Plane Geometry
- In a ∆ ABC, D and E are points on AC and BC respectively, AB and DE are perpendicular to BC. If AB = 9 cm , DE = 3 cm and AC = 24 cm, then AD is :
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According to question , we draw a figure triangle ABC
In ∆ ABC and ∆ DEC
AB || DE
Given that , AB = 9 cm , DE = 3 cm and AC = 24 cm
∴ ∠ABC = ∠DEC = 90°
∠CAB = ∠CDE
By AA-similarity theorem,
∆ ABC ~ ∆DEC∴ AB = AC DE CD ⇒ 9 = 24 3 CD ⇒ 3 = 24 CD
Correct Option: B
According to question , we draw a figure triangle ABC
In ∆ ABC and ∆ DEC
AB || DE
Given that , AB = 9 cm , DE = 3 cm and AC = 24 cm
∴ ∠ABC = ∠DEC = 90°
∠CAB = ∠CDE
By AA-similarity theorem,
∆ ABC ~ ∆DEC∴ AB = AC DE CD ⇒ 9 = 24 3 CD ⇒ 3 = 24 CD ⇒ CD = 24 = 8 cm. 3
∴ AD = AC – CD = 24 – 8 = 16 cm
- The lengths of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The size of angle C is 90°. Find the approximate length of side AC.
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As per the given in question , we draw a figure right angle triangle ACB
∠C = 90°
Here , AB = 12 cm., BC = 8 cm.
∴ AC = √AB² - BC²Correct Option: B
As per the given in question , we draw a figure right angle triangle ACB
∠C = 90°
Here , AB = 12 cm., BC = 8 cm.
∴ AC = √AB² - BC²
AC = √12² - 8²
AC = √144 - 64 = √80 ≈ 9 cm.
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In ∆ABC, DE || BC such that AD = 3 . If AC = 5.6 cm., then AE is equal to BD 5
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According to question , we draw a figure ∆ABC
Given that , AD = 3 BD 5
DE||BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
By AA – similarity,
∆ADE ≅ ∆ABCAB = AC AD AE ⇒ AB - 1 = AC - 1 AD AE ⇒ AB - AD = AC - AE AD AE ⇒ BD = AC - AE AD AE ⇒ 5 = 5.6 - AE 3 AE
Correct Option: D
According to question , we draw a figure ∆ABC
Given that , AD = 3 BD 5
DE||BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
By AA – similarity,
∆ADE ≅ ∆ABCAB = AC AD AE ⇒ AB - 1 = AC - 1 AD AE ⇒ AB - AD = AC - AE AD AE ⇒ BD = AC - AE AD AE ⇒ 5 = 5.6 - AE 3 AE
⇒ 5AE = 16.8 – 3AE
⇒ 5AE + 3AE = 16.8
⇒ 8AE = 16.8⇒ AE = 16.8 = 2.1 cm. 8
- In a triangle PQR, PQ = PR and ∠Q is twice that of ∠P. Then ∠Q is equal to
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On the basis of question we draw a figure of triangle PQR ,
PQ = PR
∴ ∠PQR is an isosceles triangle.
∴ ∠PQR = ∠PRQ and ∠PQR = 2∠QPR
In ∆ PQR,
We know that ,∠P + ∠Q + ∠R = 180°⇒ ∠Q + ∠Q + ∠Q = 180° 2 ⇒ 5 ∠Q = 180° 2
Correct Option: A
On the basis of question we draw a figure of triangle PQR ,
PQ = PR
∴ ∠PQR is an isosceles triangle.
∴ ∠PQR = ∠PRQ and ∠PQR = 2∠QPR
In ∆ PQR,
We know that ,∠P + ∠Q + ∠R = 180°⇒ ∠Q + ∠Q + ∠Q = 180° 2 ⇒ 5 ∠Q = 180° 2 ∴ ∠Q = 180° × 2 = 72° 5
- A point P lying inside a triangle is equidistant from the vertices of the triangle. Then the triangle has P as its
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As per the given in question ,
The right bisectors of the sides of a triangle meet at the point called circumcentre.Correct Option: D
As per the given in question ,
The right bisectors of the sides of a triangle meet at the point called circumcentre. It is equidistant from the vertices of the triangle
