- In figure 5.22, ABCD is a parallelogram, P and Q are midpoints of side AB and DC…
- Using opposite angles test for parallelogram, prove that every rectangle is a…
- In figure 5.23, G is the point of concurrence of medians of ΔDEF. Take point H on ray…
- Prove that quadrilateral formed by the intersection of angle bisectors of all angles of…
- 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.
