If AD is median of `DeltaABC` and P is a point on AC such that
`ar(DeltaADP):ar(DeltaABD)=4:5` then
`ar(DeltaPDC):ar(DeltaABC)` is
A
`1:10`
B
`10:1`
C
`1:3`
D
`3:1`
Text Solution
Verified by Experts
The correct Answer is:
A
Topper's Solved these Questions
AREAS OF PARALLELOGRAMS AND TRIANGLES
SCIENCE OLYMPIAD FOUNDATION |Exercise Achievers Section (HOTS) |2 Videos
CIRCLES
SCIENCE OLYMPIAD FOUNDATION |Exercise Achievers Section (HOTS) |3 Videos
Similar Questions
Explore conceptually related problems
AD is one of the medians of a ABC*X is any point on AD. Show that ar (ABX)=ar(ACX)
In DeltaABC,AD is the bisector of angleA . Then , (ar(DeltaABD))/(ar(DeltaACD))=
In triangle ABC ,AD is a median and P is a point on AD such that AP : PD = 3 : 4. Then ar (triangle BPD): ar (triangle ABC) is equal to: त्रिभुज ABC में, AD एक माध्यिका है और P, AB पर स्थित ऐसा बिंदु है कि AP : PD = 3 : 4 है| ar (triangle BPD): ar (triangle ABC) ज्ञात करें |
In triangle ABC , AD is median and P is the point on AD such that AP : PD = 3 : 4, then ar(triangle APB) :ar(triangle ABC) is equal to : त्रिभुज ABC में, AD मध्यिका है तथा P, AD पर स्थित ऐसा बिंदु है कि AP : PD = 3 : 4 है, तो ar(triangle APB) :ar(triangle ABC) का मान किसके बराबर होगा ?
In DeltaABC , P is a point on BC such that BP : PC = 2 : 3 and Q is the midpoint of BP. Then ar (DeltaABQ): ar(DeltaABC) is equal to :
If DeltaABC and DeltaDEF are two tnangles such that (AB)/(EF)=(BC)/(FD)=(CA)/(DE)=3/4 then find ar(DeltaDEF): ar(DeltaABC) .
A point D is taken on the side BC of a DeltaABC such that BD = 2DC. Prove that ar(DeltaABD) = 2 ar (DeltaADC)
In the DeltaABC , MN||BC and AM : : MB = 1/3 . Then, (ar(DeltaAMN))/(ar(DeltaABC)) =?
SCIENCE OLYMPIAD FOUNDATION -AREAS OF PARALLELOGRAMS AND TRIANGLES -Achievers Section (HOTS)