ICSE /Class 10
Maths MCQ Based On Similarity
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ICSE Class 10 Maths Similarity
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∠ B = ∠ C [Angles opposite to equal sides are equal]
In ABC ~ DEF,
AB/DE = BC/EF = 2BP/2EQ = BP/EQ ...............(i)
In triangles APB and DQE, we have
AB/DE = BP/EQ and B=E
So, by SAS-criterion of similarity, we have
APB ~ DQE
So, we have BP/EQ = AP/DQ ..................(ii)
From (i) and (ii),
AB/DE = AP/DQ
Therefore, area of ABC/area of DEF = AB2/DE2 = AP2/DQ2
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
Use the Basic Proportionality Theorem.
Draw AL BC and DN BC.
In ALO and DNO, we have
ALO =DNO = 90° and AOL = DON
Therefore, ALO ~ DNO
⇒ AL/DN = AO/DO
So, area of ABC/area of DBC = {(1/2)BC AL}/{(1/2)BC DN}
= AL/DN =AO/DO
P=B=90°
AOB=POQ
So, by AA- criterion of similarity,
AOB is similar to POQ
So, area of AOB/area of POQ =OA2/OQ2
AREA OF POQ=(100/49) 490 = 1,000cm2
Use similarity of triangles
Since DE || QR, PD/QD=PE/ER and hence â–³ PQR ∼ â–³ PDE
DE = QC/2 only if D and E are the mid points of PQ and PR respectively. So this may not be true always.
Use similarity of triangles
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