Class 9: Rectilinear Figures – Exercise 12.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: In a $\triangle ABC, D, E$ and $I$ are, respectively, the mid-points of $BC, CA$ and $AB$ . If the lengths of side $AB, BC$ and $CA$ are $7 \ cm, \ 8 \ cm$ and $9 \ cm$ , respectively, find the perimeter of $\triangle DEF$ .

Answer:

Given: $AB = 7 \ cm, BC = 8 \ cm$ and $AC = 9 \ cm$

Since $D$ and $E$ are mid points of $BC$ and $AC$ respectively,

$\displaystyle ED = \frac{1}{2} = 3.5 \ cm$

Similarly $\displaystyle FD = \frac{1}{2} AC = 4.5 \ cm$

and $\displaystyle FE = \frac{1}{2} BC = 4 \ cm$

Therefore perimeter of $\triangle DEF = 3.5 + 4.5 + 4 = 12 \ cm$

$\\$

Question 2: In a triangle $\angle ABC, \angle A = 50^o, \angle B = 60^o$ and $\angle C= 70^o$ . Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

Answer:

We know, $DE \parallel AB \Rightarrow \angle EDC = 60^o$

$\Rightarrow DEC = 180^o - 60^o - 70^o = 50^o$

$FE \parallel BC \Rightarrow \angle AEF = 70^o$

$\Rightarrow \angle AFE = 180^o - 50^o - 70^o = 60^o$

$DF \parallel AC \Rightarrow BFD = 50^o$

$\Rightarrow \angle BDF = 180^o - 60^o - 50^o = 70^o$

Therefore $\angle FDE = 180^o - 70^o - 60^o = 50^o$

$\angle DEF = 180^o - 50^o - 70^o = 60^o$

$\angle EFD = 180^o - 60^o - 50^o = 70^o$

$\\$

Question 3: In a triangle, $P, Q$ and $R$ are the mid-points of sides $BC, CA$ and $AB$ respectively. If $AC = 21 \ cm, BC = 29 \ cm$ and $AB = 30 \ cm$ , find the perimeter of the quadrilateral $ARPQ$ .

Answer:

$\displaystyle RP = \frac{1}{2} AC = 10.5 \ cm$

$\displaystyle PQ = \frac{1}{2} AB = 15 \ cm$

$\displaystyle AQ = \frac{1}{2} AC = 10.5 \ cm$

$\displaystyle AR = \frac{1}{2} AB = 15 \ cm$

Therefore perimeter of $ARPQ = 10.5 + 15 + 10.5 + 15 = 51 \ cm$

$\\$

Question 4: In a $\triangle ABC$ median $AD$ is produced to $X$ such that $AD = DX$ . Prove that $ABXC$ is a parallelogram.

Answer:

Given $BD = DC$ and $AD = XD$

Since the diagonals of a parallelogram bisects each other we can say that $ABXC$ is a parallelogram.

$\\$

Question 5: In a $\triangle ABC, E$ and $F$ are the mid-points of $AC$ and $AB$ respectively. The altitude $AP$ to $BC$ intersects $FE$ at $Q$ . Prove that $AQ = QP$ .

Answer:

Given $AF = FB$ (since $F$ is mid point of $AB$ )

$AE = EC$ (since $E$ is the mid point of $AC$ )

Since $F$ and $E$ are mid point, $FE \parallel BC$

Since $FE \parallel BC$ and $F$ is the mid point of $AB, Q$ is the mid point of $AP$

Therefore $AQ = QP$

$\\$

Question 6: In a $\triangle ABC, BM$ and $CN$ are perpendiculars from $B$ and $C$ respectively on any line passing through $A$ . If $L$ is the mid-point of $BC$ , prove that $ML=NL$ .

Answer:

Consider the diagram shown.

Construction :- Draw $OL \perp MN$

Now, If a transversal makes equal intercepts on three or more parallel lines, then any other transversal intersecting them will also make equal intercepts.

$BM, CN$ and $OL$ are perpendicular to $MN$ Therefore, $BM \parallel CN \parallel OL$

Now, $BM \parallel CN \parallel OL$ and $BC$ is the transversal making equal intercepts i.e. $BL = LC$ .

Therefore, transversal $MN$ will also make equal intercepts $\Rightarrow OM = ON$

Consider $\triangle LMO$ and $\triangle LNO$ ,

$OM = ON$

$\angle LOM = \angle LON = 90^o$

$OL$ is common

Therefore $\triangle LMO \cong \triangle LNO$ (By S.A.S criterion)

Therefore $LM = LN$

$\\$

Question 7: In the adjoining figure, $\triangle ABC$ is right-angled at $B$ . Given that $AB = 9 \ cm, AC = 15 \ cm$ and $D, E$ are the mid-points of the sides $AB$ and $AC$ respectively, calculate i) The length of $BC$ ii) The area of $\triangle ADE$ .

Answer:

i) $BC = \sqrt{15^2 - 9^2} = \sqrt{225-81} = 12 \ cm$

ii) We know: $\displaystyle \frac{DE}{BC} = \frac{AD}{AB}$

Therefore $\displaystyle DE = 12 \times \frac{4.5}{9} = 6 \ cm$

Therefore Area of $\displaystyle \triangle ADE = \frac{1}{2} \times 6 \times 4.5 = 13.5 \ cm^2$

$\\$

Question 8: In adjoining figure, $M, N$ and $P$ are the mid-points of $AB, AC$ and $BC$ respectively. If $MN = 3 \ cm, NP = 3.5 \ cm$ and $MP = 2.5 \ cm$ , calculate $BC, AB$ and $AC$ .

Answer:

Given $MN = 3 \ cm, NP = 3.5 \ cm$ and $MP = 2.5 \ cm$ and $M, N$ and $P$ are the mid-points of $AB, AC$ and $BC$ respectively

$AB = 2 \times NP = 7 \ cm$

$AC = 2 \times MP = 5 \ cm$

$BC = 2 \times MN = 6 \ cm$

$\\$

Question 9: $ABC$ is a triangle and through $A, B, C$ lines are drawn parallel to $BC, CA$ and $AB$ respectively intersecting at $P, Q$ and $R$ . Prove that the perimeter of $\triangle PQR$ is double the perimeter of $\triangle ABC$ .

Answer:

To prove: $2 \times Perimeter (\triangle ABC) = Perimeter (\triangle PQR)$

Consider, $BCQA$ and $BCAP$ . Both are parallelograms.

Hence $BC = AQ$ and $BC = AR$

$\Rightarrow AQ = AR \Rightarrow A - \ midpoint \ of \ RQ$

Similarly, we can prove that $C$ is mid point of $RP$ and $C$ is mid point of $PQ$ .

Hence $PQ = 2BC, QR = 2 AB$ and $PR = 2AC$

Therefore $PQ + QR + PR = 2 (BC + AB + AC)$

$\Rightarrow Perimeter (\triangle PQR) = 2 \times Perimeter (\triangle ABC)$

$\\$

Question 10: In the adjoining figure, $BE \perp AC$ . $AD$ is any line from $A$ to $BC$ intersecting $BE$ in $H. P, Q$ and $R$ are respectively the mid-points of $AH, AB$ and $BC$ . Prove that $\angle PQR = 90^o$

Answer:

Since $Q \ \& \ R$ are midpoints of $AB$ and $BC$ respectively,

In $\triangle ABC, QR \parallel AC$

Similarly, Since $Q \ \& \ P$ are midpoints of $AB$ and $AH$ respectively,

In $\triangle ABH, QP \parallel BH$

$\Rightarrow QP \parallel BE$

Given $AC \perp BE \Rightarrow QP \perp QR$

Therefore $\angle PQR = 90^o$

$\\$

Question 11: In adjoining figure, $AB = AC$ and $CP \perp BA$ and $AP$ is the bisector of exterior $\angle CAD$ of $\triangle ABC$ . Prove that (i) $\angle PAC = \angle BCA$ (ii) $ABCP$ is a parallelogram.

Answer:

i) $\angle BAC = 180^o - x - x$

Therefore $\angle DAP = 180^o - (180 - 2x) = 2x$

Since $\angle DAP = \angle PAC = x$

Hence $\angle PAC = \angle BCA$

ii) If $AP$ were to be $\parallel$ to $BC$ then

$\angle ABC + \angle BAP = 180^o$

$\Rightarrow x + 180^o - 2x + x = 180^o$

$\Rightarrow 180^o = 180^o$

Therefore $AP \parallel BC$

Already given that $AB \parallel PC$ . Hence $ABCP$ is a parallelogram.

$\\$

Question 12: $ABCD$ is a kite having $AB = AD$ and $BC = CD$ . Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.

Answer:

Given: $AB = AD \ \& \ BC = CD$

Construction: Join $BD \ \& \ AC$

To Prove: $EH \parallel FG$ and $EF \parallel HG$

Also $EH = FG \ and \ EF = HG$

Proof: Consider $\triangle ABD$

Since $E$ and $H$ are midpoints of $AB$ and $AD$ respectively, $EH \parallel BD$

Similarly, Since $F$ and $G$ are midpoints of $BC$ and $CD$ respectively, $FG \parallel BD$

$\Rightarrow EH \parallel FG$

Using the same logic,

Since $F$ and $E$ are midpoints of $BC$ and $AB$ respectively, $EF \parallel AC$

Similarly, Since $G$ and $H$ are midpoints of $CD$ and $AD$ respectively, $HG \parallel AC$

$\Rightarrow EF \parallel HG$

Also $\displaystyle EH = \frac{1}{2} BD \ and \ FG = \frac{1}{2} BD \Rightarrow EH = FG$

Similarly, $\displaystyle FE = \frac{1}{2} AC \ and \ HG = \frac{1}{2} AC \Rightarrow EF = HG$

Now consider $\triangle ABO$ and $\triangle ADO$

$AB = AD$ (given)

$OB = OD$ (diagonals bisect each other)

$AO$ is common

Therefore $\triangle ABO \cong \triangle ADO$

$\Rightarrow \angle AOB = \angle AOD = 90^o$

$\Rightarrow EF \perp EH$

Similarly, $FG \perp HG$

Hence we can say that $EFHG$ is a rectangle because opposites sides are equal and parallel and the sides are perpendicular.

$\\$

Question 13: Let ABC be an isosceles triangle in which $AB = AC$ . lf $D, E, F$ be the mid-points of the sides $BC, CA$ and $AB$ respectively, show that the segment $AD$ and, $EF$ bisect each other at right angles.

Answer:

Given: $AB = AC$

Since $F$ is the midpoint of $AB$ , therefore $AF = FB$

Similarly, $E$ is the midpoint of $AC$ , therefore $AE = EC$

Also $FE \parallel BC$

$D$ is also midpoint of $BC$ , therefore $BD = DC$

$AB = AC$ (given)

Therefore $\displaystyle \frac{1}{2} AB = \frac{1}{2} AC \Rightarrow AF = EF$

Since $ABC$ is isosceles triangle, $\angle ABC = \angle ACB$

Because $EF \parallel BC$

$\Rightarrow \angle AFO = \angle AEO$

Consider $\triangle AOF$ and $\triangle AOE$

$AO$ is common

$\angle AFO = \angle AOE$

$AF = AE$

Therefore $\triangle AOF \cong \triangle AOE$ (By S.A.S criterion)

Now, $\angle AOF + \angle AOE = 180^o$

$\Rightarrow 2 \angle AOF = 180^o$

$\Rightarrow \angle AOF = 90^o$

$\\$

Question 14: $ABC$ is a triangle. $D$ is a point on $AB$ such that $\displaystyle AD = \frac{1}{4} AB$ and $E$ is a point on $AC$ such that $\displaystyle AE = \frac{1}{4} AC$ . Prove that $latex\displaystyle DE = \frac{1}{4} BC $.

Answer:

Let $P$ and $Q$ be midpoints of $AB$ and $AC$ respectively.

Therefore $\displaystyle PQ = \frac{1}{2} BC$

In $\displaystyle \triangle APQ, DE = \frac{1}{2} PQ$

$\displaystyle \Rightarrow DE = \frac{1}{2} ( \frac{1}{2} BC) = \frac{1}{4} BC$

$\\$

Question 15: In the adjoining figure, $ABCD$ is a parallelogram in which $P$ is the mid-point of $DC$ and $Q$ is a point on $AC$ such that $\displaystyle CQ = \frac{1}{4} AC$ . If $PQ$ produced meets $BC$ at $R$ , prove that $R$ is a mid-point of $BC$ .

Answer:

To prove: $BR = RC$

Given: $\displaystyle QC = \frac{1}{4} AQ$

$DP = PC$

Let $QC = x \Rightarrow OC = 2x$

Therefore $Q$ is midpoint of $OC$

In $\triangle DOC, P$ and $Q$ are midpoint of $DC$ and $OC$ respectively.

Therefore $PQ \parallel OD$

Since, $Q$ is the midpoint of $OC$ and $PQ \parallel BD$

$\Rightarrow R$ is the midpoint of $BC$

$\\$

Question 16: In adjoining figure, $ABCD$ and $PQRC$ are rectangles and $Q$ is the mid-point of $AC$ . Prove that (i) $DP = PC$ (ii) $\displaystyle PP = \frac{1}{2} AC$

Answer:

i) Since $Q$ is the midpoint of $AC$ and $QP \parallel AD, P$ is the midpoint of $DC$

$\Rightarrow DP = PC$

ii) Similarly, $R$ is the midpoint of $BC$ and $P$ is the midpoint of $DC$

$\displaystyle \Rightarrow PR = \frac{1}{2} BD$

$\displaystyle \Rightarrow PR = \frac{1}{2} AC$ (since diagonals of a rectangle are equal)

$\\$

Question 17: $ABCD$ is a parallelogram, $E$ and $F$ are the mid-points of $AB$ and $CD$ respectively. $GH$ is any line intersecting $AD , EF$ and $BC$ at $G, P$ and $H$ respectively. Prove that $GP=PH$ .