Plane Geometry
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In ∆ABC, ∠A = 90°, BP and CQ are two medians. Then the value of BP² + CQ² BC²
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On the basis of given in question , we draw a figure triangle ABC ,
In ∆ AQC,
∠A = 90°
⇒ CQ² = AC² + QA²
⇒ 4CQ² = 4AC² + 4QA²
⇒ 4CQ² = 4AC² + (2QA)²
⇒ 4CQ² = 4AC² + AB²
[∵AB = 2QA]
In ∆ BPA,
BP² = BA² + AP²
⇒ 4BP² = 4BA² + 4AP²
⇒ 4BP² = 4BA² + AC²
[∵ AC = 2AP]
∴ 4 CQ² + 4 BP² = 4 AC² + AB² + 4 AB² + AC²
⇒ 4(CQ² + BP²) = 5(AC² + AB²) = 5 BC²Correct Option: B
On the basis of given in question , we draw a figure triangle ABC ,
In ∆ AQC,
∠A = 90°
⇒ CQ² = AC² + QA²
⇒ 4CQ² = 4AC² + 4QA²
⇒ 4CQ² = 4AC² + (2QA)²
⇒ 4CQ² = 4AC² + AB²
[∵AB = 2QA]
In ∆ BPA,
BP² = BA² + AP²
⇒ 4BP² = 4BA² + 4AP²
⇒ 4BP² = 4BA² + AC²
[∵ AC = 2AP]
∴ 4 CQ² + 4 BP² = 4 AC² + AB² + 4 AB² + AC²
⇒ 4(CQ² + BP²) = 5(AC² + AB²) = 5 BC²⇒ BP² + CQ² = 5 BC² 4
- If the sides of a triangle are extended in both the sides then the sum of the exterior angles so formed in both sides is
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Firstly , We draw a figure of triangle ABC extended in both the sides ,
As we know that , ∠ABC + ∠BCA + ∠BAC = 180°
Again, ∠ACB + ∠ACD = 180°
∠ABC + ∠ABF = 180°
∠BAC + ∠EAC = 180°Correct Option: C
Firstly , We draw a figure of triangle ABC extended in both the sides ,
As we know that , ∠ABC + ∠BCA + ∠BAC = 180°
Again, ∠ACB + ∠ACD = 180°
∠ABC + ∠ABF = 180°
∠BAC + ∠EAC = 180°
∴ ∠ACD + ∠ABF + ∠CAE = 540 – 180° = 360°
∴ Required answer = 2 × 360° = 720 °
- 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°
