-
Question 1: The angles of quadrilateral are in the ratio $3 : 5 : 9 : 13$. Find all the angles of the quadrilateral.
Solution: Let angles be $3x, 5x, 9x, 13x$. Sum $= 360^\circ$.
$3x+5x+9x+13x=360 \Rightarrow 30x=360 \Rightarrow x=12$.
Angles are $36^\circ, 60^\circ, 108^\circ, 156^\circ$.
-
Question 2: Show that the diagonals of a rhombus are perpendicular to each other.
Solution: Let $ABCD$ be a rhombus. Diagonals intersect at $O$. In $\triangle AOD$ and $\triangle COD$: $OA=OC$ (Parallelogram diagonals bisect), $OD=OD$ (Common), $AD=CD$ (Sides of rhombus). By SSS, $\triangle AOD \cong \triangle COD$. Thus $\angle AOD = \angle COD$. Since they are linear pair, $\angle AOD = 90^\circ$.
-
Question 3: $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$. Show that $\triangle APB \cong \triangle CQD$.
Solution: In $\triangle APB$ and $\triangle CQD$: $\angle APB = \angle CQD = 90^\circ$, $\angle ABP = \angle CDQ$ (Alternate interior angles, $AB \parallel DC$), $AB = CD$ (Opposite sides). By AAS, $\triangle APB \cong \triangle CQD$.
-
Question 4: In a parallelogram $ABCD$, $\angle A = (3x - 20)^\circ$ and $\angle C = (x + 40)^\circ$. Find the value of $x$ and measure of each angle.
Solution: Opposite angles are equal. $3x - 20 = x + 40 \Rightarrow 2x = 60 \Rightarrow x = 30$.
$\angle A = 3(30) - 20 = 70^\circ$. $\angle C = 70^\circ$.
$\angle B = 180^\circ - 70^\circ = 110^\circ$. Angles: $70^\circ, 110^\circ, 70^\circ, 110^\circ$.
-
Question 5: Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Solution: (Mid-point Theorem) Let $D, E$ be midpoints of $AB, AC$. Extend $DE$ to $F$ such that $DE=EF$. Join $CF$. $\triangle ADE \cong \triangle CFE$ (SAS). So $AD=CF$ and $\angle A = \angle ECF$ (Alt angles $\Rightarrow AB \parallel CF$). Since $AD=BD$, $BD=CF$ and $BD \parallel CF$. Thus $BCFD$ is a parallelogram. $DF \parallel BC \Rightarrow DE \parallel BC$.
-
Question 6: $ABCD$ is a trapezium in which $AB \parallel DC$ and $AD = BC$. Show that $\angle A = \angle B$.
Solution: Draw a line through $C$ parallel to $AD$ intersecting $AB$ produced at $E$. $ADCE$ is a parallelogram. $AD = CE$. Given $AD = BC$, so $CE = BC$. In $\triangle BCE$, $\angle CBE = \angle CEB$. Also $\angle A + \angle CEB = 180^\circ$ (Interior angles). $\angle B + \angle CBE = 180^\circ$ (Linear pair). Thus $\angle A = \angle B$.
-
Question 7: Show that the bisectors of angles of a parallelogram form a rectangle.
Solution: Let bisectors of $\angle A$ and $\angle B$ meet at $P$. $\frac{1}{2}\angle A + \frac{1}{2}\angle B = \frac{1}{2}(180^\circ) = 90^\circ$. In $\triangle APB$, $\angle P = 180^\circ - 90^\circ = 90^\circ$. Similarly all other angles formed by bisectors are $90^\circ$. Hence it is a rectangle.
-
Question 8: $l, m$ and $n$ are three parallel lines intersected by transversals $p$ and $q$ such that $l, m$ and $n$ cut off equal intercepts $AB$ and $BC$ on $p$. Show that they cut off equal intercepts $DE$ and $EF$ on $q$.
Solution: Join $A$ to $F$ intersecting $m$ at $G$. In $\triangle ACF$, $B$ is midpoint of $AC$ and $BG \parallel CF$, so $G$ is midpoint of $AF$. In $\triangle AFD$, $G$ is midpoint of $AF$ and $GE \parallel AD$, so $E$ is midpoint of $DF$. Thus $DE = EF$.
-
Question 9: $ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$. Show that $ABCD$ is a square.
Solution: Since $ABCD$ is a rectangle, $\angle A = \angle C = 90^\circ$. $AC$ bisects them, so $\angle DAC = \angle DCA = 45^\circ$. In $\triangle ADC$, sides opposite equal angles are equal, so $AD = CD$. A rectangle with adjacent sides equal is a square.
-
Question 10: In quadrilateral $ABCD$, the line segments bisecting $\angle C$ and $\angle D$ meet at $E$. Prove that $\angle A + \angle B = 2\angle CED$.
Solution: In $\triangle CED$, $\angle CED = 180^\circ - \frac{1}{2}(\angle C + \angle D)$.
Multiply by 2: $2\angle CED = 360^\circ - (\angle C + \angle D)$.
In quadrilateral, $\angle A + \angle B + \angle C + \angle D = 360^\circ \Rightarrow \angle C + \angle D = 360^\circ - (\angle A + \angle B)$.
Substitute: $2\angle CED = 360^\circ - (360^\circ - (\angle A + \angle B)) = \angle A + \angle B$.