Mathematics Questions and Answers – Inequalities in a Triangles

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This set of Mathematics 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

Answer: a
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).
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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

Answer: c
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

Answer: b
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).
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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

Answer: d
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

Answer: b
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).
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6. Which among the following is the shortest?

a) PQ3
b) PQ1
c) PQ4
d) PQ2
View Answer

Answer: a
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, PQ3 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

Answer: c
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).
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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

Answer: a
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

Answer: b
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).
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter