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Consider the following statements:
1) The perimeter of a triangle is greater than the sum of its three medinas.
2) In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.
Which of the above statements is/are correct?
1) The perimeter of a triangle is greater than the sum of its three medinas.
2) In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.
Which of the above statements is/are correct?
- a)1 only
- b)2 only
- c)Both 1 and 2
- d)Neither 1 nor 2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Consider the following statements:1) The perimeter of a triangle is gr...
1. Perimeter of triangle is greater than the sum of 3 medians
Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)
Hence in ΔABD, AD is a median
⇒ AB + AC > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
∴ AB + BC + AC > AD + BE + CF
2.
To prove: AB + BC + CA > 2AD
Construction: AD is joined
Proof: In triangle ABD, AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the third side] ---- 1
In triangle ADC, AC + DC > AD [because, the sum of any two sides of a triangle is always greater than the third side] ---- 2
Adding 1 and 2 we get,
AB + BD + AC + DC > AD + AD
=> AB + (BD + DC) + AC > 2AD
=> AB + BC + AC > 2AD
Hence proved
Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)
Hence in ΔABD, AD is a median
⇒ AB + AC > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
∴ AB + BC + AC > AD + BE + CF
2.
To prove: AB + BC + CA > 2AD
Construction: AD is joined
Proof: In triangle ABD, AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the third side] ---- 1
In triangle ADC, AC + DC > AD [because, the sum of any two sides of a triangle is always greater than the third side] ---- 2
Adding 1 and 2 we get,
AB + BD + AC + DC > AD + AD
=> AB + (BD + DC) + AC > 2AD
=> AB + BC + AC > 2AD
Hence proved
Most Upvoted Answer
Consider the following statements:1) The perimeter of a triangle is gr...
1. Perimeter of triangle is greater than the sum of 3 medians
Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)
Hence in ΔABD, AD is a median
⇒ AB + AC > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
∴ AB + BC + AC > AD + BE + CF
2.
To prove: AB + BC + CA > 2AD
Construction: AD is joined
Proof: In triangle ABD, AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the third side] ---- 1
In triangle ADC, AC + DC > AD [because, the sum of any two sides of a triangle is always greater than the third side] ---- 2
Adding 1 and 2 we get,
AB + BD + AC + DC > AD + AD
=> AB + (BD + DC) + AC > 2AD
=> AB + BC + AC > 2AD
Hence proved
Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)
Hence in ΔABD, AD is a median
⇒ AB + AC > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
∴ AB + BC + AC > AD + BE + CF
2.
To prove: AB + BC + CA > 2AD
Construction: AD is joined
Proof: In triangle ABD, AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the third side] ---- 1
In triangle ADC, AC + DC > AD [because, the sum of any two sides of a triangle is always greater than the third side] ---- 2
Adding 1 and 2 we get,
AB + BD + AC + DC > AD + AD
=> AB + (BD + DC) + AC > 2AD
=> AB + BC + AC > 2AD
Hence proved
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Consider the following statements:1) The perimeter of a triangle is greater than the sum of its three medinas.2) In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.Which of the above statements is/are correct?a)1 onlyb)2 onlyc)Both 1 and 2d)Neither 1 nor 2Correct answer is option 'C'. Can you explain this answer?
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