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Triangles

Short Answer Type

171. In the figure given below, ABCD is a quadrilateral with AC = AD and AB bisects ∠A. Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

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172.

In figure, AB ⊥ AE, BC ⊥ AB, CE = DE and ∠AED = 120°. Find
(a)     ∠EDC
(b)     ∠DEC
(c)    Hence prove that EDC is an equilateral triangle.

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Fill In the Blanks

173. All circles are __________(congruent, similar)

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174. All squares are __________ (similar, congruent)

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175. All _________ triangles are similar,(isosceles, equilateral)

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176. Two polygons of the same number of sides are similar, if (a) their corresponding angles are ________ and (b) their corresponding sides are _________(equal, proportional).

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Short Answer Type

177.

Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.

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178. State whether the following quadrilaterals are similar or not:

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179. In Fig. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

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Long Answer Type

180.
E and F are points on the sides PQ and PR respectively of a ∆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.
(ii) PE = 4 cm, QE = 4.5 cm. PF = 8 cm and RF = 9 cm.
(iii) PQ = 1.28 cm. PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

(i) We have,

space PE over EQ equals fraction numerator 3.9 over denominator 3 end fraction equals fraction numerator 1.3 over denominator 1 end fraction space space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis space PF over FR equals fraction numerator 3.6 over denominator 2.4 end fraction equals 3 over 2 equals fraction numerator 1.5 over denominator 1 end fraction space space space space space space... left parenthesis ii right parenthesis

From (i) and (ii), we have

PE over EQ not equal to space PF over FR

Therefore, EF is not parallel to QR [By using converse of Basic proportionality theorem]
(ii) We have,

space PE over QE equals fraction numerator 4 over denominator 4.5 end fraction equals 40 over 45 equals 8 over 9 space space space space... left parenthesis straight i right parenthesis space PF over RF space equals space 8 over 9 space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis

From (i) and (ii), we have

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Therefore,

EF parallel to QR

[Using converse of Basic proportionality theorem]

(iii) We have,

PE over PQ equals fraction numerator 0.18 over denominator 1.28 end fraction equals 18 over 128 equals 9 over 64 space space space space... left parenthesis straight i right parenthesis PF over PR equals fraction numerator 0.36 over denominator 2.56 end fraction equals 36 over 256 equals 9 over 64 space space space space space... left parenthesis ii right parenthesis

From (i) and (ii), we have

PE over PQ equals PF over PR

Therefore,

EF parallel to QR

[Using converse of Basic proportionality theorem]

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