Plane Geometry
- In ∆ ABC the straight line parallel to the side BC meets AB and AC at the points P and Q respectively. If AP = QC, the length of AB is 12 units and the length of AQ is 2 units, then the length (in units) of CQ is
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On the basis of question we draw a figure of triangle ABC ,
Here , AB = 12 units and AQ = 2 units
∠P = ∠B
∠Q = ∠C
∴ By AA — similarity,
∆ APQ ~ ∆ ABC∴ AP = AQ AB AC ⇒ AB = AC AP AQ ⇒ AB - 1 = AC - 1 = AC - AQ AP AQ AQ ⇒ AB - 1 = QC AP AQ ⇒ 12 - 1 = QC QC 2 ⇒ 12 - 1 = y y 2
Correct Option: A
On the basis of question we draw a figure of triangle ABC ,
Here , AB = 12 units and AQ = 2 units
∠P = ∠B
∠Q = ∠C
∴ By AA — similarity,
∆ APQ ~ ∆ ABC∴ AP = AQ AB AC ⇒ AB = AC AP AQ ⇒ AB - 1 = AC - 1 = AC - AQ AP AQ AQ ⇒ AB - 1 = QC AP AQ ⇒ 12 - 1 = QC QC 2 ⇒ 12 - 1 = y y 2 ⇒ 12 - y = y y 2
⇒ y² + 2y – 24 = 0
⇒ y² + 6y – 4y – 24 = 0
⇒ y (y + 6) – 4(y + 6) = 0
⇒ (y – 4) (y + 6) = 0
⇒ y = 4 because y ≠ –6
- ABC is a triangle in which DE || BC and AD : DB = 5 : 4. Then DE : BC is
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On the basis of question we draw a figure of triangle ABC in which DE || BC ,
Given that , AD : DB = 5 : 4.
DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
By AA – similarly,
∆ ADE ~ ∆ ABC∴ AB = BC AD DE ∴ DB = 4 AD 5 ⇒ DB + 1 = 4 + 1 AD 5 ⇒ DB + AD = 4 + 5 AD 5 ⇒ AB = 9 = BC AD 5 DE
Correct Option: D
On the basis of question we draw a figure of triangle ABC in which DE || BC ,
Given that , AD : DB = 5 : 4.
DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
By AA – similarly,
∆ ADE ~ ∆ ABC∴ AB = BC AD DE ∴ DB = 4 AD 5 ⇒ DB + 1 = 4 + 1 AD 5 ⇒ DB + AD = 4 + 5 AD 5 ⇒ AB = 9 = BC AD 5 DE ∴ DE = 5 = 5 : 9 BC 9
- In a right-angled triangle, the product of two sides is equal to half of the square of the third side i.e., hypotenuse. One of the acute angle must be
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On the basis of question we draw a figure of triangle ABC ,
AB × BC = AC² 2
⇒ AC² = 2 AB × BC
⇒ AB² + BC² = 2 AB × BC
Correct Option: C
On the basis of question we draw a figure of triangle ABC ,
AB × BC = AC² 2
⇒ AC² = 2 AB × BC
⇒ AB² + BC² = 2 AB × BC
⇒ (AB – BC)² = 0
⇒ AB = BC
∴ ∠BAC = ∠ACB = 45°
- I is the incentre of ∆ ABC and if ∠BAC =70°, then ∠BIC is
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As per the given in question , we draw a figure triangle ABC
BI and CI are bisectors of angles
∠B and ∠C.
Given , ∠BAC = 70°
∴ ∠ABC + ∠ACB = 180° – 70° = 110°
∴ 2∠IBC + 2 ∠ICB = 110°Correct Option: C
As per the given in question , we draw a figure triangle ABC
BI and CI are bisectors of angles
∠B and ∠C.
Given , ∠BAC = 70°
∴ ∠ABC + ∠ACB = 180° – 70° = 110°
∴ 2∠IBC + 2 ∠ICB = 110°
⇒ ∠IBC + ∠ICB = 55°
∴ ∠BIC = 180° – 55° = 125°
- In a triangle the length of the side opposite the angle which measures 45° is 8 cm, what is the length of the side opposite to the angle which measures 90°?
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On the basis of question we draw a figure of triangle ABC ,
According to the question,
AB = 8 cm.
∠ABC = 90°∴ sin 45° = AB AC
Correct Option: A
On the basis of question we draw a figure of triangle ABC ,
According to the question,
AB = 8 cm.
∠ABC = 90°∴ sin 45° = AB AC ⇒ 1 = 8 √2 AC
⇒ AC = 8√2 cm.
