Rectilinear Figures Class 9 ICSE Important Questions with Answers

Answer: (a) $10\text{ cm}$10 cm.

The diagonals of a rhombus bisect each other at right angles, giving half-diagonals $8$8 and $6$6 cm that meet at $90^\circ$90°. Side $=\sqrt{8^2+6^2}=10\text{ cm}$=√(8²+6²)=10 cm.

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: (b) $22$22.

The diagonals of a parallelogram bisect each other, so $OA=OC$OA=OC: $3x-4=x+6\Rightarrow x=5$3x-4=x+6⇒ x=5. Then $OA=11$OA=11 and $AC=2\times11=22$AC=2×11=22.

Answer: (b) Both A and R are true, but R is not the correct explanation of A. R describes a property of a parallelogram that is already known to be one; A is a test for proving that a quadrilateral is a parallelogram, which needs its own reason (congruent triangles formed by a diagonal).

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.$1/2∠ A+1/2∠ 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=1/2∠ A, ∠ PBA=1/2∠ 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=1/2 AB=1/2 DC, so $BC=\dfrac12 DC$BC=1/2 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=1/2∠ ADC and $\angle ECD=\dfrac12\angle BCD$∠ ECD=1/2∠ BCD, so

$\angle EDC+\angle ECD=\dfrac12(\angle ADC+\angle BCD)=\dfrac12\times180^\circ=90^\circ.$∠ EDC+∠ ECD=1/2(∠ ADC+∠ BCD)=1/2×180°=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 and 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 and 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) $AC=\sqrt{24^2+7^2}=\sqrt{625}=25\text{ m}$AC=√(24²+7²)=√625=25 m.

(ii) $BD=25\text{ m}$BD=25 m, because the diagonals of a rectangle are equal.

(iii) The diagonals bisect each other, so $OA=OB=OC=OD=\dfrac{25}{2}=12.5\text{ m}$OA=OB=OC=OD=25/2=12.5 m.

(iv) At $90^\circ$90°: the diagonals of a square are equal and meet at right angles.