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Maharashtra State BoardSSC (English Medium) 9th Standard

In the Figure, □ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = ABDC12(AB+DC).

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Question

In the Figure, `square`ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = `1/2 ("AB" + "DC")`.

Sum
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Solution

Given: `square`ABCD is a trapezium.

To prove: PQ || AB and PQ = `1/2`(AB + DC)

Construction: Extend line AQ in such a way that, on extending side DC, intersect it at point R.

Proof:

seg AB || seg DC      ...(Given)

and seg BC is their transversal.

∴ ∠ABC ≅ ∠RCB       ...(Alternate angles)

∴ ∠ABQ ≅ ∠RCQ       ...(i)   ...(B-Q-C)

In ∆ABQ and ∆RCQ,

∠ABQ ≅∠RCQ       ...[From (i)]

seg BQ ≅ seg CQ      ...(Q is the midpoint of seg BC)

∠BQA ≅ ∠CQR        ...(Vertically opposite angles)

∴ ∆ABQ ≅ ∆RCQ        ...(ASA test)

seg AB ≅ seg CR       ...(c.s.c.t.)  ...(ii)

seg AQ ≅ seg RQ      ...(c.s.c.t.)  ...(iii)

In ∆ADR,

Point P is the midpoint of line AD.       ...(Given)

Point Q is the midpoint of line AR.      ...[From (iii)]

∴ seg PQ || side DR        ...(Midpoint Theorem)

∴ seg PQ || side DC       ...(iv)    ...(D-C-R)

∴ side AB || side DC       ...(v)    ...(Given)

∴ seg PQ || side AB      ...[From (iv) and (v)]

PQ = `1/2` DR       ...(Midpoint Theorem)

= `1/2` (DC + CR)

= `1/2` (DC + AB)        ...[From (ii)]

∴ PQ = `1/2` (AB + DC)

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Chapter 5: Quadrilaterals - Problem Set 5 [Page 74]

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Balbharati Mathematics 2 [English] Standard 9 Maharashtra State Board
Chapter 5 Quadrilaterals
Problem Set 5 | Q 8 | Page 74

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