Plane Geometry
- In ∆ ABC if the median ( 1/2 ) AD = BC, then ∠BAC is equal to
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According to question , we draw a figure ∆ ABC
Point ‘D’ is the mid-point of side BC.AD = 1 BC 2
⇒ 2AD = BC
∴ AD = BD = DC
∴ AB = AC
Correct Option: A
According to question , we draw a figure ∆ ABC
Point ‘D’ is the mid-point of side BC.AD = 1 BC 2
⇒ 2AD = BC
∴ AD = BD = DC
∴ AB = AC
∴ AD ⊥ BC
∴ ∠ABD = ∠DAB = 45°
∴ ∠BAC = 90°
- In ∆ ABC two medians BE and CF intersect at the point O and P, Q are the midpoints of BO and CO respectively. If the length of PQ = 3cm, then the length of FE will be
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On the basis of question we draw a figure of triangle ABC ,
The line joining the mid-points of two sides of a triangle is parallel to the third side and half of that one.∴ FE = 1 BC (From ∆ ABC) 2
Correct Option: A
On the basis of question we draw a figure of triangle ABC ,
The line joining the mid-points of two sides of a triangle is parallel to the third side and half of that one.∴ FE = 1 BC (From ∆ ABC) 2 PQ = 1 BC (From ∆ BOC) 2
∴ FE = PQ = 3 cm.
- In a triangle PQR, S and T are the points on PQ and PR respectively, such that ST || QR and PS / SQ = 3 / 5 , PR = 6 cm, then PT is
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We draw a figure triangle PQR whose S and T are the points on PQ and PR respectively,
In ∆PST and ∆PQR,
∠PST = ∠PQR (∵ ST || QR)
∠PTS = ∠PRQ
∴ By AA– similarity,
∆ PST ~ ∆PQR∴ PS = PT PQ PR ⇒ PQ = PR PS PT ⇒ PS + SQ = 6 PS PT ⇒ 1 + SQ = 6 PS PT ⇒ 1 + 5 = 6 3 PT ⇒ 8 = 6 3 PT
⇒ 8 PT = 6 × 3
Correct Option: B
We draw a figure triangle PQR whose S and T are the points on PQ and PR respectively,
In ∆PST and ∆PQR,
∠PST = ∠PQR (∵ ST || QR)
∠PTS = ∠PRQ
∴ By AA– similarity,
∆ PST ~ ∆PQR∴ PS = PT PQ PR ⇒ PQ = PR PS PT ⇒ PS + SQ = 6 PS PT ⇒ 1 + SQ = 6 PS PT ⇒ 1 + 5 = 6 3 PT ⇒ 8 = 6 3 PT
⇒ 8 PT = 6 × 3⇒ PT = 6 × 3 = 9 8 4
∴ PT = 2.25 cm.
- An exterior angle of a triangle is 115° and one of the interior opposite angles is 45°. Then the other two angles are
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The exterior angle of a triangle is equal to the sum of two remaining opposite angles.
Given that , ∠ACD = 115° and ∠A = 45°
we know that , ∠ACD = ∠A + ∠B
⇒ 115° = 45° + ∠B
⇒ ∠B = 115° – 45° = 70°
⇒ ∠ACB = 180° – 115° = 65°
OR
∠ACB = 180° – ∠ACDCorrect Option: A
The exterior angle of a triangle is equal to the sum of two remaining opposite angles.
Given that , ∠ACD = 115° and ∠A = 45°
we know that , ∠ACD = ∠A + ∠B
⇒ 115° = 45° + ∠B
⇒ ∠B = 115° – 45° = 70°
⇒ ∠ACB = 180° – 115° = 65°
OR
∠ACB = 180° – ∠ACD
∠ACB = 180° – 115° = 65°
∴ ∠B = 180° – 65° – 45° = 70°
- In a ∆ ABC, ∠A + ∠B = 75° and ∠B + ∠C = 140°, then ∠B is
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Given that , ∠A + ∠B = 75° and ∠B + ∠C = 140°
In ∆ ABC,
We know that , ∠A + ∠B + ∠C = 180°
∴ ∠B = (∠A + ∠B + ∠B + ∠C) – (∠A + ∠B + ∠C)Correct Option: B
Given that , ∠A + ∠B = 75° and ∠B + ∠C = 140°
In ∆ ABC,
We know that , ∠A + ∠B + ∠C = 180°
∴ ∠B = (∠A + ∠B + ∠B + ∠C) – (∠A + ∠B + ∠C)
∠B = 75° + 140° – 180° = 35°
