Chapter 7.2, Problem 17PSC

a

To determine

To find:The most descriptive name of the figure formed by connecting consecutive midpoints of the sides of the rhombus

Answer to Problem 17PSC

The figure formed by joining consecutive midpoints of rhombus will be rectangle

Explanation of Solution

Given information:

A rhombus whose midpoints are joined

Let ABCD be rhombus and P, Q, R and S are the midpoints of sides AB, BC, CD and DA.

  

In triangles ABD and BDC, we have

  $\begin{array}{l}\text{SP}\parallel \text{BDand SP}=\frac{1}{2}\text{BD}\mathrm{....}\text{by midpoint theorem}\mathrm{....}\text{(i)}\\ \text{RQ}\parallel \text{BDand RQ}=\frac{1}{2}\text{BD}\mathrm{....}\text{by midpoint theorem}\mathrm{....}\text{(ii)}\end{array}$

From (i) and (ii), we have

  $\text{SP}\parallel \text{RQ}$

  $\Rightarrow$ PQRS is a parallelogram

Further, since diagonals of a rhombus bisect each other at right angles

  $\text{AC}\parallel \text{BD}$

Since, SP $\parallel$ AD, PQ $\parallel$ AC and AC $\perp$ BD

  $\begin{array}{l}\Rightarrow \text{SP}\perp \text{PQ}\\ \Rightarrow \angle \text{QPS=90}°\\ \end{array}$

  $\therefore$ PQRS is a rectangle

b

To determine

To find: The most descriptive name of the figure formed by connecting consecutive midpoints of the sides of the kite

Answer to Problem 17PSC

The figure formed by joining consecutive midpoints of kite will be rectangle

Explanation of Solution

Given information:

A kite whose midpoints are joined

Let ABCD be kite and P, Q, R and S are the midpoints of sides AB, BC, CD and DA.

Join BD and AC.

  

Now, In isosceles triangle ABD, P is the midpoint of AB and Q is the midpoint of AD, so we have

  $\text{PQ}=\frac{1}{2}\text{BD and PQ}\parallel \text{BD}$

Similarly, in triangle BCD,

  $\text{RS}=\frac{1}{2}\text{BD and RS}\parallel \text{BD}$ and

Hence,

  $\text{PQ}=\text{ RS and PQ}\parallel \text{RS}$

Similarly, we can prove that

  $\text{PR}=\text{ QS and PR}\parallel \text{QS}$

Now, using SSS postulate, we can prove that

  $\Delta \text{ABC}\cong \Delta \text{DAC}$

Therefore, AC is a bisector of isosceles triangle ABD and

Hence $\text{AC}\perp \text{BD and PQ}\perp \text{PR}$

Hence, PQRS is a rectangle

c

To determine

To find:The most descriptive name of the figure formed by connecting consecutive midpoints of the sides of the square

Answer to Problem 17PSC

The figure formed by joining consecutive midpoints of rhombus will be square

Explanation of Solution

Given information:

A square whose midpoints are joined

Let ABCD be square and P, Q, R and S are the midpoints of sides AB, BC, CD and DA.

  $\Rightarrow \text{AB}=\text{BC}=\text{CD}=\text{DA}$ ….. sides of square are equal

  

In triangle ADC, we have

  $\text{SR}\parallel \text{ACand SR}=\frac{1}{2}\text{AC}\mathrm{....}\text{by midpoint theorem}\mathrm{....}\text{(i)}$

In triangle ABC, we have

  $\text{PQ}\parallel \text{ACand PQ}=\frac{1}{2}\text{AC}\mathrm{....}\text{by midpoint theorem}\mathrm{....}\text{(i)}$

From (i) and (ii), we have

  $\text{SR}\parallel \text{PQ and SR}=\text{PQ}=\frac{1}{2}\text{AC}\mathrm{......}\text{(iii)}$

Similarly,

  $\begin{array}{l}\text{SP}\parallel \text{BD and BD}\parallel \text{RQ}\\ \Rightarrow \text{SP}\parallel \text{RQ and SP=}\frac{1}{2}\text{BD}\\ \text{and RQ}=\frac{1}{2}\text{BD}\end{array}$

Since diagonals of a square bisect each other at right angle

  $\Rightarrow$ AC = BD

  $\Rightarrow$

  $\text{SP}=\text{RQ}=\frac{1}{2}\text{AC}\mathrm{.....}\text{(iv)}$

From (iii) and (iv), we have

PQ = QR = RS = SP

Since diagonals of a square bisect each other at right angles

  $\Rightarrow \angle \text{EOF}=90°$

  $\begin{array}{l}\text{Now,}\\ \text{RQ}\parallel \text{DB,RE}\parallel \text{FO}\\ \text{Also, SR}\parallel \text{AC}\Rightarrow \text{FR}\parallel \text{OE}\\ \therefore \text{OERF is a parallelogram}\end{array}$

  $\text{So, }\angle \text{FRE=}\angle \text{EOF=90}°$

….. opposite angles are equal

Thus, PQRS is a parallelogram with

  $\begin{array}{l}\angle \text{R}=90°\text{ and SR}=\text{PQ}=\text{SP}=\text{RQ}\\ \therefore \text{PQRS is a square}\end{array}$

d

To determine

To find:The most descriptive name of the figure formed by connecting consecutive midpoints of the sides of the rectangle

Answer to Problem 17PSC

The figure formed by joining consecutive midpoints of rectangle will be rhombus

Explanation of Solution

Given information:

A rectangle with midpoints of its sides

Let ABCD is a rectangle and E, F, G and H are the midpoints of AB, BC, CD and DA.

Given, AE = EB, BF = FC, CG = GD and DH = HA

AB = DC … opposite sides of rectangle are equal

  $\therefore \frac{1}{2}\text{AB}=\frac{1}{2}\text{CD}\Rightarrow \text{AE = DG}$

Now,

  $\angle \text{EAH}\cong \angle \text{GDH}$ …. All interior angles are equal

  $\therefore \Delta \text{EHA}\cong \Delta \text{GHD}$ ….. SAS postulate

  $\Rightarrow \text{EH}=\text{GH}$ ….. corresponding sides in congruent triangles

  $\therefore$ EFGH is a parallelogram

EH = FG; EF = HG….. opposite sides of parallelogram are equal

EH = GH = FG = FE

  $\therefore$ EFGH is a rhombus…. By definition

e

To determine

To find:The most descriptive name of the figure formed by connecting consecutive midpoints of the sides of the parallelogram

Answer to Problem 17PSC

The figure formed by joining consecutive midpoints of parallelogram will be parallelugram

Explanation of Solution

Given information:

A parallelogram with midpoints of its sides

  

Here, in triangle ADB

We can conclude that

  $\frac{\text{AP}}{\text{AD}}=\frac{\text{AM}}{\text{AB}}$

So, by converse of Basic Proportionate theorem, we have

PM $\parallel$ DB…… (i)

Similarly, ON $\parallel$ DB…… (ii)

From (i) and (ii), we have

PM $\parallel$ ON

Similarly, we can prove that

P0 $\parallel$ MN

Hence, PONM is a parallelogram.

f

To determine

To find:The most descriptive name of the figure formed by connecting consecutive midpoints of the sides of the quadrilateral

Answer to Problem 17PSC

The figure formed by joining consecutive midpoints of quadrilateral will be parallelogram

Explanation of Solution

Given information:

A quadrilateral with midpoints of its sides

  

Let ABCD is a quadrilateral

Join A and C

Let M and N are the midpoints of AB and BC

Then in triangle ABC,

MN $\parallel$ AC and $\text{MN}=\frac{1}{2}\text{AC}$ …. By midpoint theorem…… (i)

Similarly, OP $\parallel$ AC and $\text{OP}=\frac{1}{2}\text{AC}$ ….. by midpoint theorem …..(ii)

From (i) and (ii),

  $\text{MN}\parallel \text{OP and MN = OP}$

  $\because$ one pair of opposite sides are parallel and equal

  $\therefore$ MNOP is a parallelogram.

g

To determine

To find:The most descriptive name of the figure formed by connecting consecutive midpoints of the sides of the isosceles trapezoid

Answer to Problem 17PSC

The figure formed by joining consecutive midpoints of isosceles trapezoid will be rhombus

Explanation of Solution

Given information:

A quadrilateral with midpoints of its sides

The diagonals AC and BD are in the trapezium

By midpoint theorem the opposite sides of the quad. obtained by joining the midpoints will come equal to each other and half of the diagonal in between the opposite sides

Then by congruency the two diagonals will come equal and all the sides will come equal.

Hence proved it is a rhombus.