½ AD

View Answer

Explanation: In ΔACD, AD = AC

⇒ ∠ADC = ∠ACD ————–(i) (Angles opposite to equal sides are equal)

⇒ ∠ADC = ∠ACB ————–(ii)

Since, exterior angle of a triangle is greater than each of the interior opposite angle,

∠ADC > ∠ABD

⇒ ∠ACB > ∠ABC (From equation ii)

⇒ AB > AC (Side opposite to larger angle is the longer side in a triangle)

⇒ AB > AC (AD = AC).

6. Which among the following is the shortest?

a) PQ_{3}

b) PQ_{1}

c) PQ_{4}

d) PQ_{2}

View Answer

Explanation: Among all the segments that can be drawn to a given line from a given point (not on the line), the perpendicular line segment is the shortest. Hence, PQ

_{3}which is perpendicular to line l is the shortest line segment.

7. Which among the following relation holds true if AD, BE and DF are the altitudes of the triangle?

a) AD + BE + CF > AB + BC + AC

b) AD + BE + CF = AB + BC + AC

c) AD + BE + CF < AB + BC + AC

d) AD + BE + CF > 2(AB + BC + AC)

View Answer

Explanation: Among all the segments that can be drawn to a given line from a given point (not on the line), the perpendicular line segment is the shortest.

So, In ΔABC, AD ⊥ BC ⇒ AB > AD and AC > AD

⇒ AB + AC > 2AD ———–(i)

BE ⊥ AC ⇒ BC > BE and BA > BE

⇒ BC + BA > 2BE ———–(ii)

CF ⊥ AB ⇒ CA > CF and CB > CF

⇒ CA + CB > 2CF ———–(iii)

Adding equation i, ii and iii, AB + AC + BC + BA + CA + CB > 2AD + 2BE + 2CF

⇒ 2(AB + BC + AC) > 2(AD + BE + CF)

⇒ (AB + BC + AC) > (AD + BE + CF).

8. Find the correct relation among the following.

a) PQ + PR > 2PS

b) PQ + PR < 2PS

c) PQ + PR < PS

d) PQ + PR = ½ PS

View Answer

Explanation: From Figure, QS = QR. So, PS is the median from P to side QR.

As sum of any two sides of a triangle is greater than twice the median drawn to the third side,

PQ + PR > 2PS.

9. Identify the correct relation among the following.

a) AB + BC + CD + AD = AC + BD

b) AB + BC + CD + AD > AC + BD

c) AB + BC + CD + AD < 2(AC + BD)

d) AB + BC + CD + AD = ¼ (AC + BD)

View Answer

Explanation: Since sum of any two sides of a triangle is greater than the third side,

In ΔABC, AB + BC > AC ———- (i)

In ΔACD, AD + CD > AC ———- (ii)

In ΔBCD, BC + CD > BD ———- (iii)

In ΔABD, AB + AD > BD ———- (iv)

Adding equation i, ii, iii and iv,

AB + BC + AD + CD + BC + CD + AB + AD > AC + AC + BD + BD

⇒ 2(AB + BC + CD + AD) > 2(AC + BD)

⇒ (AB + BC + CD + AD) > (AC + BD).

**Sanfoundry Global Education & Learning Series – Mathematics – Class 9**.

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