myhelper: Exercise 12

Exercise 12

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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 a right angled triangle

(iii) 1.4 cm, 4.8 cm, 5 cm
and (1.4)²+(5)²=1.96+25=26.96

and (4.8)²=23.04

∵(1.4)²+(5)²≠4.8²

∴It is not a right angled triangle

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Question 2

Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. 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

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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 tight?

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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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

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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

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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+(BC+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-BC²

⇒AC²+BC²=4AD²-3AC²

But BC²+AC²=AB²[from (i)]

∴AB²=4AD²-3AC²

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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

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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

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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

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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=$\left(\frac{BC}{2}\right)^2$+AD²$\left[\because \mathrm{BD}=\frac{\mathrm{BC}}{2}\right]$

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

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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

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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

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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

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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

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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

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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 : QX = 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

S.chand Class 8 Maths Solution Chapter 12 Quadrilaterals Exercise 12

Exercise 12

Question 1

ML Aggarwal Solution Class 9 Chapter 12 Pythagoras Theorem Exercise 12

Exercise 12

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question P.Q.

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