Ex – 6.3 Que No. 5
Ex – 6.3 Que No. 5 Q.5) S and T are points on the PR and QR sides of the △PQR, such that ∠P = ∠RTS. Show that △RPQ ~ △RTS. Solution:-
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Ex – 6.3 Que No. 5 Q.5) S and T are points on the PR and QR sides of the △PQR, such that ∠P = ∠RTS. Show that △RPQ ~ △RTS. Solution:-
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Ex – 6.3 Que No. 4 Q.4) In Fig. 6.36, QR/ QS = QT/ PR = and ∠1 = ∠2. Show that △PQS ~ △TQR. Solution:-
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Ex – 6.3 Que No. 3 Q.3) Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD. Solution:-
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Ex – 6.3 Que No. 2 Q.2) In Fig. 6.35, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB. Solution:-
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Ex – 6.3 Que No. 1 Q.1) State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (i). Solution:- (ii). Solution:- (iii). Solution:- (iv). Solution:- (v). Solution:- (vi). Solution:-
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Ex – 6.2 Que No. 2 Q.2) E and F are points on the sides PQ and PR respectively of a D PQR. For each of the following cases, state whether EF || QR : (i). PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm Solution:- (ii).
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Ex – 6.2 Que No. 1 Q.1) In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i). DE || BC. Find EC Solution:- (ii). DE || BC. Find AD Solution:-
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Ex – 6.1 Que No. 3 Q.3) State whether the following quadrilaterals are similar or not: Solution:-
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Ex – 6.1 Que No. 2 Q.2) Give two different examples of pair of (i). Similar figures are :- Solution:- (a). Any two squares are :- (b). Any two rectangles are :- (ii). Non – similar figures are :- Solution:- (a). Circle and Triangle are :- (b). Rectangle and Square are :-
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Ex – 6.1 Que No. 1 Q.1) Fill in the blanks using the correct word given in brackets : (i). All circles are _________ (congruent, similar). Solution:- Both circles have the same shape but not the same size because all the circles have the same shape but the same size is not necessary. Hence all
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