ICSE /Class 10
Maths MCQ Based On Similarity
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ICSE Class 10 Maths Similarity
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area of PQR/area of DEF = 64/144 = PQ2/DE2
PQ/DE = 8/12 = 2/3
In AOB and COD,
AOB=COD ....vert. opp.angles
OAB=OCD......corresponding angles
So, by AA-criterion of similarity, we AOB ~ COD
⇒ Area of AOB/Area of COD = AB2/CD2 = (2DC)2/DC2 = 4/1
In OCD and OAB, we have
AOB=COD................(vert. opp. angles)
OAB=OCD.................( alternate angles)
So, OCD ~ OAB
Now, area of OCD/area of OAB = CD2/AB2 = CD2/(3CD)2 = 1/9
Use the Basic Proportionality theorem th find the missing sides.
Use Pythagoras Theorem
Use the Basic Proportionality theorem to find the ratio of sides.
Since ABCD is a square
AB=BC=CD=DA and AC= BC
Given, BCE ~ ACF
⇒ area of BCE/area of ACF = BC2/AC2=BC2/( BC)2=1/2
Area (ΔM) / Area (ΔN) = (h1)2 / (h2)2
(frac {Perimeter , of , triangle MNO}{Perimeter , of , triangle PQR} = frac {MN}{PQ})
basic proportionality theorem
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