Plane Geometry
- If angle bisector of a triangle bisect the opposite side, then what type of triangle is it ?
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As per the given in question , we draw a figure of an isosceles triangle ABC
AB = AC
BD = DCCorrect Option: D
As per the given in question , we draw a figure of an isosceles triangle ABC
AB = AC
BD = DC
∠ADB = 90°
The triangle will be either isosceles or equilateral.
- In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of the half of the vertex angle of the triangle is
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On the basis of given in question , we draw a figure of an isosceles triangle ABC ,
In ∆ ABC,
AB = AC
∴ ∠ABC = ∠ACB = y°
∴ ∠BAC = 4y
As we know that , ∠A + ∠B + ∠C = 180°
∴ 4y + y + y = 180°
⇒ 6y = 180°
⇒ y = 30°Correct Option: A
On the basis of given in question , we draw a figure of an isosceles triangle ABC ,
In ∆ ABC,
AB = AC
∴ ∠ABC = ∠ACB = y°
∴ ∠BAC = 4y
As we know that , ∠A + ∠B + ∠C = 180°
∴ 4y + y + y = 180°
⇒ 6y = 180°
⇒ y = 30°
∴ ∠BAC = 4 × 30° = 120°∠BAD = 1 ∠BAD = 60° 2
- ∆ABC is an isosceles triangle with AB = AC = 10 cm, AD = 8 cm is the median on BC from A. The length of BC is
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We draw a figure of an isosceles triangle ABC with AB = AC = 10 cm, AD = 8 cm is the median on BC from A ,
D, is the mid–point of BC.
Given that , AB = AC = 10 cm.
AD ⊥ BC
From ∆ ABD,
BD = √AB² - AD²
BD = √10² - 8²
BD = √100 - 64Correct Option: B
We draw a figure of an isosceles triangle ABC with AB = AC = 10 cm, AD = 8 cm is the median on BC from A ,
D, is the mid–point of BC.
Given that , AB = AC = 10 cm.
AD ⊥ BC
From ∆ ABD,
BD = √AB² - AD²
BD = √10² - 8²
BD = √100 - 64
BD = √36 = 6 cm.
∴ BC = 2BD = 2 × 6 = 12 cm.
- ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 cm and BC = 24 cm then the radius of circle is
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On the basis of given in question , we draw a figure of an isosceles triangle ABC inscribed in a circle ,
AD ⊥ BC
Given , BC = 24 cm
BD = DC = 12 cm.
OC = OA = Circum-radius = r cm.
AD = √AB² - BD²
AD = √(12√5)² - (12)²
AD = √144 × 5 = 144
AD = √144(5 –1)
AD = √144 × 4 = 24 cm.
In ∆OCD,
OD = (24 – r) cm.Correct Option: B
On the basis of given in question , we draw a figure of an isosceles triangle ABC inscribed in a circle ,
AD ⊥ BC
Given , BC = 24 cm
BD = DC = 12 cm.
OC = OA = Circum-radius = r cm.
AD = √AB² - BD²
AD = √(12√5)² - (12)²
AD = √144 × 5 = 144
AD = √144(5 –1)
AD = √144 × 4 = 24 cm.
In ∆OCD,
OD = (24 – r) cm.
∴ OC² = OD² + CD²
⇒ r² = (24 – r)² + 12²
⇒ r² = 576 – 48r + r² + 144
⇒ 48r = 720
⇒ r = 720 ÷ 48 = 15 cm.
- In a ∆ ABC, DE || BC. D and E lie on AB and AC respectively. If AB = 7 cm and BD = 3 cm, then find BC : DE.
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We draw a figure triangle ABC in which DE || BC. D and E lie on AB and AC respectively,
DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
∴ By AA - similarity theorem,
∆ ADE ~ ∆ ABCAB = BC AD DE ⇒ AB = BC AB - BD DE ⇒ 7 = BC 7 - 3 DE
Correct Option: C
We draw a figure triangle ABC in which DE || BC. D and E lie on AB and AC respectively,
DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
∴ By AA - similarity theorem,
∆ ADE ~ ∆ ABCAB = BC AD DE ⇒ AB = BC AB - BD DE ⇒ 7 = BC 7 - 3 DE ⇒ BC = 7 = 3.5 = 3.5 : 2 DE 4 2
