Question 1: In the adjoining figure, \triangle ABC is an obtuse triangle, obtuse at \angle B . If AD \perp CB , prove that AC^2 = AB^2 + BC^2 + 2 BC \times BD .2018-10-13_15-21-36

Answer:

Given \triangle ABC is an obtuse triangle, \angle B is obtuse angle

In \triangle ADB , we have 

AB^2 = AD^2 + DB^2          (by Pythagoras theorem)

In \triangle ADC , we have

AC^2 = AD ^2 + DC^2

\Rightarrow AC^2 = AD^2 + (DB + BC)^2

\Rightarrow AC^2 = (AD^2 + DB^2) + BC^2 + 2 DB \times BC

\Rightarrow AC^2 = AB^2 + BC^2 + 2  DB \times BC

Hence proved.

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Question 2: In the adjoining figure, \angle B of \triangle ABC is acute angle and AD \perp BC . Prove that AC^2 = AB^2 + BC^2 - 2 BC \times BD

Answer:2018-10-13_15-21-48

In \triangle ADB, \angle D = 90^o

Therefore AB^2 = AD^2 + BD^2          (by Pythagoras theorem)

Similarly, in \triangle ADC , we have

AC^2 = AD^2 + DC^2          (by Pythagoras theorem)

\Rightarrow AC^2 = AD^2 + (BC-BD)^2

\Rightarrow AC^2 = AD^2 + BC^2 + BD^2 - 2 BC \times BD

\Rightarrow AC^2 = AB^2 + BC^2 - 2 BC \times BD

Hence proved.

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Question 3: Prove in any triangle, the sum of the square of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.2018-10-13_15-21-13

Answer:

Construction: Altitude, \angle AED = 90^o

The \triangle ADB , is obtuse angle triangle, and \triangle ADC is an acute angle triangle.

In \triangle ABD, \angle D is obtuse and AE \perp BD .

Therefore AB^2 = AD^2 + BD^2 + 2 BD \times DE … … … … … i)

Similarly, in \triangle ADC is acute at \angle D and AE \perp CD

Therefore AC^2 = AD^2 + DC^2 - 2 DC \times DE

\Rightarrow AC^2 = AD^2 + BD^2 - 2 BD \times DE (since CD = BD ) … … … … … ii)

Adding i) and ii) we get

AB^2 + AC^2 = 2 AD^2 + 2 BD^2

\Rightarrow AB^2 + AC^2 = 2 (AD^2 + BD^2) … … … … … ii)

Putting BD = \frac{1}{2} BC    in iii) we get

AB^2 + AC^2 = 2 AD^2 + 2 (\frac{1}{2} BC^2)

Hence proved.

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Question 4: Prove that three times the sum of the square of the sides of a triangle is equal to 4 times the sum of square of medians of the triangle.2018-10-13_15-21-26

Answer:

Given \triangle ABC and AD, BE and CF are medians as shown in the diagram.

In \triangle ABC , with AD as the median of BC , we have

AB^2 + AC^2 = 2 (AD^2 + BD^2)

\Rightarrow AB^2 + AC^2 = 2 \{AD^2 + (\frac{BC}{2})^2 \}

\Rightarrow 2(AB^2 + AC^2 ) = 4 AD^2 + BC^2 … … … … … i)

Similarly with BE bisecting AC and CF bisecting AB , we get

2 (AB^2 + BC^2) = 4 BE^2 + AC^2

and 2 (AC^2 + BC^2) = 4CF^2 + AB^2

Adding i), ii) and iii) we get

4(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2) + BC^2 + CA^2 + AB^2

\Rightarrow 3 (AB^2 + BC^2 + CA^2) = 4 (AD^2 + BE^2 + CF^2)

Hence proved.

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Question 5: P and Q are mid points of side CA and CB respectively in \triangle ABC . \angle C is 90^o . Prove that (i) 4 AQ^2 = 4 AC^2 + BC^2 (ii) 4 BP^2 = 4 BC^2 + AC^2 (iii) 4 (AQ^2 + BP^2) = 5 AB^2 2018-10-13_15-20-45

Answer:

(i) In \triangle AQC, \angle C = 90^o

AQ^2 = AC^2 + QC^2 (by Pythagoras theorem)

\Rightarrow 4 AQ^2 = 4 AC^2 + 4 QC^2 (multiply by 4 on both sides)

\Rightarrow 4AQ^2 = 4 AC^2 + (2BC^2) (since BC = 2 QC )

(ii) In \triangle BPC , \angle C= 90^o

BP^2 = BC^2 + CP^2

\Rightarrow 4 BP^2 = 4 BC^2 + 4 CP^2

\Rightarrow 4 BP^2 = 4 BC^2 + AC^2 (since AC = 2CP )

(iii) From i) and ii) we have

4 AQ^2 = 4 AC^2 + BC^2

4 BP^2 = 4 BC^2 + AC^2

Adding, we get

4 AQ^2 + 4 BP^2 = (4 AC^2 + BC^2) + (4 BC^2 + AC^2)

\Rightarrow 4 (AQ^2 + BP^2) = 5 (AC^2 + BC^2) (since we have AB^2 = AC^2 + BC^2 )

\Rightarrow 4(AQ^2 + BP^2) = 5 AB^2

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Question 6: In the adjoining figure, \triangle ABC is right angles at \angle B . AD and CE are two medians drawn from vertices A and C respectively. If AC is 5 \ cm and AD = \frac{3\sqrt{5}}{2} \ cm , find the length of CE .2018-10-13_15-20-54

Answer:

In \triangle ABD

AD^2 = AB^2 + BD^2

\Rightarrow AD^2 = AB^2 + (\frac{BC}{2})^2

\Rightarrow AD^2 = AB^2 + \frac{1}{4} BC^2

Similarly in \triangle BCE

CE^2 = BC^2 + \frac{1}{4} AB^2

\Rightarrow AD^2 + CE^2 = AB^2 + \frac{1}{4} BC^2 + BC^2 + \frac{1}{4} AB^2

\Rightarrow  AD^2 + CE^2 = \frac{5}{4} (AB^2 + BC^2)

\Rightarrow AD^2 + CE^2 = \frac{5}{4} AC^2

\Rightarrow ( \frac{3 \sqrt{5}}{2} )^2 + CE^2 = \frac{5}{4} (5)^2

\Rightarrow CE^2 = \frac{125}{4} - \frac{45}{4} = 20

\Rightarrow CE = 2\sqrt{5} cm

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Question 7: In \triangle ABC, \angle C = 90^o . Let BC = a, CA = b and AB = c and let p be the length of the perpendicular from C to AB . Prove that

(i) cp = ab

(ii) \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} 2018-10-13_15-20-21

Answer:

Consider the figure

(i) CD \perp AB , then CD = p

Area of \triangle ABC = \frac{1}{2} \times AB \times CD = \frac{1}{2} cp

Also the area of \triangle ABC = \frac{1}{2} \times BC \times AC = \frac{1}{2} ab

Therefore \frac{1}{2} cp = \frac{1}{2} ab

\Rightarrow cp = ab

(ii) In \triangle ABC, \angle C = 90^o , therefore

AB^2 = BC^2 + AC^2

\Rightarrow c^2 = a^2 +b^2

(since cp = ab \Rightarrow c = \frac{ap}{p} )

\Rightarrow ( \frac{ab}{p} )^2 = a^2 + b^2

\Rightarrow \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}

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Question 8: In an equilateral triangle with side a , prove that (i) Altitude = \frac{a\sqrt{3}}{2}    (ii) Area = \frac{\sqrt{3}}{4} a^2 2018-10-13_15-20-33

Answer:

Consider adjoining figure.

(i) Draw AD \perp BC

D is the mid point since \triangle ABC is equilateral with AB = AC = BC

\Rightarrow AB = a, BD = \frac{1}{2} BC = \frac{a}{2}

Since \triangle ABD is right angled at \angle D

\therefore AB^2 = AD^2 + BD^2

\Rightarrow a^2 = AD^2 + ( \frac{a}{2} )^2

\Rightarrow AD^2 =a^2 - ( \frac{a}{2} )^2

\Rightarrow AD^2 = a^2 - \frac{a^2}{4} = \frac{3}{4} a^2

\Rightarrow AD = \frac{\sqrt{3}}{2} a

(ii) Now Area of \triangle ABC = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} a^2 

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Question 9: ABC is a triangle in which AB = AC and D   is any point on BC . Prove that AB^2 - AD^2 = BD.CD 2018-10-13_15-19-51

Answer:

Draw AE \perp BC

In \triangle AEB and \triangle AEC , we have

AB = AC

AE is common

and \angle B = \angle C (since AB = AC )

\therefore \triangle AEB \cong \triangle AEC

\Rightarrow BE = CE

Since \triangle AED and \triangle ABE are right angles at E

\therefore AD^2 = AE^2 + DE^2

AB^2 = AE^2 + BE^2

\Rightarrow AB^2 - AD^2 = BE^2 - DE^2

\Rightarrow AB^2 - AD^2 = (BE - DE)(BE + DE)

\Rightarrow AB^2 - AD^2 = (BE - DE)(CE + DE)

\Rightarrow AB^2 - AD^2 = BD .CD

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Question 10: From a point O in the interior of \triangle ABC , perpendiculars OD, OE and OF are drawn to side BC, CA and AB respectively. Prove:

i) AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2

ii) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2 2018-10-13_15-20-03

Answer:

Let O be the point inside \triangle ABC and let OD \perp BC, OE \perp CA and OF \perp AB

i) In \triangle OFA, \triangle ODB and \triangle OEC we have

OA^2 = AF^2 + OF^2

OB^2 = BD^2 + OD^2

and OC^2 = CE^2 + OE^2

Adding the above three

OA^2 + OB^2 + OC^2 = AF^2 + OF^2 + BD^2 + OD^2 + CE^2 + OE^2

AF^2 +BD^2 +CE^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2

ii) In right triangle \triangle ODB and \triangle ODC we have

OB^2 = OD^2 + BD^2

OC^2 = OD^2 + CD^2

OB^2 - OC^2 = (OD^2 + BD^2)-(OD^2 + CD^2)

\Rightarrow OB^2 - OC^2 = BD^2 - CD^2

Similarly, OC^2 - OA^2 = CE^2 - AE^2

OA^2 - OB^2 = AF^2 - BF^2

Adding i) , ii) and iii) 

(OB^2 - OC^2) + (OC^2 - OA^2) + (OA^2 - OB^2) = (BD^2 - CD^2) + (CE^2 - AE^2) + (AF^2 - BF^2)

\Rightarrow (BD^2 + CE^2 + AF^2) - (AE^2 + CD^2 + BF^2) = 0

\Rightarrow AF^2 + BD^2 + CE^2 = AE^2 + BF^2 + CD^2

Hence proved.

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Question 11: A point O inside a rectangle ABCD is joined with each of the vertices A, B, C and D . Prove that OB^2 + OD^2 = OC^2 + OA^2

Answer:2018-10-13_15-19-22

Join O to A, B, C and D

Draw a line EF \parallel AB through O

In \triangle OEA and \triangle OFC we have

OA^2 = OE^2 + AE^2

OC^2 = OF^2 + CF^2

\Rightarrow OA^2 + OC^2 = OE^2 + AE^2 + OF^2 + CF^2

\Rightarrow OA^2 + OC^2 = OE^2 + OF^2 + AE^2 + CF^2 … … … … … i)

Similarly, in \triangle OFB and \triangle OFC we have

OB^2 = OF^2 + FB^2

OD^2 = OE^2 + DE^2

\Rightarrow OB^2 + OD^2 = (OF^2 + FB^2) + (OE^2 + DE^2)

\Rightarrow OB^2 + OD^2 = OE^2 + OF^2 + DE^2 + BF^2

\Rightarrow OB^2 + OD^2 = OE^2 + OF^2 + CF^2 + AE^2 … … … … … ii)

Since DE = CF and AE = BF

Therefore from i) and ii) we get

OA^2 + OC^2 = OB^2 + OD^2

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Question 12: In \triangle ABC, AC > AB, D is mid point  of BC and AE \perp BC . Prove:

i) AC^2 = AD^2 + BC.DE + \frac{1}{2} BC^2

ii) AB^2 = AD^2 - BC.DE + \frac{1}{4} BC^2

iii) AB^2 + AC^2 = 2 AD^2 + \frac{1}{2} BC^2

Answer:2018-10-13_15-19-36

Given \angle AED = 90^o

\therefore \angle ADE < 90^o and \angle ADC > 90^o

i) In \triangle ADC, \angle ADC is an obtuse angle

\therefore AC^2 = AD^2 + DC^2 + 2 DC.DE

\Rightarrow AC^2 = AD^2 + ( \frac{1}{2} BC)^2 + 2 . \frac{1}{2} BC .DE

\Rightarrow AC^2 = AD^2 + \frac{1}{4} BC^2 + BC.DE

\Rightarrow AC^2 =AD^2 + BC.DE + \frac{1}{4} BC^2

ii) In \triangle ABD, \angle ADE is an acute angle

\therefore AB^2 = AD^2 + BD^2 - 2 BD .DE

\Rightarrow AB^2 = AD^2 + (\frac{1}{2} BC)^2 + 2 . \frac{1}{2} BC.DE

\Rightarrow AB^2 = AD^2 + \frac{1}{4} BC^2 - BC .DE

\Rightarrow AB^2 = AD^2 - BC.DE + \frac{1}{4} BC^2

iii) From i) and ii) we get

AB^2 + AC^2 = 2 AD^2 + \frac{1}{2} BC^2

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Question 13: In an equilateral \triangle ABC, BC is trisected at D . Prove that 9 AD^2 = 7 AB^2 2018-10-13_15-19-51

Answer:

BD = \frac{1}{3} BC

Draw AE \perp BC

Join AD

In \triangle AEB and \triangle AEC , we have AB = AC

\angle AEB = \angle AEC = 90^o

and AE is common

\therefore \triangle AEB \sim \triangle AEC

Therefore we have

BD = \frac{1}{2} BC

DC = \frac{2}{3} BC

BE = EC = \frac{1}{2} BC

Since  \angle C = 60^o (equilateral triangle)

Therefore \triangle ADC is an acute triangle

\therefore AD^2 = AC^2 + DC^2 - 2 DC \times EC

\Rightarrow AD^2 = AC^2 + ( \frac{2}{3} BC)^2 - 2 . \frac{2}{3} BC  \times \frac{1}{2} BC

\Rightarrow AD^2 = AC^2 + \frac{4}{9} BC^2 - \frac{2}{3} BC^2 (since AB = AC = BC )

\Rightarrow AD^2 = AB^2 + \frac{4}{9} AB^2 - \frac{2}{3} AB^2

\Rightarrow AD^2 = \frac{7}{9} AB^2

\Rightarrow 9AD^2 = 7 AB^2

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Question 14: In adjoining figure \triangle ABC is a right angled triangle.  \angle B = 90^o and point D and E trisect BC . Prove  8AE^2 = 3AC^2 + 5AD^2 2018-10-13_15-18-52

Answer:

Given BD = DE = CE

Let BD = DE = CE = x

\Rightarrow BE = 2x and BC = 3x

In \triangle ABD, \triangle ABE and \triangle ABC we have

AD^2 = AB^2 + BD^2

\Rightarrow AD^2 = AB^2 + x^2

AE^2 = AB^2 + BE^2

\Rightarrow AE^2 = AB^2 + 4x^2

and AC^2 = AB^2 + BC^2

\Rightarrow AC^2 = AB^2 + 9x^2

Now 8AE^2 - 3 AC^2 - 5 AD^2 = 8(AB^2 + 4x^2) - 3(AB^2 + 9x^2) - 5(AB^2 + x^2)

\Rightarrow 8AE^2 - 3 AC^2 - 5 AD^2 = 0

\Rightarrow 8AE^2 = 3 AC^2 + 5AD^2

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Question 15: ABC is a right angled triangle. \angle A = 90^o . A circle is inscribed in it. The lengths of the two sides containing the right angle are 6 \ cm and 8 \ cm . Find the radius of the circle.2018-10-13_15-18-24

Answer:

In \triangle ABC , we have

BC^2 = AB^2 + AC^2

BC^2 = 6^2 + 8^2 = 100

\Rightarrow BC = 10 \ cm

Now, Area of \triangle ABC = Area \ of \  \triangle OAB + Area \ of \  \triangle OBC + Area \ of \  \triangle OCA

\Rightarrow \frac{1}{2} AB \times AC = \frac{1}{2} AB \times r  + \frac{1}{2} BC \times r  + \frac{1}{2} CA \times r 

\Rightarrow 48 = 6r + 10r+8r

\Rightarrow r = 48 = 24 r 

\Rightarrow r = 2 \ cm

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Question 16:  In adjoining figure, D  is the mid point of BC  and AE \perp BC  . If BC = a, AC = b, AB = c, ED = x, AD = p  and AE = h  , prove that: 2018-10-14_10-07-44.jpg

i) b^2 = p^2 + ax + \frac{a^2}{4}

ii) c^2 = p^2 - ax + \frac{a^2}{4}

iii) b^2 + c^2 = 2p^2 + \frac{a^2}{2}

Answer:

In \triangle AEB: h^2 + ( \frac{a}{2} -x)^2 = c^2 … … … … … … i)

In \triangle AED: h^2 +x^2 = p^2 … … … … … … ii)

In \triangle AEC: h^2 + (x + \frac{a}{2} )^2 = b^2 … … … … … … iii)

(ii) From i) and ii)

p^2 -x^2 + ( \frac{a}{2} -x)^2 = c^2

p^2 -x^2 + \frac{a^2}{4} + x^2 - ax = c^2

p^2 + \frac{a^2}{4} - ax = c^2

or c^2 = p^2 - ax + \frac{a^2}{4}

(i) from ii) and iii) we get

p^2 -x^2 + (x + \frac{a}{2} )^2 = b^2

p^2 -x^2 + x^2 + \frac{a^2}{4} + ax = b^2

p^2 + \frac{a^2}{4} + ax = b^2

(iii) Adding (ii) and (iii) we get

b^2 + c^2 = p^2 + ax + \frac{a^2}{4} + p^2 -ax + \frac{a^2}{4}

b^2 + c^2 = 2p^2 + \frac{a^2}{2}

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Question 17: In the adjoining figure, \angle B < 90^o, AD \perp BC . Prove:

i) b^2 = h^2 + a^2 + x^2 -2ax 2018-10-13_15-17-44

ii) b^2 = a^2 + c^2 -2ax

Answer:

In \triangle ADB: h^2 + x^2 = c^2 … … … … … … i)

In \triangle ADC: h^2 + (a-x)^2 = b^2 … … … … … … ii)

(i) We know, AC^2 = AB^2 + BC^2 - 2 BC \times BD

b^2 = c^2 +a^2 -2.a.x

b^2 = h^2 + x^2 + a^2 -2ax

or b^2 = h^2 + a^2 + x^2 - 2ax

(ii) From  i and ii)

c^2 - x^2 + (a-x)^2 = b^2

c^2 - x^2 + a^2 + x^2 - 2ax = b^2

c^2 + a^2 -2ax = b^2

or b^2 = a^2 + c^2 -2ax 

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Question 18: \triangle ABD is right angled. \angle A = 90^o and AC \perp BD . Prove that:

i) AB^2 = BC.DC

ii) AC^2 = BC.DC

iii) AD^2 = BD.CD

iv) \frac{AB^2}{AC^2} = \frac{BD}{DC}

Answer:

In \triangle ABC: AB^2 = AC^2 + BC^2 … … … … … … i)

In \triangle ACD: AD^2 = AC^2 + DC^2 … … … … … … ii)

In \triangle ABD: DB^2 = AB^2 + AD^2 … … … … … … iii)

(ii) AB^2 + AD^2 = 2 AC^2 + BC^2 + DC^2

DB^2 = 2 AC^2 + BC^2 + DC^2

(DC + BC)^2 =  2 AC^2 + BC^2 + DC^2

DC^2 + BC^2 + 2 DC.BC =  2 AC^2 + BC^2 + DC^2

AC^2 = DC.BC

(i) AB^2 = DC.BC + BC^2

AB^2 = BC (DC + BC)

AB^2 = BC.DB

(iii) AD^2 = DC.BC + DC^2

AD^2 = DC (BC + DC)

AD^2 = DC.DB

(iv) Dividing i) by ii) we get

\frac{AB^2}{AC^2} = \frac{BC.DB}{DC.BC} = \frac{BD}{DC}

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