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)
- ∆ABC is an isosceles right angled triangle having ∠ C = 90°. If D is any point on AB, then AD2 + BD2 is equal to
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We draw a figure of an isosceles right angled triangle having ∠ C = 90° ,
AC² + CB² = AB²
⇒ 2BC² = (AD + DB)²
⇒ 2BC² = AD² + DB² + 2AD.BD ..... (i)
∆ CEB and ∆CED are right angles.
CD² = CE² + ED²
and, BC² = CE² + BE²
BC² – CD² = BE² – DE²
BC² – CD² = (BE + DE) (BE – DE)
BC² – CD² = (AE + DE) (BE – DE)Correct Option: B
We draw a figure of an isosceles right angled triangle having ∠ C = 90° ,
AC² + CB² = AB²
⇒ 2BC² = (AD + DB)²
⇒ 2BC² = AD² + DB² + 2AD.BD ..... (i)
∆ CEB and ∆CED are right angles.
CD² = CE² + ED²
and, BC² = CE² + BE²
BC² – CD² = BE² – DE²
BC² – CD² = (BE + DE) (BE – DE)
BC² – CD² = (AE + DE) (BE – DE)
BC² – CD² = AD . BD ..... (ii)
∴ From equations (i) and (ii)
AD² + DB² = 2CD²
- The vertical angle A of an isosceles triangle ∆ABC is three times the angle B of it. The measure of the angle A is
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As per the given in question , we draw a figure of an isosceles triangle ABC
In ∆ABC,
AB = AC
∴ ∠B = ∠C
∵ ∠A + ∠B + ∠C = 180°∴ ∠A + ∠A + ∠A = 180° 3 3 ⇒ 3∠A + ∠A + ∠A = 180° 3 ⇒ 5∠A = 180° 3
Correct Option: B
As per the given in question , we draw a figure of an isosceles triangle ABC
In ∆ABC,
AB = AC
∴ ∠B = ∠C
∵ ∠A + ∠B + ∠C = 180°∴ ∠A + ∠A + ∠A = 180° 3 3 ⇒ 3∠A + ∠A + ∠A = 180° 3 ⇒ 5∠A = 180° 3 ⇒ ∠A = 180° × 3 = 108° 5
- ∆ABC is isosceles having AB = AC and ∠A = 40°. Bisectors PO and OQ of the exterior angles ∠ABD and &Ang;ACE formed by producing BC on both sides, meet at O. Then the value of ∠BOC is
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On the basis of given in question , we draw a figure of an isosceles triangle ABC which bisectors PO and OQ of the exterior angles ∠ABD and &Ang;ACE formed by producing BC on both sides, meet at O
Given , AB = AC
∴ ∠ABC = ∠ACB = 140 ÷ 2 = 70°
∴ ∠ABD = ∠ACE = 180° - 70° = 110°Correct Option: A
On the basis of given in question , we draw a figure of an isosceles triangle ABC which bisectors PO and OQ of the exterior angles ∠ABD and &Ang;ACE formed by producing BC on both sides, meet at O
Given , AB = AC
∴ ∠ABC = ∠ACB = 140 ÷ 2 = 70°
∴ ∠ABD = ∠ACE = 180° - 70° = 110°
∴ ∠PBD = 55° = ∠CBO
∠QCE = ∠BCO = 55°
&there4 ∠BOC = 180° – 2 × 55° = 180° – 110° = 70°
