Exercise-8.1
Maths - Exercise
Q1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
\(\textbf{Given:} \\ ABCD \small\text{ is a parallelogram with } \\AC = BD. \\\) \(\textbf{To Prove:} \\ \quad ABCD \text{ is a rectangle.} \\\) \(\textbf{Proof:} \\ \small\text{Since } ABCD \text{ is a parallelogram:} \) \[\begin{aligned}AB &= CD \\ AD &= BC \\ AB &\parallel CD\\ AD &\parallel BC \end{aligned}\] Let diagonals \(AC\) and \(BD\) intersect at \(O\)
Diagonals of a parallelogram bisect each other
\[\therefore AO = OC\\ BO = OD \] Given: \[AC = BD, \text{ so} \\ AO = OC = BO = OD \] Consider \(\triangle ABO\) and \(\triangle CDO\): \[\begin{aligned} AO &= OC \\ BO &= OD \\ AB &= CD \\ \implies \triangle ABO &\cong \triangle CDO \\ (\text{by SSS criterion}) \end{aligned}\] So,
\(\angle OAB = \angle OCD \)
Similarly, all four angles at the vertices are split equally.
In parallelogram \(ABCD:\)
\[\begin{aligned}\angle BAD + \angle ABC &= 180^\circ \\ \text{But } \angle BAD &= \angle ABC\\ \implies 2\angle BAD &= 180^\circ \\\implies \angle BAD &= 90^\circ \end{aligned} \] Thus, each angle of \(ABCD\) is \(90^\circ.\)
therefore, \(ABCD\) is a rectangle.
Q2. Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Given:
\(ABCD\) is a square with\[ AB = BC = CD = DA \]
To Prove:
\[ AC = BD\\AO = OC\\BO = OD \] Diagonals bisect each other at \(90^\circ\).Proof:
Step 1: Diagonals are equal. Since all sides of the square are equal and each angle is \(90^\circ\), consider triangle \(ABC\): \[ AC^2 = AB^2 + BC^2 \] If \(AB = BC = a\): \[\begin{aligned} AC^2 &= a^2 + a^2 \\&= 2a^2 \\\implies AC &= \sqrt{2}a \end{aligned}\] Similarly, \(BD = \sqrt{2}a\). Thus, the diagonals are equal: \[ AC = BD \] Step 2: Diagonals bisect each other. In any square, the diagonals bisect each other, so \[ AO = OC,\quad BO = OD \] Step 3: Diagonals intersect at right angles.Consider \(\triangle AOD\) and \(\triangle COD\): \[\begin{aligned}AD &= CD\text{ (sides of square)}\\ OD &= OD\text{ (common side)}\\ AO &= CO\text{ (half of equal diagonals)}\end{aligned} \] \[ \triangle AOD \cong \triangle COD\\\text{By SSS-Rule} \] So, \[ \angle AOD = \angle COD \] Since these angles form a straight line along diagonal \(AC\), \[ \angle AOD + \angle COD = 180^\circ \] Therefore, \[\begin{aligned} 2\angle AOD &= 180^\circ \\\implies \angle AOD &= 90^\circ \end{aligned}\] Diagonals bisect each other at \(90^\circ\).
Conclusion:
The diagonals of a square are equal, bisect each other, and intersect at \(90^\circ\).
Q3. Diagonal \(AC\) of a parallelogram \(ABCD\) bisects
\(\angle A\) (see Fig. 8.11). Show that
(i) it bisects \(\angle C\) also,
(ii) \(ABCD\) is a rhombus.
Solution:
\(\textbf{Given:}\\\\\) \(ABCD\) is a parallelogram in which the diagonal \(AC\) bisects \(\angle
A\).
\[
\begin{aligned}
AB &= CD, \\
BC &= AD, \\
\angle DAC &= \angle CAB.
\end{aligned}
\]
\(\textbf{To prove:}\)
\[
\text{(i) } AC \text{ bisects } \angle C, \\
\text{(ii) } ABCD \text{ is a rhombus.}
\]
\(\textbf{Proof:}\)
Consider triangles \( \triangle ACD \) and \( \triangle ABC \).
\[
AB = CD \\ \small\text{(Opposite sides of a parallelogram)} \\
BC = AD \\ \small\text{(Opposite sides of a parallelogram)} \\
AC = AC \\ \small\text{(Common side)}
\]
\[\therefore \triangle ACD \cong \triangle ABC \\ \small(\text{By SSS congruence})\]
By \(CPCT\) (Corresponding Parts of Congruent Triangles),
\[
\angle DCA = \angle BCA.
\]
Thus, the diagonal \(AC\) bisects \(\angle C\).
\[
\text{Hence, (i) is proved.}
\]
Now, since \(AC\) bisects \(\angle A\),
\[
\angle DAC = \angle CAB.
\]
From the congruence of triangles
\( \triangle ACD \) and \( \triangle ABC \),
\[
\begin{aligned}
AD &= DC \quad \text{and} \\ AB &= BC.
\end{aligned}
\]
\(\therefore AB = BC = CD = DA.\)
\[
\boxed{\text{Hence, } ABCD \text{ is a rhombus.}}
\]
Q4. \(ABCD\) is a rectangle in which diagonal \(AC\) bisects
\(\angle A\) as well as \(\angle C.\) Show that:
(i) ABCD is a square
(ii) diagonal \(BD\) bisects \(\angle B\) as well as \(\angle D.\)
Solution:
Given: ABCD is a rectangle in which the diagonal \(AC\) bisects both \(\angle A\) and \(\angle C\).
To prove:
(i) ABCD is a square
(ii) Diagonal \(BD\) bisects \(\angle B\) as well as \(\angle D\)
Proof:
Since ABCD is a rectangle, all its angles are right angles, that is,
\(\angle A = \angle B = \angle C = \angle D = 90^{\circ}\)
Because \(AC\) bisects \(\angle A\), we have
\(\angle CAB = \angle DAC = \frac{1}{2} \times 90^{\circ} = 45^{\circ}\)
In triangle \(\triangle ABC\),
\(\angle ABC = 90^{\circ}\) and \(\angle CAB = 45^{\circ}\).
Therefore, the remaining angle \(\angle ACB = 45^{\circ}\).
Hence, triangle \(\triangle ABC\) is a right isosceles triangle, and so
\[AB = BC\tag{1}\]Similarly, since \(AC\) bisects \(\angle C\),
\(\angle BCA = \angle ACD = 45^{\circ}\).
In triangle \(\triangle BCD\), which is right-angled at C, we have
\(\angle BCD = 90^{\circ}\) and \(\angle BCA = 45^{\circ}\).
Therefore, triangle \(\triangle BCD\) is also right isosceles, and hence
\[BC = CD\tag{2}\]From (1) and (2), we get
\(AB = BC = CD\).
Since opposite sides of a rectangle are equal, \(AD = BC\).
Therefore, all sides are equal:
\(AB = BC = CD = DA\).
Hence, ABCD is a square. (Part \((i)\) proved)
(ii) To prove: Diagonal \(BD\) bisects \(\angle B\) and \(\angle D\).
In square ABCD, consider triangles \(\triangle ABD\) and \(\triangle CBD\).
We have:
\[\scriptsize\begin{aligned} AB &= CB\text{ (all sides of a square are equal)}\\ AD &= CD\text{ (all sides of a square are equal)}\\ BD &= BD\text{ (common side)} \end{aligned}\] Therefore, \[\triangle ABD \cong \triangle CBD\\\text{ (by SSS congruence)}\] By CPCT, \[\angle ABD = \angle DBC\]Hence, diagonal \(BD\) bisects \(\angle B\).
Similarly, it can be shown that \(BD\) also bisects \(\angle D\).
Therefore, diagonal \(BD\) bisects both \(\angle B\) and \(\angle D\).
Hence proved.
Q5. In parallelogram \(ABCD\), two points \(P\) and \(Q\) are
taken on diagonal \(BD\) such that \(DP = BQ\)
(see Fig. 8.12). Show that:
(i) \(\triangle APD \cong \triangle CQB\)
(ii) \(AP = CQ\)
(iii) \(\triangle AQB \cong \triangle CPD\)
(iv) \(AQ = CP\)
(v) \(APCQ\) is a parallelogram
Solution:
Given:
\[ \begin{aligned} DP &= BQ,\\ AD &= BC,\\ AB &= CD \end{aligned} \]To Prove: (i) to (v) as stated above.
Proof:
(i) To prove
\(\triangle APD \cong \triangle CQB\)
In triangles \(APD\) and \(CQB\):
\[ \begin{aligned} DP &= BQ \text{ (Given)}\\ \angle APD &= \angle CBQ\; (\text{Alternate angles, since } \\AD &\parallel BC\; BD \text{ is a transversal})\\ AD &= BC \text{ (Given)} \end{aligned} \] \[ \therefore \triangle APD \cong \triangle CQB \quad (\text{By SAS rule}) \] \[ \Rightarrow AP = CQ \quad (\text{By CPCT}) \]Hence proved (i) and (ii).
(iii) To prove:
\(\triangle AQB \cong \triangle CPD\)
In triangles \(AQB\) and \(CPD\):
\[ \begin{aligned} BQ &= DP \text{ (Given)}\\ \angle ABQ &= \angle CDP \end{aligned}\](Alternate angles, since \(AB \parallel CD\) and \(BD\) is a transversal)
\[ AB = CD \](Opposite sides of a parallelogram are equal)
\[ \therefore \triangle AQB \cong \triangle CPD \\(\text{By SAS rule}) \] \[ \Rightarrow AQ = CP \quad (\text{By CPCT}) \]Hence proved (iii) and (iv).
(v) To prove
\(APCQ\) is a parallelogram
From the above results, we have
\[ \begin{aligned} AP &= CQ,\\ AQ &= CP. \end{aligned} \]When both pairs of opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram.
\[ \therefore APCQ \text{ is a parallelogram.} \]Hence proved.
Q6. \(ABCD\) is a parallelogram and \(AP\) and \(CQ\) are
perpendiculars from vertices \(A\) and \(C\) on diagonal
BD (see Fig. 8.13). Show that
(i) \(\triangle APB \cong \triangle CQD\)
(ii) \(AP = CQ\)
Solution:
Given:
\(ABCD\) is a parallelogram and \(AP\) and \(CQ\) are perpendiculars to diagonal \(BD\).
To Prove: \(\triangle APB \cong \triangle CQD\) and \(AP = CQ\)
Proof:
In triangles \(APB\) and \(CQD\):
\[ AB = CD \\ \small\text{(Opposite sides of a parallelogram are equal)}\\\\ \angle APB = \angle CQD = 90^\circ \\ \small\text{(Each perpendicular to } BD\text{)}\\\\ \angle PBA = \angle QDC \\ \small\text{(Alternate angles, since } \\AB \parallel CD \small\text{ and } BD \text{ is a transversal)} \] \[ \therefore\triangle APB \cong \triangle CQD \\ (\text{By AAS congruence rule}) \] \[ \Rightarrow AP = CQ \quad (\text{By CPCT}) \]Hence proved.
Q7. \(ABCD\) is a trapezium in which \(AB || CD\) and
\(AD = BC\) (see Fig. 8.14). Show that
(i) \(\angle A = \angle B\)
(ii) \(\angle C = \angle D\)
(iii) \(\triangle ABC \cong \triangle BAD\)
(iv) diagonal \(AC =\) diagonal \(BD\)
Solution:
Given:
\[ \begin{aligned} AB &\parallel CD,\\ AD &= BC \end{aligned} \]To Prove:
\[ \begin{aligned} (i)\ &\angle A = \angle B, \\ (ii)\ &\angle C = \angle D, \\ (iii)\ &\triangle ABC \cong \triangle BAD, \\ (iv)\ &AC = BD \end{aligned} \]Proof:
(i) To prove
\(\angle A = \angle B\)
Since \(AB \parallel CD\), and \(AD\) and \(BC\) act as transversals, we have
\[ \begin{align} \small\angle DAB + \angle CBA &= \small180^\circ\tag{1}\\ \small\angle CDA + \angle DCB &= \small180^\circ\tag{2} \end{align} \]Given that \(AD = BC\), the trapezium \(ABCD\) is isosceles.
In an isosceles trapezium, the angles adjacent to each base are equal. Hence,
\[ \angle DAB = \angle CBA \] \[\therefore \angle A = \angle B\]Hence proved.
(ii) To prove
\(\angle C = \angle D\)
Similarly, since \[AB \parallel CD\\AD = BC\] the same property applies to the other pair of base angles. Thus,
\[ \angle ADC = \angle BCD \] \[\therefore \angle C = \angle D\]Hence proved
(iii) To prove
\(\triangle ABC \cong \triangle BAD\)
In triangles \(ABC\) and \(BAD\):
\[ \begin{aligned} AB &= AB \text{ (Common side)}\\ AD &= BC \text{ (Given)}\\ \angle A &= \angle B \text{ (Proved earlier)} \end{aligned} \] \[ \therefore \triangle ABC \cong \triangle BAD \\ (\text{By SAS congruence rule}) \]Hence proved.
(iv) To prove:
\(AC = BD\)
From the congruence of triangles \(ABC\) and \(BAD\), we have by CPCT:
\[ AC = BD \]Hence proved.
