Plane Geometry
- A ship after sailing 12 km towards south from a particular place covered 5 km more towards east. Then the straightway distance of the ship from that place is
-
View Hint View Answer Discuss in Forum
According to question , we draw a figure triangle ABC
From right angle triangle , we have
∴ OB = √OA² + AB²
OB = √12² + 5²
OB = √144 + 25Correct Option: D
According to question , we draw a figure triangle ABC
From right angle triangle , we have
∴ OB = √OA² + AB²
OB = √12² + 5²
OB = √144 + 25
OB = √169 = 13 km.
- In ∆ ABC, two points D and E are taken on the lines AB and BC respectively in such a way that AC is parallel to DE. Then ∆ ABC and ∆ DBE are
-
View Hint View Answer Discuss in Forum
According to question , we draw a figure triangle ABC in which two points D and E are taken on the lines AB and BC respectively in such a way that AC is parallel to DE ,
In ∆ ABC and ∆ DBE,
DE || AC
∴ ∠BAC = ∠BDE
∠BCA = ∠BED
∴ By AA–similarityCorrect Option: C
According to question , we draw a figure triangle ABC in which two points D and E are taken on the lines AB and BC respectively in such a way that AC is parallel to DE ,
In ∆ ABC and ∆ DBE,
DE || AC
∴ ∠BAC = ∠BDE
∠BCA = ∠BED
∴ By AA–similarity
∆ ABC ~ ∆ DBE
From above it is clear that ∆ ABC and ∆ DBE are congruent triangles .
- Inside a triangle ABC, a straight line parallel to BC intersects AB and AC at the point P and Q respectively. If AB = 3 PB, then PQ : BC is
-
View Hint View Answer Discuss in Forum
As per the given in question , we draw a figure triangle ABC
Let PB = x ⇒ AB = 3x and AP = 3x – x = 2x
In ∆ ABC, PQ || BC
⇒ ∆ APQ ~ ∆ ABCCorrect Option: D
As per the given in question , we draw a figure triangle ABC
Let PB = x ⇒ AB = 3x and AP = 3x – x = 2x
In ∆ ABC, PQ || BC
⇒ ∆ APQ ~ ∆ ABC⇒ AP = PQ = 2x = 2 = 2 : 3 AB BC 3x 3
- In ∆ ABC, D and E are points on AB and AC respectively such that DE || BC and DE divides the ∆ ABC into two parts of equal areas. Then ratio of AD and BD is
-
View Hint View Answer Discuss in Forum
On the basis of question we draw a figure of triangle ABC in which D and E are points on AB and AC respectively such that DE || BC ,
DE ||BC
∠ADE = ∠ABC
∠AED = ∠ACB
∴ ∆ADE ~ ∆ABCNow, ☐ BDEC = 1 ∆ ADE 1
[DE divides ∆ into two equal parts]⇒ ☐ BDEC + 1 = 1 + 1 ∆ ADE ⇒ ∆ ABC = 2 = AB² ∆ ADE AD² ⇒ AB = √2 AD ⇒ AB - 1 = √2 - 1 AD ⇒ BD = √2 - 1 AD
Correct Option: B
On the basis of question we draw a figure of triangle ABC in which D and E are points on AB and AC respectively such that DE || BC ,
DE ||BC
∠ADE = ∠ABC
∠AED = ∠ACB
∴ ∆ADE ~ ∆ABCNow, ☐ BDEC = 1 ∆ ADE 1
[DE divides ∆ into two equal parts]⇒ ☐ BDEC + 1 = 1 + 1 ∆ ADE ⇒ ∆ ABC = 2 = AB² ∆ ADE AD² ⇒ AB = √2 AD ⇒ AB - 1 = √2 - 1 AD ⇒ BD = √2 - 1 AD ⇒ AD = 1 BD √2 - 1
∴ AD : BD = 1 : 2 √2
- In ∆ ABC, DE || AC. D and E are two points on AB and CB respectively. If AB = 10 cm and AD = 4 cm, then BE : CE is
-
View Hint View Answer Discuss in Forum
According to question , we draw a figure of triangle ABC
Given , AB = 10 cm and AD = 4 cm
DE || AC
∆ BDE ~ ∆ BAC⇒ BD = BE DA EC
Correct Option: D
According to question , we draw a figure of triangle ABC
Given , AB = 10 cm and AD = 4 cm
DE || AC
∆ BDE ~ ∆ BAC⇒ BD = BE DA EC ⇒ 6 = BE 4 EC
⇒ BE : CE = 3 : 2
