Plane Geometry
- In ∆ ABC, ∠B = 60°, ∠C = 40°, AD is the bisector of ∠A and AE is drawn perpendicular on BC from A. Then the measure of ∠EAD is
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As per the given in question , we draw a figure of triangle ABC whose AD is the bisector of ∠A and AE is drawn perpendicular on BC from A ,
Given that , ∠B = 60°, ∠C = 40°
∠BAC = 180° – 60° – 40° = 80°
∠BAD = ∠DAC = 40°
In ∆ ABE,Correct Option: C
As per the given in question , we draw a figure of triangle ABC whose AD is the bisector of ∠A and AE is drawn perpendicular on BC from A ,
Given that , ∠B = 60°, ∠C = 40°
∠BAC = 180° – 60° – 40° = 80°
∠BAD = ∠DAC = 40°
In ∆ ABE,
∠BAE = 90° – 60° = 30°
∠EAD = 40° – 30° = 10°
- In a ∆ ABC, AD ,BE and CF are three medians. Then the ratio (AD + BE + CF) :
(AB + AC + BC) is
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As per the given in question , we draw a figure of triangle ABC
In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it.
∴ AB² + AC² = 2 (AD² + BD²)⇒ AB² + AC² = 2 
AD² + BC² 
4
⇒ 2(AB² + AC²) = 4AD² + BC²
Similarly,
2(AB² + BC²) = 4 BE² + AC²
2 (AC² + BC²) = 4 CF² + AB²
Correct Option: E
As per the given in question , we draw a figure of triangle ABC
In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it.
∴ AB² + AC² = 2 (AD² + BD²)⇒ AB² + AC² = 2 AD² + BC² 4
⇒ 2(AB² + AC²) = 4AD² + BC²
Similarly,
2(AB² + BC²) = 4 BE² + AC²
2 (AC² + BC²) = 4 CF² + AB²
On adding all three, we get
4 (AB² + BC² + AC²) = 4 (AD² + BE² + CF²) + BC² + AC² + AB²
⇒ 3 (AB² + BC² + AC²) = 4 (AD² + BE² + CF²)
- The internal bisectors of the angles B and C of a triangle ABC meet at I. If ∠BIC = (∠A/2) + X, then X is equal to
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On the basis of given in question , we draw a figure triangle ABC whose I is incentre,
In ∆ ABC,
We know that , ∠A + ∠B + ∠ = 180°
∴ ∠B + ∠ = 180° – ∠A∴ ∠1 (∠B + ∠C) = 90° - ∠A 2 2
In ∆ BIC,∠B + ∠C + ∠BIC = 180° 2 2 ∴ 90° - ∠A + ∠BIC = 180° 2 ⇒ ∠BIC = 180° - 90° + ∠A 2
Correct Option: C
On the basis of given in question , we draw a figure triangle ABC whose I is incentre ,
In ∆ ABC,
We know that , ∠A + ∠B + ∠ = 180°
∴ ∠B + ∠ = 180° – ∠A∴ ∠1 (∠B + ∠C) = 90° - ∠A 2 2
In ∆ BIC,∠B + ∠C + ∠BIC = 180° 2 2 ∴ 90° - ∠A + ∠BIC = 180° 2 ⇒ ∠BIC = 180° - 90° + ∠A 2 ∠BIC = 90° + ∠A 2
∴ X = 90°
- The measure of the angle between the internal and external bisectors of an angle is
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As per the given in question , we draw a figure of triangle ABC
From figure ,∠ACF = ∠FCB = ∠C 2 ∠ACE = ∠ECD = 180° - ∠C 2 ∠ACE = 90° - ∠C 2
∴ ∠FCE = ∠FCA + ∠ACE
Correct Option: D
As per the given in question , we draw a figure of triangle ABC
From figure ,∠ACF = ∠FCB = ∠C 2 ∠ACE = ∠ECD = 180° - ∠C 2 ∠ACE = 90° - ∠C 2
∴ ∠FCE = ∠FCA + ∠ACE∠FCE = ∠C + 90° - ∠C = 90° 2 2
- The sum of three altitudes of a triangle is
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As per the given in question , we draw a figure of triangle ABC
AP < AB
BQ < BC
CR < ACCorrect Option: B
As per the given in question , we draw a figure of triangle ABC
AP < AB
BQ < BC
CR < AC
∴ AP + BQ + CR < AB + BC + AC
⇒The sum of three altitudes of a triangle is less than the sum of sides.
