ML Aggarwal Solution Class 9 Chapter 12 Pythagoras Theorem Exercise 12

  Exercise 12

Question 1

Lengths of sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:

(i) 3 cm, 8 cm, 6 cm

(ii) 13 cm, .12 cm, 5 cm

(iii) 1.4 cm, 4.8 cm, 5 cm

Sol :

We use Pythagoras Theorem's converse :

(i) Sides of a triangle are 3cm, 8cm, 6cm

⇒3²+6²=9+36=45

and 8²=64

∵45≠64

∴It is not a right triangle

(ii) Sides are 13 cm, 12 cm and 5 cm
⇒12²+5²=144+25=169
and 13²=169
∵12²+5²=13²

∴It is not a right angled triangle


(iii) 1.4 cm, 4.8 cm, 5 cm
and (1.4)²+(4.8)²=1.96+23.04=25
and (5)²=25

∵(1.4)²+(4.8)²=5²

∴It is not a right angled triangle

Question 2

Foot of a 10 m long ladder leaning against a vertical well is 6 m away from the base of the wail. Find the height of the point on the wall where the top of the ladder reaches.

Sol :

Let Ab be wall and AC be the ladder

Ladder AC=10 cm

BC=6 cm

Let height of wall AB=h

By Pythagoras Theorem

⇒AC²=BC²+AB²

⇒10²=6²+h²
⇒100=36+h²
⇒h²=100-36=64=(8)²
∵h=8

∴Height of wall=8 cm

Question 3

A guy attached a wire 24 m long to a vertical pole of height 18 m and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taught?

Sol :

Let AB be the pole and AC be the wire attached

AB=18 m and AC=24 m

In right ΔABC,

⇒AC²=BC²+AB² (Pythagoeas Theorem)

⇒24=BC²+18² 
⇒BC²=24²-18²
⇒BC=√576-324=√252

⇒√4×9×7=2×3√7=6√7 m

Question 4

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Sol :

To poles AB and CD are 12m apart

AB=6 m , CD=11 m

From A, draw AE||BD

Then AE=BD=12 m

CE=CD-ED=CD-AB

=11-6=5 m

Now in right ΔACE

⇒AC²=AE²+CE² (Pythagoras Theorem)

⇒12²+5²=144+25=169=(13)²

∴AC=13 m

∴Distance between their tops=13 m

Question 5

In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is 4:3, find the sides.

Sol :

In right angled triangle hypotenuse=20 cm

ratio of other two sides=4 : 3

Let first side=4x

then Second side=3x

By Pythagoras theorem'

⇒(Hypotenuse)²=(First side)²+(Second side)² 

∴(20)²=(4x)²+(3x)² 

⇒(20)²=16x²+9x² 

⇒400=25x²

⇒x²$=\frac{400}{25}$

⇒x²=16

⇒x=√16=4

∴First side=4x=4×4 cm=16 cm
Second side=3x=3×4 cm=12 cm

Hence , other two sides of right angled triangle=16 cm and 12 cm

Question 6

If the sides of a triangle are in the ratio 3:4:5, prove that it is right-angled triangle.

Sol :

Let three sides of given triangle ABC is AB

BC and CA=3 : 4 : 5 

Let AB=3x , BC=4x and CA=5x

Here (AB)²+(BC)²

=(3x)²+(4x)²

=9x²+16x²=25x²

Also, (CA)²=(5x)²=25x²

i.e. (AB)²+(BC)²=(CA)²

Hence ,ABC is right angled triangle

Question 7

For going to a city B from city A, there is route via city C such that AC ⊥ CB, AC = 2x km and CB=2(x+ 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of highway.

Sol :

In right ΔABC, ∠C=90°

⇒(2x)²+[2(x+7)]²=26²

⇒4x²+4(x²+14x+49)=676

⇒4x²+4x²+56x+196-676=0

⇒8x²+56x-480=0

⇒x²+7x-60=0  (Dividing by 8)

⇒x(x+12)-5(x+12)=0

⇒(x+12)(x-5)=0

Either x+12=0, then x=-12 which is not possible being negative 

or x-5=0, then x=5

Now distance between AC=2x

=2×5=10km

and between BC=2(x+7)=2(5+7)

=2×12=24

∴Distance from A to C and B to C=10+24=34 km

∴Distance saved=34-26=8km

Question 8

The hypotenuse of right triangle is 6m more than twice the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle.

Sol :

Let the shortest side of right angled triangle= x m

Hypotenuse=(2x+6) m

Third side=[(2x+6)-2] m

By Pythagoras theorem,

⇒(2x+6)²=x²+[(2x+6)-2]²

⇒4x²+36+24x=x²+(2x+4)²

⇒4x²+36+24x=x²+4x²+16+16x

⇒36+24x=x²+16+16x

⇒0=x²+16+16x-36-24x

⇒0=x²-8x-20

⇒x²-8x-20=0

⇒x-10x+2x-20=0

⇒x(x-10)+2(x-10)=0

Either x+2=0 or x-10=0

x=-2 (Which is not possible)

or x=10

Hence , shortest=x=10 m

Hypotenuse=(2x+6)=(2×10+6)=26 m

Third side=(2x+6)-m=26m-24m=24 m

Question 9

ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

Sol :

ΔABC is an isosceles right triangle, right angle at C, AC=BC

To prove : AB²=AC²

Proof : In right ΔABC

⇒∠C=90°

⇒AB²=AC²+BC² (Pythagoras Theorem)

=AC²+BC² (∵BC=AC)

=2AC²

Question 10

In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².

Sol :

In ΔABC, AD⟂BC

To prove : AB²+CD²=AC²+BD²

Proof : In ΔABC , AD⊥BC

∴ΔABD and ΔACD are right triangle

In right ΔADB

⇒AB²=AD²+BC² (Pythagoras theorem)

⇒AD²=AB²-BD²...(i)

Similarly in right ΔADB

⇒AC²=AD²+CD²

⇒AD²=AC²-CD²..(ii)

From (i) and (ii)

⇒AB²-BD²=AC²-CD²

⇒AB²+CD²=AC²+BD²

Question 11

In ∆PQR, PD ⊥ QR, such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a – b) = (c + d) (c – d).

Sol :

In ΔPQR, PQ⟂QR

PQ=a , PR=b, QD=c, DE=d

To prove : (a+b)(a-b)=(c+d)(c-d)

Proof : In ΔPQR, PQ⟂QR

Now in right ΔPQD

⇒PQ²=PD²+QD² (Pythagoras theorem)

⇒PD²=PQ²-QD²=a²-c²..(i)

Similarly in right ΔPDR

⇒PR²=PD²+DR²

⇒PD²=PR²-DR²

⇒b²-d²...(ii)

From (i) and (ii)

⇒a²-c²=b²-d²

⇒a²-b²=c²-d²

⇒(a+b)(a-b)=(c+d)(c-d)

Question 12

ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. Find the altitude on BC and Hence, calculate its area.

Sol :

To find : Altitude on BC i.e. value of AD 

In isosceles triangle perpendicular from vertex bisects the base

∴BD=DC

∴BD$=\frac{1}{2}\times 8$=4cm

In right angled triangle ABD

By Pythagoras theorem

⇒AD²+BD²=AB²

⇒AD²+(4)²=(12)²

⇒AD²+16=144

⇒AD²=128

⇒AD=√128=√64×2=8√2

∴Altitude of ΔABC$=\frac{1}{2}\times (base)\times (altitude)$

$=\frac{1}{2}\times 8\times 8\sqrt{2}cm^2$

=4×8√2 cm²

=32√2 cm²

Question 13

Find the area and the perimeter of a square whose diagonal is 10 cm long.

Sol :

Let ABCD be a square whose diagonal AC=10 cm

Let length of sides of squared=x cm

In ΔABC

By Pythagoras theorem

⇒AC²+AB²+BC²

⇒(10)²=x²+x²

⇒2x²=100

⇒x²$=\frac{100}{2}=50$

⇒x=√50

⇒x=√25×2

⇒x=5√2

Area of square=side×side

=5√2×5√2

=25×2 cm²

Perimeter of square=4×side

=4×5√2 cm

=20√2 cm

Question 14

(a) In fig. (i) given below, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, ∠ ABD = ∠BCD = 90°. Calculate the length of AB.

(b) In fig. (ii) given below, ABCD is a quadrilateral in which AB = AD, ∠A = 90° =∠C, BC = 8 cm and CD = 6 cm. Find AB and calculate the area of ∆ ABD

Sol :

(a) Given : ABCD is a quadrilateral in which AD=13 cm, DC=12 cm, BC=3 cm, and ∠ABD=∠BCD=90°

To calculate : The length of AB

In right angled triangle BCD

By Pythagoras theorem

⇒BD²=BC²+DC²

⇒BD²=(3)²+(12)²

⇒BD²=9+144

⇒BD²=153

Now, In right angled ΔABD,

By Pythagoras theorem

⇒AD²=AB²+BD²

⇒AB²=AD²-BD²

⇒(13)²-153  (∵BD²=153)

⇒169-153=16

⇒AB=√16=4

Hence, Length of AB=4 cm

(b) In right angled triangle BCD

By Pythagoras theorem

⇒BD²=BC²+CD²

⇒(8)²+(6)²

⇒64+36=100

⇒BD=√100=10 cm

∴BD=10 cm

In right angled triangle ABD,

⇒BD²=AB²+AD²

⇒BD²=AB²+AB² (∵AB=AD (given))

⇒(10)²=2AB²

⇒2AB²=100

⇒AB²$=\frac{100}{2}=50$

⇒AB=√50

⇒√25×2=5√2

∴AB=5√2 cm

Area of ΔABD$=\frac{1}{2}\times AB\times AD$

$=\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}~cm^2$  (∵AB=AD)

$=\frac{25\times 2}{2}$

=25 cm²

Question 15

(a) In figure (i) given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm.Calculate the length of BD.

(b) In figure (ii) given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

(c) In figure (iii) given below, ∠ D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.

Sol :

(a) Here AB=12 cm, AC=13 cm , CE=10 cm and DE=6 cm

To calculate the length of BD

In right angled ΔABC

By Pythagoras theorem,

⇒AC²=AB²+BC²

⇒(13)²=(12)²+BC²

⇒BC²=(13)²-(12)²

⇒BC²=169-144

⇒BC²=25

⇒BC=√25=5

∴BC=5 cm...(1)

In right angled ΔCED

By Pythagoras theorem

⇒CE²=CD²+DE²

⇒(10)²=CD²+(6)²

⇒CD²=100-36

⇒CD²=64

⇒CD=√64

⇒CD=8...(2)

∴CD=8 cm

Hence, length of BD=BC+CD

=5 cm+8 cm  [Putting from (1) and (2)]

=13 cm

(b) Here ∠PSR=90°

PQ=10 cm, QS=6 cm and RQ=9 cm

To calculate the length of PR

In right angled ΔPQS

By Pythagoras theorem

⇒PQ²=PS²+QS²

⇒(10)²=PS²+(6)²

⇒(10)²-(6)²=PS²

⇒100-36=PS²

⇒PS²=64

⇒PS=√64=8

∴PS=8 cm

Now, in right angled ΔPSR

By Pythagoras theorem

⇒PR²=PS²+RS²

⇒PR²=(8)²+(15)²  (RS=RQ+QS)

⇒PR²=64+225

=(9+6)cm=15cm

⇒PR²=289

⇒PR=√289=17

∴PR=17 cm

(c) Here ∠D=90°

⇒AB=16 cm, BC=12 cm and CA=6 cm

To find CD

Let the value of CD= x cm

By Pythagoras theorem

⇒AB²=AD²+BD²

⇒(16)²=AD²+(BC+CD)²

⇒(16)²=AD²+(12+x)²

⇒AD²=(16)²-(12+x)²...(1)

Now, in right angle ΔACD

By Pythagoras theorem

⇒AC²=AD²+CD²

⇒(6)²=[(16)²-(12+x)]²+x² (∵Find (1) putting the value of AD)

⇒36=256-(144+x²+24x)+x²

⇒36=256-144-x²-24x+x²

⇒36=256-144-24x

⇒24x=256-144-36

⇒24x=76

⇒$x=\frac{76}{24}=\frac{19}{6}=3\frac{1}{6}$

Hence , CD$=3\frac{1}{6}$ cm

Question 16

(a) In figure (i) given below, BC = 5 cm,

∠B =90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.

(b) In the figure (ii) given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB2 = 4AD² – 3AC².

Sol :

(a) Here BC=5 cm , ∠B=90° , AB=5AE , CD=2AE, AC=ED

To calculate the lengths of EA, CD , AB and AC

In right angled ΔABC

By Pythagoras theorem

⇒AC²=AB²+BC²...(i)

Also, in right angled ΔBED

⇒ED²,in right angled ΔBED

⇒ED²=BE²+BD²...(ii)

But AC=ED

⇒AC²=ED²...(iii)

From (i) , (ii) and (iii)

⇒AB²+BC²=BE²+BD²

⇒(5EA)²+(5)²=(4EA)²+(BE+CD)²

(∵BE=AB-EA=5EA-EA=4EA)

⇒25EA²+25=16EA²+(5+2EA)²

(∵CD=2EA)

⇒25EA²+25-16EA²=25+4EA²+20EA

⇒25x²+25-16x²=25+4x²+20x (Let EA=x cm)

⇒9x²-4x²=20x

⇒5x²=20x

⇒x=4 cm  (∵x≠0)

∴EA=4 cm

CD=2AE=2×4 cm=8 cm

AB=5AE=5×4 cm=20 cm

In right angled ΔABC

By Pythagoras theorem

⇒AC²=AB²+BC²

⇒AC²=(20)²+(5)²

⇒AC²=400+25=425

⇒AC=√425=√25×17=5√17

Hence , AC=5√17

(b) In right ΔABC, ∠C=90°

D is mid point of BC

To prove : AB²=4AD²-3AC²

Proof : In right ΔABC,∠C=90° 

⇒AB²=AC²+BC²...(i)

(Pythagoras theorem)

But in right ΔADC

⇒AD²=AC²+DC²

⇒AC²=AD²-DC²...(ii)

From (i) and (ii)

⇒AC²=AD²$-\left(\frac{BC}{2}\right)^2$

(∵D is mid point of BC)

⇒AC²=AD²$-\frac{BC^2}{4}$

⇒4AC²=4AD²-BC²

⇒AC²+3AC²=4AD²-3AC²

But BC²+AC²=AB²  [from (i)]

∴AB²=4AD²-3AC²

Question 17

In ∆ABC, AB = AC = x, BC = 10 cm and the area of ∆ ABC is 60 cm². Find x.

Sol :

Given : In ∆ABC, AB=AC=x, BC=10 cm and area of ∆ABC=60 cm²

In isosceles triangle ABC

⇒BD$=\frac{1}{2}\times$ BC

⇒BD$=\frac{1}{2}\times 10$ cm=5 cm

In
right angled ABD

By Pythagoras theorem

⇒AB²=BD²+AD²

⇒x²=(5)²+AD²

⇒AD²=x²-25²

⇒AD$=\sqrt{x^2-25}$

Area of ΔABC$=\frac{1}{2}\times base \times height$

⇒$60=\frac{1}{2} \times 10 \times \sqrt{x^{2}-25}$

⇒$\frac{60 \times 2}{10}=\sqrt{x^{2}-25}$

⇒$12=\sqrt{x^{2}-25}$

Squaring both sides , we get

⇒$(12)^{2}=\left(\sqrt{x^{2}-25}\right)^{2}$

⇒144=x²-25

⇒144+25=x²

⇒x²=169

⇒x=√169=13

∴Hence , x=13 cm

Question 18

In a rhombus, If diagonals are 30 cm and 40 cm, find its perimeter.

Sol :

Given : AC=30 cm and BD=40 cm where AC and BD are diagonals of rhombus ABCD

Required : Side of rhombus 

We know that in rhombus diagonals are bisect each other also perpendicular to each other

∴AO$=\frac{1}{2}$AC$=\frac{1}{2}\times 30$ cm=15 cm

and BO$=\frac{1}{2}BD$ $=\frac{1}{2}\times 40$=20 cm

In right angled ΔAOB

By Pythagoras theorem

⇒AB²=AO²+BO²

⇒(15)²+(20)²

⇒225+400=625

⇒AB=√625=25

Side of rhombus (a)=25 cm

Perimeter of rhombus=4a=4×25=100 cm

Question 19

(a) In figure (i) given below, AB || DC, BC = AD = 13 cm. AB = 22 cm and DC = 12cm. Calculate the height of the trapezium ABCD.

(b) In figure (ii) given below, AB || DC, ∠ A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.

(c) In figure (iii) given below, ABCD is a square of side 7 cm. if

AE = FC = CG = HA = 3 cm,

(i) prove that EFGH is a rectangle.

(ii) find the area and perimeter of EFGH.

Sol :

(a) Given : AB||DC , BC=AD=13 cm, AB=22 cm and DC=12 cm

Required : Height of trapezium ABCD

Here CD=MN=12 cm

Also, AM=BN

∴AB=AM+MN+BN

⇒22=AM+12+AM

⇒22-12=2AM

⇒10=2AM

⇒AM$=\frac{10}{2}=5$

∴AM=5 cm

In right angled ΔAMD

⇒AD²=AM²+DM²

⇒(13)²=(5)²+DM²

⇒DM²=(13)²-(5)²

⇒DM²=169-25=144

⇒DM²=√144=12 cm

Hence , height of trapezium=12 cm

(b) Given : AB||DC ,∠A=90°, DC=7 cm

AB=7 cm and AC=25 cm

Required : BC

In right angled triangle

⇒AC²=AD²+CD² (By Pythagoras theorem)

⇒(25)²=AD²+(7)² 

⇒AD²=625-49=576 

⇒AD=√576=24

∴AD=24 cm

Also, DM=MC=24 cm  (∵AB||DC)

Also, Am=DC=7 cm

i.e. AM=7 cm

∴BM=AB-AM=10 cm

In right angled triangle

⇒BC²=MC²+BM² (By Pythagoras theorem)

⇒(24)²+(10)²

⇒576+100=676=(26)²

⇒BC=26

∴BC=26 cm

(c) Given : ABCD is a square of side=7 cm

AE=FC=CG=HA=3 cm

To prove : (i) EFGH is a rectangle

(ii) To find the area and perimeter of EFGH

Proof : BE=BF=DG=DH=7-3=4 cm

In right angled ΔAEH

⇒HE²=HA²+AE² (By Pythagoras theorem)

⇒HE²=(3)²+(3)² 

⇒9+9=18

⇒HE=√18=3√2 cm

∴HE=GF=3√2 cm

Again In right angled ΔEBF

⇒EF²=EB²+BF²

⇒(4)²+(4)²

⇒16+16=32

EF=√32=√16×2=4√2 cm

∴EF=HG=4√2 cm

Join EG

In ΔEFG

⇒EF²+GF²

=(3√2)²+(4√2)²

=18+32=50

Also, ⇒EH²+HG²=(3√2)²+(4√2)²

=18+32=50

∴EF²+GF²=EH²+HG²

i.e. EG²=HF²

i.e. EG=HF

i.e.Diagonals of quadrilaterals are equal

∴EFGH is a rectangle 

Area of rectangle EFGH=HE×EF

=3√2×4√2 cm²

=24 cm²

Perimeter of rectangle EFGH=2(EF+HE)

=2(4√2+3√2)

=2×7√2 cm

=14√2 cm

Question 20

AD is perpendicular to the side BC of an equilateral Δ ABC. Prove that 4AD² = 3AB².

Sol :

Given : ABC is an equilateral triangle and AD⟂BC

AD⟂BC

To prove : 4AD² = 3AB².

Proof : Since ABC is an equilateral triangle 

∴AB=BC=CA

In right angled triangle ABD

⇒AB²=BD²+AD² (by Pythagoras theorem)

⇒AB²=$\left(\frac{BC}{2}\right)^2$+AB²  $\left[\because \mathrm{BD}=\frac{\mathrm{BC}}{2}\right]$

⇒$\mathrm{AB}^{2}=\frac{(\mathrm{AB})^{2}}{4}+\mathrm{AD}^{2}$  [∵AB=BC]

⇒$A B^{2}-\frac{A B^{2}}{4}=A D^{2}$

⇒$\frac{4 \mathrm{AB}^{2}-\mathrm{AB}^{2}}{4}=\mathrm{AD}^{2}$

⇒$\frac{3 \mathrm{AB}^{2}}{4}=\mathrm{AD}^{2}$

⇒$3 \mathrm{AB}^{2}=4 \mathrm{AD}^{2}$

⇒$4 \mathrm{AD}^{2}=3 \mathrm{AB}^{2}$

Hence ,the result is proved

Question 21

In figure (i) given below, D and E are mid-points of the sides BC and CA respectively of a ΔABC, right angled at C.

Sol :

Prove that :

(i) 4AD²=4AC²+BC²

(ii) 4BE²=4BC²+AC²

(iii) 4(AD²+BE²)=5AB²

(a) Given : In ΔABC , right angled at C. D and E are mid points of the sides BC and CA respectively

To prove : (i) 4AD²=4AC²+BC²

(ii) 4BE²=4BC²+AC²

(iii) 4(AD²+BE²)=5AB²

Proof : In right angle ΔACD,

⇒AD²=AC²+CD² (by Pythagoras theorem)

⇒4AD²=4AC²+4BD²

(Multiplying both sides by 4)

⇒4AD²=4AC²+(2BD)²

⇒4AD²=4AC²+BC²...(1)

(∵2BD=BC ∴D is mid points of BC)

(ii) In right angled ΔBCE

⇒BE²=BC²+CE²  (By Pythagoras theorem)

⇒4BE²=4BC²+4CE² (Multiplying both sides by 4)

⇒4BE²=4BC²+(2CE)²

⇒4BE²=4BC²+AC²...(1)

(∵2CE=AC ∴E is mid points of AC)

Adding (1) and (2) , we get

⇒4AD²+4BE²=4AC²+BC²+AC²

⇒4(AD²+BE²)=5AC²+5BC²

⇒5(AC²+BC²)

⇒5(AB²)

(∵ In right angled ΔABC, AC²+BC²=AB²)

Hence , 4(AD²+BE²)=5AB²

Question 22

If AD, BE and CF are medians of ΕABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

Sol :

Given : AD, BE and CF are medians of ΔABC.

To prove : 3(AB²+BC²+CA²)=4(AD²+BE²+CF²)

Construction : Draw AP⟂BC

Proof : In right angled ΔAPB

⇒AB²=AP²+BP²

⇒AP²+(BD-PD)²

⇒AP²+BD²+PD²-2BD.PD

⇒(AP²+PD²)+BD²-2BD.PD

⇒$\mathrm{AD}^{2}+\left(\frac{1}{2} \mathrm{BC}\right)^{2}-2 \times\left(\frac{1}{2} \mathrm{BC}\right) \cdot \mathrm{PD}$

(∵AP²+PD²=AD² and $\mathrm{BD}=\frac{1}{2} \mathrm{BC}$)

⇒$A D^{2}+\frac{1}{4} B C^{2}-B C . P D$...(1)

Now , in ΔAPC

⇒AC²=AP²+PC² (By Pythagoras theorem)

⇒AP²+(PD+DC)²

⇒AP²+PD²+DC²+2PD.DC

⇒$\left(\mathrm{AP}^{2}+\mathrm{PD}^{2}\right)+\left(\frac{1}{2} \mathrm{BC}\right)^{2}+2 \mathrm{PD} \times\left(\frac{1}{2} \mathrm{BC}\right)$ $\left(\because \mathrm{DC}=\frac{1}{2} \mathrm{BC}\right)$

⇒$\mathrm{AD}^{2}+\frac{1}{4} \mathrm{BC}^{2}+\mathrm{PD} \cdot \mathrm{BC}$...(2)

Adding (1) and (2) 

∴$A B^{2}+A C^{2}=2 A D^{2}+\frac{1}{2} B C^{2}$...(3)

Similarly . Draw the perpendicular from B and C on AC and AB respectively, we get

⇒$\mathrm{BC}^{2}+\mathrm{CA}^{2}=2 \mathrm{CF}^{2}+\frac{1}{2} \mathrm{AB}^{2}$...(4)

⇒$\mathrm{AB}^{2}+\mathrm{BC}^{2}=2 \mathrm{BE}^{2}+\frac{1}{2} \mathrm{AC}^{2}$...(5)

Adding (3), (4) and (5), we get

⇒2(AB²+BC²+CA²)

⇒$2\left(\mathrm{AD}^{2}+\mathrm{BE}^{2}+\mathrm{CF}^{2}\right)+\frac{1}{2}\left(\mathrm{BC}^{2}+\mathrm{AB}^{2}+\mathrm{AC}^{2}\right)$

⇒$2\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)-\frac{1}{2}\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\right.$$\left.\mathrm{CA}^{2}\right)=2\left(\mathrm{AD}^{2}+\mathrm{BE}^{2}+\mathrm{CF}^{2}\right)$

⇒$\frac{3}{2}\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)=2\left(\mathrm{AD}^{2}+\mathrm{BE}^{2}+\mathrm{CF}^{2}\right)$

∴3(AB²+BC²+CA²)=4(AD²+BE²+CF²)

Hence proved

Question 23

(a) In fig. (i) given below, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that

AB² + CD² = AD² + BC².

(b) In figure (ii) given below, OD⊥BC, OE ⊥CA and OF ⊥ AB. Prove that :

(i) OA² + OB² + OC² = AF² + BD² + CE² + OD² + OE² + OF².

(ii) OAF² + BD² + CE² = FB² + DC² + EA².

Sol :

Given : In quadrilateral ABCD the diagonals AC and BD intersects at O at right angles

To prove : AB²+CD²=AD²+BC²

Proof : In right angled ΔAOB

⇒AB²=AO²+OB²...(1)

(By Pythagoras theorem)

In right angled ΔCOD

⇒CD²=OD²+OC²...(2)

Adding (1) and (2) 

⇒AB²+CD²=(AO+OB)²+(OD²+OC²)

⇒AB²+CD²=(OA²+OD²)+(OB²+OC²)..(3)

Now , in right angled triangle AOD and BOC

By Pythagoras theorem

⇒OA²+OD²=AD²...(4)

⇒OB²+OC²=BC²...(4)

From (3),(4) and (5), we get

⇒AB²+CD²=AD²+BC²

Hence , the result

(b) Given : OD⊥BC, OE⊥CA and OF⊥AB

To prove : 

(i) OA²+OB²+OC²=AF²+BD²+CE²+OD²+OE²+OF²

(ii) AF²+BD²+CE²=FB²=DC²+EA²

Proof : 
In right angled ΔAOF

⇒OA²=AF²+OF²...(1)
In right angled ΔBOD 

⇒OB²=BD²+OD²...(2)

In right angled ΔCOE

⇒OC²=CE²+OE²...(3)

Adding (1),(2) and (3) , we get
⇒OA²+OB²+OC²=AF²+BD²+CE²+OD²+OE²+OF²...(proved (i) part)

(ii) Also, OA²+OB²+OC²
=AF²+BD²+CE²=OA²+OB²+OC²-OD²-OE²-OF²...(4)

Again in ΔBOF, ΔCOD, ΔAOE, 
⇒BF²=OB²-OF²
⇒DC²=OC²-OD²
and EA²=OA²-OE²

Adding above , we get

⇒BF²+DC²+EA²=OB²-OF²+OC²-OD²+OA²-OF²

⇒BF²+DC²+EA²=OA²+OB²+OC²-OD²-OE²-OF²...(5)

From (4) and (5)

⇒AF²+BD²+CE²=BF²+DC²+EA²

Hence , the result

Question 24

In a quadrilateral, ABCD ,∠B = 90° = ∠D. Prove that 2 AC² – BC2 = AB² + AD² + DC².

Sol :

Given : In quadrilateral ABCD , ∠B = 90° and ∠D = 90°

To prove : 2AC² – BC2 = AB² + AD² + DC².

Construction : Join AC

Proof : In right angled ΔACD

⇒AC²=AB²+BC²...(1)

(by Pythagoras theorem)

In right angled ΔACD

⇒AC²=AD²+DC²...(2)

(by Pythagoras theorem)

Adding (1) and (2) , we get

⇒AC²+AC²=AB²+BC²+AD²+DC²

⇒2AC²=AB²+BC²+AD²+DC²

⇒2AC²-BC²=AB²+AD²+DC²

Hence , the result

Question 25

In a ∆ABC, ∠A = 90°, CA = AB and D is a point on AB produced. Prove that :

DC² – BD² = 2AB. AD.

Sol :

Given : ∆ABC in which ∠A = 90° , CA=AB and D is point on AD produced

To prove : DC²-BD²=2AB.AD

Proof : In right angled ΔACD

⇒DC²=AC²+AD²
⇒DC²=AC²+(AB+BD)²

⇒DC²=AC²+AB²+BD²+2AB.BD

⇒DC²-BD²=AC²+AB²+2AB.BD

But AC=AB (given)

⇒DC²-BD²=AB²+AB²+2AB.BD

⇒DC²-BD²=2AB²+2AB.BD

⇒DC²-BD²=2AB(AB+BD)

⇒DC²-BD²=2AB.AD

Hence , the result

Question 26

In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD² = AC² + BD.CD.

Sol :

Given : Isosceles ΔABC such that AB=AC.

D is mid point on BC produced

To prove : AD²=AC²+BD.CD

Construction : Draw AP⊥BC

Proof : In right angled ΔAPD

⇒AD²=AP²+PD²

⇒AD²=AP²+(PC+CD)²

⇒AD²=AP²+PC²+CD²+2PC.CD

In right angled ΔAPC

⇒AC²=AP²+PC²

∴AD²=AC²+CD²+2PC.CD

But ΔABC is isosceles triangle and AP⟂BC
∴$\mathrm{PC}=\frac{1}{2} \mathrm{BC}$
∴AD²=AC²+CD²+$2\times \frac{1}{2}$BC.CD
⇒AD²=AC²+CD²+BC.CD

⇒AD²=AC²+CD.BD

i.e. AD²=AC²+BD.CD

Hence, the result

Question P.Q.

(a) In figure (i) given below, PQR is a right angled triangle, right angled at Q. XY is parallel to QR. PQ = 6 cm, PY = 4 cm and PX : OX = 1:2. Calculate the length of PR and QR.

(b) In figure (ii) given below, ABC is a right angled triangle, right angled at B.DE || BC.AB = 12 cm, AE = 5 cm and AD : DB = 1: 2. Calculate the perimeter of A ABC.

(c)In figure (iii) given below. ABCD is a rectangle, AB = 12 cm, BC – 8 cm and E is a point on BC such that CE = 5 cm. DE when produced meets AB produced at F.

(i) Calculate the length DE.

(ii) Prove that ∆ DEC ~ AEBF and Hence, compute EF and BF.

Sol :

(a) Given : In right angled ΔPQR, XY||QR , PQ=6 cm , PY=4 cm and PX : QX=1 : 2

Required : The length of PR and QR

PX : QX=1: 2

Let PX=x cm
then QX=2x cm

∴PQ=PX+QX

⇒6=x+2x

⇒3x=6
⇒$x=\frac{6}{3}=2$

∴PX=2 cm and QX=2×2cm=4 cm

In right angled ΔPXY

⇒PY²=PX²+XY² (by Pythagoras theorem)

⇒(4)²=(2)²+XY² 
⇒XY² =(4)²-4

⇒XY² =12

⇒XY=√12=2√3

Also ,XY||QR

⇒$\frac{P X}{P Q}=\frac{X Y}{Q R}$

⇒$\frac{2}{6}=\frac{2 \sqrt{3}}{\mathrm{QR}}$

⇒6√3

Also , $\frac{P X}{P Q}=\frac{P Y}{P R}$

⇒$\frac{2}{6}=\frac{4}{P R}$

⇒PR$=\frac{6 \times 4}{2}=\frac{24}{2}$=12 cm

Hence, PR=12 cm and QR=6√3 cm

(b) Given : In right angled ΔABC

∠B=90° , DE||BC , AB=12 cm, AE=5 cm and 

AD : DB = 1 : 2

Required : The perimeter of ΔABC

AD :  DB= 1 : 2

Let AD=x cm

then DB=2x cm

∴AB=AD+DB

⇒12=x+2x

⇒3x=12

⇒$x=\frac{12}{3}=4$

∴AD=x=4 cm and DB=2x=2×4=8cm

In right angled ΔADE

⇒AE²=AD²+DE²  (by Pythagoras theorem)

⇒(5)²=(4)²+DE² 

⇒25=16+DE² 

⇒DE²=25-16 

⇒DE²=9 

⇒DE=√9=3 cm

Now, DE||BC (given)

∴$\frac{A D}{A B}=\frac{D E}{B C}$

⇒$\frac{4}{12}=\frac{3}{\mathrm{BC}}$

⇒$\mathrm{BC}=\frac{12 \times 3}{4}=$

⇒3×3=9 cm

Also , $\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$

⇒$\frac{4}{12}=\frac{5}{\mathrm{AC}}$

⇒$\mathrm{AC}=\frac{12 \times 5}{4}$

⇒3×5=15

Perimeter of ΔABC=AB+BC+AC

=12cm+9cm+15cm=36cm

(c) Given : ABCD is rectangle , AB=12cm , BC=8 cm, and E is a point on BC such that CE=5 cm

Required : (i) The length of DE

(ii) To prove : ΔDEC~ΔEBF and Hence, find EF and BF

(i) In right angled ΔCDE

DE²=CD²+CE²

DE²=AB²+CE² [CD=AB]

DE²=(12)²+(5)²

DE²=144+25=169

DE=√169=13 cm

(ii) In ΔDEC and ΔEBF

∠DEC=∠BEF (vertically opposite angles)

∠DCE=∠EBF (each 90°)

∴ΔDCE ~ ΔEBF (by A.A axiom of similarity)

∴$\frac{C E}{B E}=\frac{D E}{E F}$

⇒$\frac{5}{3}=\frac{13}{\mathrm{EF}}$  (∵BE=8 cm-5 cm=3 cm)

⇒5×EF=13×3

⇒EF$=\frac{13 \times 3}{5}=\frac{39}{5}$=7.8 cm

Also ,$\frac{C E}{B E}=\frac{D E}{B F}$

⇒$\frac{5}{3}=\frac{12}{\mathrm{BF}}$

(∵BF=8.5 cm-5 cm=3 cm also CD=AB=12 cm)

⇒BF×5=12×3

⇒BF$=\frac{12 \times 3}{5}=\frac{36}{5}$=7.2 cm

Hence , DE=13 cm, EF=7.8 cm and BF=7.2 cm