Plane Geometry
- In ∆ABC, AD is the internal bisector of ∠A, meeting the side BC at D. If BD = 5 cm, BC = 7.5 cm, then AB : AC is
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As per the given in question , we draw a figure of a triangle ABC in which AD is the internal bisector of ∠A, meeting the side BC at D ,
AD is the internal bisector of ∠A.
Here , BD = 5 cm, BC = 7.5 cm∴ AB = BD = 5 AC DC 7.5 - 5
Correct Option: A
As per the given in question , we draw a figure of a triangle ABC in which AD is the internal bisector of ∠A, meeting the side BC at D ,
AD is the internal bisector of ∠A.
Here , BD = 5 cm, BC = 7.5 cm∴ AB = BD = 5 AC DC 7.5 - 5 AB : AC = 5 = 2 : 1 2.5
- O and C are respectively the orthocentre and circumcentre of an acute-angled triangle PQR. The points P and O are joined and produced to meet the side QR at S. If ∠PQS = 60° and ∠QCR = 130°, then ∠RPS=
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On the basis of given in question , we draw a figure with O and C are respectively orthocentre and circumcentre of an acute-angled triangle PQR ,
∠ PQS = 60°
∠ QCR = 130°∴ ∠QPR = 1 × 130° = 65° 2
⇒ ∠QRP =180° – 60° – 65° = 55°
⇒ ∠PCQ = 110°
∴ In ∆ QCR,
QC = CR
⇒ ∠CQR = ∠CRQ = 25°
[∵ ∠CQR + ∠CRQ = 50°]
Correct Option: B
On the basis of given in question , we draw a figure with O and C are respectively orthocentre and circumcentre of an acute-angled triangle PQR ,
∠ PQS = 60°
∠ QCR = 130°∴ ∠QPR = 1 × 130° = 65° 2
⇒ ∠QRP =180° – 60° – 65° = 55°
⇒ ∠PCQ = 110°
∴ In ∆ QCR,
QC = CR
⇒ ∠CQR = ∠CRQ = 25°
[∵ ∠CQR + ∠CRQ = 50°]
∴ ∠PQC = ∠QPC = 35°
[∵ ∠PQC + ∠QPC = 70°]
Similarly, ∠ CPR = 30°
∴ ∠RPS = 35°
- The sides of a triangle are in the ratio 3 : 4 : 6. The triangle is :
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Here , The sides of a triangle are in the ratio 3 : 4 : 6
Let the sides of the triangle be 3k, 4k and 6k units.
Clearly, (3k)² + (4k)² < (6k)²Correct Option: C
Here , The sides of a triangle are in the ratio 3 : 4 : 6
Let the sides of the triangle be 3k, 4k and 6k units.
Clearly, (3k)² + (4k)² < (6k)²
∴ The triangle will be obtuse angled.
- In an isosceles triangle ∆ ABC, AB = AC and ∠A = 80°. The bisector of∠B and ∠ C meet at D. The ∠BDC is equal to
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As per the given in question , we draw a figure of an isosceles triangle ABC
Given that , AB = AC
∴ ∠ABC = ∠ACB
∠A = 80°
∴ ∠B + ∠C = 180° – 80° = 100°
∴ ∠B = 100 ÷ 2 = 50° = ∠CCorrect Option: C
As per the given in question , we draw a figure of an isosceles triangle ABC
Given that , AB = AC
∴ ∠ABC = ∠ACB
∠A = 80°
∴ ∠B + ∠C = 180° – 80° = 100°
∴ ∠B = 100 ÷ 2 = 50° = ∠C
∴ ∠DBC = ∠DCB = 50 ÷ 2 = 25°
∴ ∠BDC = 180° – (∠DBC + ∠DCB) = 180° – 50° = 130°
- In the figure (not drawn to scale) given below, if AD = DC = BC and ∠BCE = 96°, then ∠DBC is :
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From given figure ,
Let ∠ACD = a = ∠DAC
∴ ∠CDB = 2a = ∠CBD
The angles of the base of an isosceles triangle are equal.
∴ ∠ACB = 180° – 96° = 84°
⇒ ∠ACD + ∠DCB = 84°
⇒ a + 180° – 4a = 84°
⇒ 180° – 3a = 84°Correct Option: C
From given figure ,
Let ∠ACD = a = ∠DAC
∴ ∠CDB = 2a = ∠CBD
The angles of the base of an isosceles triangle are equal.
∴ ∠ACB = 180° – 96° = 84°
⇒ ∠ACD + ∠DCB = 84°
⇒ a + 180° – 4a = 84°
⇒ 180° – 3a = 84°
⇒ 3a = 180° – 84° = 96°
⇒ a = 96 ÷ 3 = 32°
⇒ ∠DBC = 2a = 64°
