Plane Geometry
- In ∆ ABC, AD ⊥ BC and AD² = BD.DC. The measure of ⊥BAC is:
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According to question , we draw a figure triangle ABC
In right angled ∆ ABD and ∆ ADC,
AB² = AD² + BD² and, AC² = AD² + DC²
On adding, we get
AB² + AC² = 2AD² + BD² + CD²
⇒ AB² + AC² = 2BD × CD + BD² + CD²
[∵ AD² = BD × CD]Correct Option: C
According to question , we draw a figure triangle ABC
In right angled ∆ ABD and ∆ ADC,
AB² = AD² + BD² and, AC² = AD² + DC²
On adding, we get
AB² + AC² = 2AD² + BD² + CD²
⇒ AB² + AC² = 2BD × CD + BD² + CD²
[∵ AD² = BD × CD]
⇒ AB² + AC² = (BD + CD)² = BC²
∆ ∠BAC = 90°
- ⊥A of ∆ ABC is a right angle. AD is perpendicular on BC. If BC = 14 cm and BD = 5 cm, then measure of AD is :
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In ∆ ABC and ∆ DAC,
∠ADC = ∠BAC , ∠C = ∠C
By AA – similarity criterion,
∆ ABC ~ ∆ DAC∴ AB = BC = AC DA AC DC ⇒ CB = AC CA CD
⇒ CA² = CB × CD
= BC (BC – BD)
= 14 × (14 – 5) = 14 × 9
= 126 sq. cm.
∴ From ∆ ADC
CA² = AD² + C’D²
Correct Option: C
In ∆ ABC and ∆ DAC,
∠ADC = ∠BAC , ∠C = ∠C
By AA – similarity criterion,
∆ ABC ~ ∆ DAC∴ AB = BC = AC DA AC DC ⇒ CB = AC CA CD
⇒ CA² = CB × CD
= BC (BC – BD)
= 14 × (14 – 5) = 14 × 9
= 126 sq. cm.
∴ From ∆ ADC
CA² = AD² + C’D²
⇒ 126 = AD² + 9²
⇒ AD² = 126 – 81 = 45
⇒ AD = √45 = 3√5cm.
- In ∆ ABC, ∠B = 90°, AB = 8 cm and BC = 15 cm, then sinC = ?
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On the basis of question we draw a figure of triangle ABC ,
Given that , ∠B = 90°, AB = 8 cm and BC = 15 cm
∴ AC = √AB² + BC²
AC = √8² + 15²
AC = √64 + 225
AC = √289 = 17 cm.Correct Option: B
On the basis of question we draw a figure of triangle ABC ,
Given that , ∠B = 90°, AB = 8 cm and BC = 15 cm
∴ AC = √AB² + BC²
AC = √8² + 15²
AC = √64 + 225
AC = √289 = 17 cm.∴ sin c = AB = 8 AC 17
- In ∆ ABC, AB = BC = k, AC = √2k, then ∆ ABC is a :
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According to question ,
AB = BC = k, AC = √2k
Now , AB² + BC² = k² + k² = 2k² = AC²Correct Option: B
According to question ,
AB = BC = k, AC = √2k
Now , AB² + BC² = k² + k² = 2k² = AC²
∴ ∆ ABC is a right angled triangle.
- The sides of a right triangle ABC are a, b and c, where c is the hypotenuse. What will be the radius of the in circle of this triangle?
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As per the given in question , we draw a figure right-angled triangle ACB
In radius = r units
Area of (∆OAC + ∆OBC + ∆OAB) = Area of ∆ABC⇒ 1 br + 1 ar + 1 cr = 1 ab 2 2 2 2
r (a + b + c) = ab⇒ r = ab . . . . . (i) a + b + c
In right angled triangle ∆ABC,
a² + b² = c²
⇒ (a + b)² – 2ab = c²
⇒ (a + b)² – c² = 2ab
⇒ (a + b + c) (a + b – c) = 2ab
∴ From equation (i),r = ab a + b + c
Correct Option: B
As per the given in question , we draw a figure right-angled triangle ACB
In radius = r units
Area of (∆OAC + ∆OBC + ∆OAB) = Area of ∆ABC⇒ 1 br + 1 ar + 1 cr = 1 ab 2 2 2 2
r (a + b + c) = ab⇒ r = ab . . . . . (i) a + b + c
In right angled triangle ∆ABC,
a² + b² = c²
⇒ (a + b)² – 2ab = c²
⇒ (a + b)² – c² = 2ab
⇒ (a + b + c) (a + b – c) = 2ab
∴ From equation (i),r = ab a + b + c r = (a + b + c) (a + b - c) 2(a + b + c) r = a + b - c 2
