Rectilinear Figures (Quadrilaterals) — ICSE Class 9 Maths Important Questions

Answer: (b) $720^\circ$720^.

Sum $=(n-2)\times180^\circ=(6-2)\times180^\circ=4\times180^\circ=720^\circ$=(n-2)×180^=(6-2)×180^=4×180^=720^.

Answer: (b) $110^\circ$110^.

Adjacent angles of a parallelogram are supplementary, so $\angle B=180^\circ-70^\circ=110^\circ$B=180^-70^=110^.

Answer: (c) Rhombus.

Diagonals that bisect each other at right angles indicate a rhombus. If they were also equal it would be a square, but here they are unequal, so it is a rhombus.

Answer: (c) $12$12.

Each exterior angle $=180^\circ-150^\circ=30^\circ$=180^-150^=30^. Number of sides $=\dfrac{360^\circ}{30^\circ}=12$=360^/30^=12.

Answer: (b) Both statements are true: a rectangle's diagonals are equal, and a rectangle is indeed a parallelogram with a right angle. However R states the definition of a rectangle, which does not by itself explain why the diagonals are equal (that needs a congruence argument). So R is not the correct explanation of A.

The sum of the interior angles of a quadrilateral is $360^\circ$360^.

Let the fourth angle be $x$x.

$80^\circ+95^\circ+112^\circ+x=360^\circ$80^+95^+112^+x=360^

$287^\circ+x=360^\circ$287^+x=360^

$x=360^\circ-287^\circ=73^\circ.$x=360^-287^=73^.

The fourth angle is $73^\circ$73^.

Let the angles be $3x$3x, $4x$4x, $5x$5x and $6x$6x.

Sum of angles of a quadrilateral $=360^\circ$=360^:

$3x+4x+5x+6x=360^\circ$3x+4x+5x+6x=360^

$18x=360^\circ\Rightarrow x=20^\circ.$18x=360^ x=20^.

The angles are $3x=60^\circ$3x=60^, $4x=80^\circ$4x=80^, $5x=100^\circ$5x=100^, $6x=120^\circ$6x=120^.

In parallelogram $ABCD$ABCD, $AD\parallel BC$AD BC, so co-interior (adjacent) angles are supplementary:

$\angle A+\angle B=180^\circ.$A+ B=180^.

Dividing by $2$2:

$\dfrac12\angle A+\dfrac12\angle B=90^\circ.$12 A+12 B=90^.

Since $AP$AP bisects $\angle A$A and $BP$BP bisects $\angle B$B:

$\angle PAB=\dfrac12\angle A,\qquad \angle PBA=\dfrac12\angle B.$PAB=12 A, PBA=12 B.

So $\angle PAB+\angle PBA=90^\circ$PAB+ PBA=90^.

In $\triangle APB$APB, the angle sum is $180^\circ$180^:

$\angle APB=180^\circ-(\angle PAB+\angle PBA)=180^\circ-90^\circ=90^\circ.$APB=180^-( PAB+ PBA)=180^-90^=90^.

Hence $\angle APB=90^\circ$APB=90^. Proved.

Given: Parallelogram $ABCD$ABCD with diagonals $AC=BD$AC=BD.

To prove: $ABCD$ABCD is a rectangle, i.e. one angle is $90^\circ$90^.

Proof: In $\triangle ABC$ABC and $\triangle BAD$BAD:

$AB=BA$AB=BA (common)

$BC=AD$BC=AD (opposite sides of a parallelogram)

$AC=BD$AC=BD (given)

By SSS, $\triangle ABC\cong\triangle BAD$ABC BAD.

Hence $\angle ABC=\angle BAD$ABC= BAD (c.p.c.t.).

But $AD\parallel BC$AD BC, so $\angle ABC+\angle BAD=180^\circ$ABC+ BAD=180^ (co-interior angles).

Since the two angles are equal and supplementary, each is $90^\circ$90^:

$\angle BAD=90^\circ.$BAD=90^.

A parallelogram with one right angle is a rectangle. Hence $ABCD$ABCD is a rectangle. Proved.

Since $AB\parallel DC$AB DC and $CE$CE is a transversal, $\angle BEC=\angle DCE$BEC= DCE (alternate angles).

But $CE$CE bisects $\angle BCD$BCD, so $\angle DCE=\angle BCE$DCE= BCE.

Therefore $\angle BEC=\angle BCE$BEC= BCE, making $\triangle BEC$BEC isosceles with $BE=BC$BE=BC.

Since $E$E is the mid-point of $AB$AB, $BE=\dfrac12 AB=\dfrac12 DC$BE=12 AB=12 DC, so $BC=\dfrac12 DC$BC=12 DC, i.e. $DC=2BC$DC=2BC. (This confirms the configuration is consistent.)

Now consider $DE$DE. Draw the bisector: since $AB\parallel DC$AB DC, $\angle AED=\angle EDC$AED= EDC (alternate angles) and $DE$DE bisects $\angle ADC$ADC (by a symmetric argument $AE=AD$AE=AD).

In parallelogram $ABCD$ABCD, $\angle ADC+\angle BCD=180^\circ$ADC+ BCD=180^ (co-interior).

$\angle EDC=\dfrac12\angle ADC$EDC=12 ADC and $\angle ECD=\dfrac12\angle BCD$ECD=12 BCD, so

$\angle EDC+\angle ECD=\dfrac12(\angle ADC+\angle BCD)=\dfrac12\times180^\circ=90^\circ.$EDC+ ECD=12( ADC+ BCD)=12×180^=90^.

In $\triangle DEC$DEC: $\angle DEC=180^\circ-(\angle EDC+\angle ECD)=180^\circ-90^\circ=90^\circ$DEC=180^-( EDC+ ECD)=180^-90^=90^. Hence proved.

Given: Parallelogram $ABCD$ABCD whose diagonals $AC$AC and $BD$BD intersect at $O$O.

To prove: $OA=OC$OA=OC and $OB=OD$OB=OD.

Proof: In $\triangle AOB$AOB and $\triangle COD$COD:

$AB=CD$AB=CD (opposite sides of a parallelogram are equal)

$\angle OAB=\angle OCD$OAB= OCD (alternate angles, since $AB\parallel DC$AB DC and $AC$AC is a transversal)

$\angle OBA=\angle ODC$OBA= ODC (alternate angles, since $AB\parallel DC$AB DC and $BD$BD is a transversal)

By ASA, $\triangle AOB\cong\triangle COD$AOB COD.

Hence, by c.p.c.t.:

$OA=OC\quad\text{and}\quad OB=OD.$OA=OC OB=OD.

Therefore the diagonals bisect each other. Proved.

(Conversely, if the diagonals of a quadrilateral bisect each other, it is a parallelogram, which is a standard test.)

Given: Parallelogram $ABCD$ABCD with $P,Q$P,Q on diagonal $BD$BD and $DP=BQ$DP=BQ.

To prove: $APCQ$APCQ is a parallelogram.

Construction / idea: Join $AC$AC meeting $BD$BD at $O$O.

Since $ABCD$ABCD is a parallelogram, its diagonals bisect each other, so

$OA=OC\quad\text{and}\quad OB=OD \quad\text{...(1)}$OA=OC OB=OD ...(1)

Given $DP=BQ$DP=BQ. From (1), $OD=OB$OD=OB. Subtracting:

$OD-DP=OB-BQ\Rightarrow OP=OQ \quad\text{...(2)}$OD-DP=OB-BQ OP=OQ ...(2)

(assuming $P$P near $D$D and $Q$Q near $B$B; the same result follows for the other configuration by adding).

Now in quadrilateral $APCQ$APCQ, the diagonals are $AC$AC and $PQ$PQ, meeting at $O$O.

From (1), $OA=OC$OA=OC; from (2), $OP=OQ$OP=OQ.

Thus the diagonals $AC$AC and $PQ$PQ of quadrilateral $APCQ$APCQ bisect each other at $O$O.

A quadrilateral whose diagonals bisect each other is a parallelogram.

Therefore $APCQ$APCQ is a parallelogram. Hence proved.

(i) Each interior angle of a regular $n$n-gon $=\dfrac{(n-2)\times180^\circ}{n}$=(n-2)×180^/n.

(ii) For $n=8$n=8: interior angle $=\dfrac{(8-2)\times180^\circ}{8}=\dfrac{6\times180^\circ}{8}=\dfrac{1080^\circ}{8}=135^\circ$=(8-2)×180^/8=6×180^/8=1080^/8=135^.

(iii) Each exterior angle $=180^\circ-135^\circ=45^\circ$=180^-135^=45^ (or $\dfrac{360^\circ}{8}=45^\circ$360^/8=45^).

(iv) Number of sides $=\dfrac{360^\circ}{\text{exterior angle}}=\dfrac{360^\circ}{24^\circ}=15$=360^/exterior angle=360^/24^=15 sides.