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:2018-12-24_20-13-15

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

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

ED = \frac{1}{2} AB = 3.5 \ cm

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

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

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

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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:2018-12-24_20-17-22

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

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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:2018-12-24_20-02-03

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

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

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

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

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

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

Answer:2018-12-24_19-59-01

Given BD = DC and AD = XD

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

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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:2018-12-24_19-56-01

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

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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:2018-12-24_19-51-46

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

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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:2018-12-24_19-48-47.jpg

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

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

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

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

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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:2018-12-24_19-36-22

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

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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:2018-12-24_19-33-17

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)

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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:2018-12-24_19-01-32

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

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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:2018-12-24_18-53-26

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.

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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:2018-12-24_18-47-54

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

Similarly, 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.

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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:2018-12-24_18-35-54

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 \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

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

Answer:2018-12-24_18-34-19

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

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

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

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

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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 CQ = \frac{1}{4} AC  . If PQ produced meets BC at R , prove that R is a mid-point of BC .

Answer:2018-12-24_18-30-06

To prove: BR = RC

Given: 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

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

Answer:2018-12-24_18-23-58

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

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

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

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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 .

Answer:2018-12-24_18-20-25

Since E is the midpoint of AB and F is the midpoint of DC

AE = BE = \frac{1}{2} AB

And CF = DF = \frac{1}{2} CD

But AB = CD

\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD \Rightarrow BE = CF

Since BC \parallel EF \Rightarrow AD \parallel EF

\Rightarrow AEFD is a parallelogram

\Rightarrow AE = GP

Since E is the midpoint of AB ,

Therefore AE = BE \Rightarrow GP = PH

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Question 18: Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Answer:2018-12-24_18-09-05

PQ = \frac{1}{2} AC \ \ \ \& \ \ \ PQ \parallel AC

RS = \frac{1}{2} AC \ \ \ \& \ \ \ RS \parallel AC

\Rightarrow PQ = RS \ \ \ \& \ \ \ PQ \parallel RS

Similarly,

RQ = \frac{1}{2} DB \ \ \ \& \ \ \ RQ \parallel DB

SP = \frac{1}{2} DB \ \ \ \& \ \ \ SP \parallel DB

\Rightarrow RQ = SP \ \ \ \& \ \ \ RQ \parallel SP

Therefore PQRS is a parallelogram

\Rightarrow PR \ \ \& \ \  SQ bisect each other as diagonals of a parallelogram bisect each other.

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Question 19: Fill in the blanks to make the following statements correct:

(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is Isosceles Triangle

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is Right Angled

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is Parallelogram

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