Plane Geometry
- Incentre of ∆ ABC is I. ∠ABC = 90° and ∠ACB = 70°. ∠AIC is
-
View Hint View Answer Discuss in Forum
As per the given in question , we draw a figure triangle ABC whose I is incentre ,
The point of intersection of internal bisectors of the angles of a triangle is incentre.
Given that , ∠ABC = 90° and ∠ACB = 70°
As we know that , ∠BAC + ∠ABC + ∠ACB = 180°
∠BAC = 180° – 90° – 70° = 20°Correct Option: E
As per the given in question , we draw a figure triangle ABC whose I is incentre ,
The point of intersection of internal bisectors of the angles of a triangle is incentre.
Given that , ∠ABC = 90° and ∠ACB = 70°
As we know that , ∠BAC + ∠ABC + ∠ACB = 180°
∠BAC = 180° – 90° – 70° = 20°∴ ∠AIC = 180 - ∠A - ∠C 2 2
∠AIC = 180° –10° – 35° = 135°
- If in ∆ ABC, DE||BC, AB = 7.5 cm, BD = 6 cm. and DE = 2cm, then the length of BC in cm is :
-
View Hint View Answer Discuss in Forum
According to question , we draw a figure
Here , AB = 7.5 cm, BD = 6 cm. and DE = 2cm
In ∆ ADE and ∆ ABC,
∵ DE | | BC
∵ ∠D = ∠B ; ∠E = ∠C
∴ By AA - similarity,
∆ ADE ~ ∆ ABC∴ AD = DE AB BC ⇒ AB - BD = DE AB BC ⇒ 7.5 - 6 = 2 7.5 BC ⇒ 1.5 = 2 7.5 BC
Correct Option: C
According to question , we draw a figure
Here , AB = 7.5 cm, BD = 6 cm. and DE = 2cm
In ∆ ADE and ∆ ABC,
∵ DE | | BC
∵ ∠D = ∠B ; ∠E = ∠C
∴ By AA - similarity,
∆ ADE ~ ∆ ABC∴ AD = DE AB BC ⇒ AB - BD = DE AB BC ⇒ 7.5 - 6 = 2 7.5 BC ⇒ 1.5 = 2 7.5 BC ⇒ 1 = 2 ⇒ BC = 2 × 5 = 10 cm. 5 BC
- In case of an acute angled triangle, its orthocentre lies
-
View Hint View Answer Discuss in Forum
According to question ,
As we know that the orthocentre of an acute angled triangle lies inside the triangle.Correct Option: A
According to question ,
As we know that the orthocentre of an acute angled triangle lies inside the triangle. Hence , option A is correct answer .
- In a triangle PQR, the side QR is extended to S. ∠QPR = 72° and ∠PRS = 110°, then the value of ∠PQR is :
-
View Hint View Answer Discuss in Forum
As per the given in question , we draw a figure triangle ABC in which the side QR is extended to S ,
Given that , ∠QPR = 72° and ∠PRS = 110°,
In exterior angle ,
∠PQR + ∠QPR = ∠PRS = 110
∴ ∠PRQ = 180° – 110° = 70°Correct Option: A
As per the given in question , we draw a figure triangle ABC in which the side QR is extended to S ,
Given that , ∠QPR = 72° and ∠PRS = 110°,
In exterior angle ,
∠PQR + ∠QPR = ∠PRS = 110
∴ ∠PRQ = 180° – 110° = 70°
∴ ∠PQR = 180° – ∠QPR – ∠PRQ
∠PQR = 180° – 72° – 70° = 38°
- In ∆ ABC, ∠B = 70° and ∠C = 60°. The internal bisectors of the two smallest angles of ∆ ABC meet at O. The angle so formed at O is
-
View Hint View Answer Discuss in Forum
We draw a figure triangle ABC in which the internal bisectors of the two smallest angles of ∆ ABC meet at O ,
Given , ∠B = 70° ; ∠C = 60°
We know that , ∠A + ∠B + ∠C = 180°
∴ ∠A = 180° – 70° – 60° = 50°
According to the question,
∠OAC = 25° ;Correct Option: A
We draw a figure triangle ABC in which the internal bisectors of the two smallest angles of ∆ ABC meet at O ,
Given , ∠B = 70° ; ∠C = 60°
We know that , ∠A + ∠B + ∠C = 180°
∴ ∠A = 180° – 70° – 60° = 50°
According to the question,
∠OAC = 25° ;
∠OCA = 30°
∴ ∠AOC = 180° – 25° – 30° = 125°
