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Practice Set 5.2, Chapter 5 - Quadrilaterals Mathematics Part II Solutions for Class 9 Math

Practice Set 5.2
  1. In figure 5.22, 􀀍ABCD is a parallelogram, P and Q are midpoints of side AB and DC…
  2. Using opposite angles test for parallelogram, prove that every rectangle is a…
  3. In figure 5.23, G is the point of concurrence of medians of ΔDEF. Take point H on ray…
  4. Prove that quadrilateral formed by the intersection of angle bisectors of all angles of…
  5. In figure 5.25, if points P, Q, R, S are on the sides of parallelogram such that AP =…
Practice Set 5.2
Question 1.

In figure 5.22, 􀀍ABCD is a parallelogram, P and Q are midpoints of side AB and DC respectively, then prove 􀀍APCQ is a parallelogram.



Answer:

Given AB ∥ to DC and AB = DC as ABCD is ∥gram.

⇒ AP ∥CQ (parts of ∥ sides are ∥) & 1/2 AB = 1/2 DC


⇒ AP = QC (P and Q are midpoint of AB and DC respectively)


⇒ AP = PB and DQ = QC


Hence APCQ is a parallelogram as the pair of opposite sides is = and ∥.




Question 2.

Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.


Answer:

Opposite angle property of parallelogram says that the opposite angles of a parallelogram are congruent.

Given a rectangle which had at least one angle as 90°.



If ∠ A is 90° and AD = BC (opposite sides of rectangle are ∥ and =)


AB is transversal


⇒ ∠ A + ∠B = 180 (angles on the same side of transversal is 180°)


But ∠B + ∠C is 180 (AD ∥ BC, opposite sides of rectangle)


⇒ ∠ A = ∠C = 90°


Since opposite ∠s are equal this rectangle is a parallelogram too.



Question 3.

In figure 5.23, G is the point of concurrence of medians of ΔDEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that 􀀍GEHF is a parallelogram.



Answer:

Given G is the point of concurrence of medians of Δ DEF so the medians are divided in the ratio of 2:1 at the point of concurrence. Let O be the point of intersection of GH AND EF.

The figure is shown below:



⇒ DG = 2 GO


But DG = GH


⇒ 2 GO = GH


Also DO is the median for side EF.


⇒ EO = OF


Since the two diagonals bisects each other


⇒ GEHF is a ∥gram.


Question 4.

Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle. (Figure 5.24)



Answer:

Given ABCD is a parallelogram

AR bisects ∠BAD, DP bisects ∠ADC , CP bisects ∠BCD and BR bisects ∠CBA


∠BAD + ∠ABC = 180° (adjacent ∠s of parallelogram are supplementary)


But 1/2 ∠BAD = ∠BAR


1/2 ∠ABC = ∠RBA


∠BAR + ∠RBA = 1/2 × 180° = 90°


⇒ Δ ARB is right angled at ∠ R since its acute interior angles are complementary.


Similarly Δ DPC is right angled at ∠ P and


Also in Δ COB , ∠BOC = 90° ⇒ ∠POR = 90° (vertically opposite angles)


Similarly in ΔADS , ∠ASD = 90° = ∠PSR (vertically opposite angles)


Since vertically opposite angles are equal and measures 90° the quadrilateral is a rectangle.



Question 5.

In figure 5.25, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS then prove that 􀀍PQRS is a parallelogram.



Answer:

Given ABCD is a parallelogram so

AD = BC and AD ∥BC


and DC = AB and DC ∥ AB


also AP = BQ = CR = DS


⇒ AS = CQ and PB = DR


in ΔAPS and Δ CRQ


∠ A = ∠C (opposite ∠s of a parallelogram are congruent)


AS = CQ


AP = CR


ΔAPS ≅ Δ CRQ( SAS congruence rule)


⇒ PS = RQ (c.p.c.t.)


Similarly PQ= SR


Since both the pair of opposite sides are equal


PQRS is ∥gram.