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ML Aggarwal Solutions For Grade 9 Maths Affiliate 12 Pythagoras Theorem helps students to master the concept of Pythagoras theorem.

These solutions provide students an advantage with practical questions.

This chapter deals with Pythagoras theorem and its different applications. ML Aggarwal Solutions can be downloaded from our website in PDF format for free. These solutions are prepared by our skilful team. The concepts covered in this affiliate are explained in a uncomplicated mode and then that whatever pupil tin can easily understand.

Pythagoras theorem is the fundamental theorem in Mathematics, which defines the relation between the hypotenuse, base and altitude of a correct angled triangle. Co-ordinate to this theorem, the square of hypotenuse is equal to the sum of squares of distance and base of a correct angled triangle.

In ML Aggarwal solutions For Class nine Maths Chapter 12, we come across solving dissimilar awarding questions.

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Practice 12

Solution:

We use the Pythagoras theorem to bank check whether the triangles are right triangles.

We have hⁱⁱ
= b²+a²
[Pythagoras theorem]

Where h is the hypotenuse, b is the base and a is the altitude.

(i)Given sides are iii cm, 8 cm and half-dozen cm

b^two+a^two
= three^two+ 6^two
= 9+36 = 45

h²
= 8²
= 64

hither 45 ≠ 64

Hence the given triangle is not a correct triangle.

(two) Given sides are thirteen cm, 12 cm and 5 cm

b²+aⁱⁱ
= 12²+ 5²
= 144+25 = 169

h²
= xiii²
= 169

here b²+a²
= h^two

Hence the given triangle is a right triangle.

Length of the hypotenuse is thirteen cm.

(iii) Given sides are 1.4 cm, iv.8 cm and 5 cm

b^two+a²
= 1.4²+ 4.8²
= 1.96+23.04 = 25

hⁱⁱ
= 5ⁱⁱ
= 25

here b²+a²
= h²

Hence the given triangle is a right triangle.

Length of the hypotenuse is 5 cm.

ii. Foot of a 10 k long ladder leaning against a vertical well is 6 m away from the base of the wall. Observe the height of the indicate on the wall where the top of the ladder reaches.

Solution:

Allow PR be the ladder and QR be the vertical wall.

Length of the ladder PR = x m

PQ = 6 m

Let height of the wall, QR = h

Co-ordinate to Pythagoras theorem,

PR^two
= PQ²+QR²

10ⁱⁱ
= 6²+QR²

100 = 36+QR²

QR²
= 100-36

QR²
= 64

Taking square root on both sides,

QR
= 8

Hence the summit of the wall where the top of the ladder reaches is 8 m.

three. A guy attached a wire 24 thou long to a vertical pole of height eighteen m and has a stake attached to the other stop. How far from the base of operations of the pole should the pale be driven and then that the wire will be tight?

Solution:

Let Air-conditioning be the wire and AB be the height of the pole.

Air-conditioning = 24 cm

AB = 18 cm

According to Pythagoras theorem,

AC^two
= ABⁱⁱ+BC^two

24^two
= 18²+BC²

576 = 324+BC²

BC²
= 576-324

BC²
= 252

Taking square root on both sides,

BC = √252

= √(iv×ix×7)

= 2×iii√7

= 6√seven cm

Hence the altitude
is 6√vii cm.

four.
2 poles of heights six m and 11 m stand up on a plane basis. If the distance between their feet is 12 g, find the distance between their tops.

Solution:

Let AB and CD be the poles which are 12 m autonomously.

AB = 6 one thousand

CD = 11 m

BD = 12 thou

Draw AE BD

CE = 11-6 = v m

AE = 12 k

Co-ordinate to Pythagoras theorem,

AC²
= AE²+CE^two

AC²
= 12²+5²

Ac²
= 144+25

Air-conditioning²
= 169

Taking square root on both sides

Air-conditioning = xiii

Hence the distance betwixt their tops is 13 m.

5.
In a correct-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is 4:iii, detect the sides.

Solution:

Given hypotenuse, h = 20 cm

Ratio of other 2 sides, a:b = 4:3

Let altitude of the triangle be 4x and base be 3x.

According to Pythagoras theorem,

h^two
= b^two+aⁱⁱ

20²
= (3x)ⁱⁱ+(4x)²

400 = 9x²+16x²

25xⁱⁱ
= 400

ten²
= 400/25

x²
= 16

Taking foursquare root on both sides

x = 4

so base of operations, b = 3x = 3×4 = 12

altitude, a = 4x = 4×four = 16

Hence the other sides are 12 cm and 16 cm.

6.
If the sides of a triangle are in the ratio iii:4:v, prove that it is right-angled triangle.

Solution:

Given the sides are in the ratio three:4:5.

Let ABC be the given triangle.

Let the sides exist 3x, 4x and hypotenuse exist 5x.

According to Pythagoras theorem,

AC^two
= BCⁱⁱ+ABⁱⁱ

BC²+AB²= (3x)²+(4x)ⁱⁱ

= 9x²+16x²

= 25x²

AC²
= (5x)²
= 25x²

ACⁱⁱ
= BC²+AB²

Hence ABC is a right angled triangle.

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=two(ten+ 7) km. It is proposed to construct a 26 km highway which straight connects the two cities A and B. Find how much distance will be saved in reaching metropolis B from urban center A after the construction of highway.

Solution:

Given AC = 2x km

CB = 2(x+7)km

AB = 26

Given Air conditioning CB.

According to Pythagoras theorem,

ABⁱⁱ
= CB^two+Ac^two

26²
= ( ii(x+7))²+(2x)²

676 = 4(x²+14x+49) + 4x²

4x²+56x+196+4x^two
= 676

8x^two+56x+196 = 676

8x²+56x +196-676 = 0

8x²+56x -480 = 0

x²+7x -60 = 0

(x-5)(x+12) = 0

(10-5) = 0 or (10+12) = 0

x = v or 10 = -12

Length cannot be negative. So x = 5

BC = two(x+7) = 2(5+7) = 2×12 = 24 km

Ac = 2x = 2×5 = 10 km

Total distance = AC + BC = 10+24 = 34 km

Distance saved = 34-26 = 8 km

Hence the distance saved is 8 km.

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

Solution:

Let the shortest side be x.

Then hypotenuse = 2x+six

Third side = 2x+6-two = 2x+4

Co-ordinate to Pythagoras theorem,

AB^two
= CB²+Air-conditioning²

(2x+6)²
= xⁱⁱ+(2x+4)²

4x^two+24x+36 = 10²+4x²+16x+16

ten²-8x-20 = 0

(x-10)(x+2) = 0

x-10 = 0 or 10+2 = 0

x = 10 or x = -2

x cannot exist negative.

So shortest side is 10 m.

Hypotenuse = 2x+6

= ii×10+6

= twenty+6

= 26 m

Third side = 2x+four

= = 2×10+4

= 20+iv

= 24 chiliad

Hence the shortest side, hypotenuse and third side of the triangle are 10 m, 26 m and 24 g respectively.

9.
ABC is an isosceles triangle right angled at C. Testify that AB² = 2AC².

Solution:

Let ABC be the isosceles right angled triangle .

C = 90˚

Ac = BC [isosceles triangle]

According to Pythagoras theorem,

AB²
= BC²+Air-conditioning^two

AB^two
= Ac^two+AC²
[∵AC = BC]

AB²
= 2ACⁱⁱ

Hence proved.

10. In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = Air conditioning² + BD².

Solution:

Given AD BC.

And so ADB and ADC are right triangles.

In ADB,

AB²
= Advertizement^two+BD^two
[Pythagoras theorem]

Advert²
= AB²– BD²
…(i)

In ADC,

AC²
= Advertizing²+CD²
[Pythagoras theorem]

ADⁱⁱ
= Air-conditioning²– CD²
…(ii)

Comparing (i) and (ii)

AB²– BD²
= Ac²– CD^two

AB²+ CD²
= AC²+ BD^two

Hence proved.

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).

Solution:

Given PQ = a, PR = b, QD = c and DR = d.

PD QR.

Then PDQ and PDR are correct triangles.

In PDQ,

PQⁱⁱ
= PD²+QD²
[Pythagoras theorem]

PD²
= PQ²– QD^two

PD²
= aⁱⁱ– cⁱⁱ
…(i) [∵
PQ = a and QD = c]

In PDR,

PR²
= PD²+DR²
[Pythagoras theorem]

PDⁱⁱ
= PRⁱⁱ– DRⁱⁱ

PD²
= b²– d²
…(ii) [∵
PR = b and DR = d]

Comparing (i) and (2)

a²– cⁱⁱ= b²– dⁱⁱ

a²– b²= c²– d²

(a+b)(a-b) = (c+d)(c-d)

Hence proved.

12. ABC is an isosceles triangle with AB = Air-conditioning = 12 cm and BC = eight cm. Find the altitude on BC and Hence, calculate its surface area.

Solution:

Permit AD exist the distance of ABC.

Given AB = AC = 12 cm

BC = 8 cm

The altitude to the base of an isosceles triangle bisects the base.

So BD = DC

BD = 8/2 = 4 cm

DC = 4 cm

ADC is a right triangle.

AB²
= BD²
+Advertising²
[Pythagoras theorem]

AD²
= AB²
-BD²

AD^two
= 12ⁱⁱ-4²

AD²
= 144-xvi

Advertizementⁱⁱ
= 128

Taking square root on both sides,

AD = √128 = √(2×64) = 8√2 cm

Area of ABC = ½ ×base of operations ×meridian

= ½ ×8×8√2

= 4×eight√2

= 32√2 cm²

Hence the area of triangle is 32√2 cm².

13. Detect the surface area and the perimeter of a square whose diagonal is 10 cm long.

Solution:

Given length of the diagonal of the foursquare is 10 cm.

Air-conditioning = x

Let AB = BC = ten [Sides of square are equal in measure]

B = 90˚ [All angles of a square are xc˚]

ABC is a right triangle.

ACⁱⁱ
= ABⁱⁱ+BC²

10^two
= x²+10²

100 = 2x²

x²
= fifty

x = √50 = √(25×two)

x = v√ii

So expanse of square = xⁱⁱ

= (5√2)²
= 50 cm²

Perimeter = 4x

= iv×5√two

= 20√2 cm

Hence surface area and perimeter of the square are fifty cm²
and 20√2 cm.

xiv.
(a) In fig. (i) given beneath, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = iii cm, ∠ ABD = ∠BCD = 90°. Calculate the length of AB.
(b) In fig. (ii) given beneath, ABCD is a quadrilateral in which AB = Advertizement, ∠A = 90° =∠C, BC = eight cm and CD = vi cm. Find AB and calculate the area of ∆ ABD.

Solution:

(i)Given Advert = 13 cm, DC = 12 one thousand

BC = 3 cm

ABD = BCD = 90˚

BCD is a right triangle.

BD^two
= BCⁱⁱ+DC²
[Pythagoras theorem]

BD^two
= 3^two+12²

BD²
= 9+144

BD²
= 153

ABD is a right triangle.

ADⁱⁱ
= AB²+BD²
[Pythagoras theorem]

xiii²
= AB²+153

169 = AB²+153

AB²
= 169-153

ABⁱⁱ
= 16

Taking square root on both sides,

AB = iv cm

Hence the length of AB is four cm.

(2)Given AB = Advert, A = ninety° = C, BC = viii cm and CD = half-dozen cm

BCD is a right triangle.

BDⁱⁱ
= BC²+DC²
[Pythagoras theorem]

BD²
= eight²+6²

BD²
= 64+36

BD²
= 100

Taking square root on both sides,

BD = 10 cm

ABD is a right triangle.

BD²
= AB²+Advertisingⁱⁱ
[Pythagoras theorem]

10²
= 2AB²
[∵AB = Advertizing]

100 = 2ABⁱⁱ

AB²
= 100/two

AB²
= 50

Taking square root on both sides,

AB = √fifty

AB = √(2×25)

AB = 5√2 cm

Hence the length of AB is 5√2 cm.

Solution:

(a)Given AB = 12 cm, AC = 13 cm, CE = x cm and DE = vi cm

ABC is a right triangle.

AC^two
= AB²+BC^two
[Pythagoras theorem]

13²
= 12^two+BC²

BCⁱⁱ
= xiii²-12²

BC²
= 169-144

BC²
= 25

Taking square root on both sides,

BC = v cm

CDE is a right triangle.

CEⁱⁱ
= CD²+DEⁱⁱ
[Pythagoras theorem]

10²
= CD^two+6²

100 = CD²+36

CD²
= 100-36

CD^two
= 64

Taking square root on both sides,

CD = viii cm

BD = BC +CD

BD = five+8

BD = xiii cm

Hence the length of BD is 13 cm.

(b) Given PSR = 90°, PQ = x cm, QS = half dozen cm and RQ = ix cm

PSQ is a right triangle.

PQ²
= PS²+QSⁱⁱ
[Pythagoras theorem]

10²
= PS²+vi²

100 = PS²+36

PS²
= 100-36

PS²
= 64

Taking square root on both sides,

PS = 8 cm

PSR is a correct triangle.

RS = RQ+QS

RS = 9+half dozen

RS = 15 cm

PR²
= PS²+RS²
[Pythagoras theorem]

PR²
= 8²+fifteen²

PR²
= 64+225

PR²
= 289

Taking square root on both sides,

PR = 17 cm

Hence the length of PR is 17 cm.

(c) D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm

ADC is a right triangle.

Ac²
= AD^two+CD^two
[Pythagoras theorem]

six²
= AD²+CD²
…..(i)

ABD is a right triangle.

ABⁱⁱ
= Advertising^two+BD²
[Pythagoras theorem]

16²
= AD²+(BC+CD)^two

16²
= AD²+(12+CD)²

256 = Advert^two+144+24CD+CD²

256-144 = Ad²+CD²+24CD

AD²+CD²
= 112-24CD

half-dozen²
= 112-24CD [from (i)]

36 = 112-24CD

24CD = 112-36

24CD = 76

CD = 76/24 = nineteen/vi

Solution:

(a)Given BC = v cm,
B =90°, AB = 5AE,

CD = 2AE and Air-conditioning = ED

ABC is a correct triangle.

AC²
= AB²+BC²
…(i) [Pythagoras theorem]

BED is a correct triangle.

ED²
= Be^two+BD²

[Pythagoras theorem]

Air-conditioning²
= BEⁱⁱ+BD^two
…(two) [∵AC = ED]

Comparing (i) and (ii)

AB²+BC²
= Be²+BD²

(5AE)ⁱⁱ+5²
= (4AE)²+(BC+CD)²
[∵BE = AB-AE = 5AE-AE = 4AE]

(5AE)²+25 = (4AE)²+(5+2AE)²
…(iii)

[∵BC = 5, CD = 2AE]

Let AE = 10. So (iii) becomes,

(5x)²+25 = (4x)²+(5+2x)^two

25xⁱⁱ+25 = 16x²+25+20x+4x²

25x²
= 20x²+20x

5x²
= 20x

x = xx/5 = 4

AE = four cm

CD = 2AE = 2×iv = 8 cm

AB = 5AE

AB = 5×iv = xx cm

ABC is a right triangle.

AC²
= AB²+BC²
[Pythagoras theorem]

Air-conditioning²
= 20²+5ⁱⁱ

Air-conditioning²
= 400+25

ACⁱⁱ
= 425

Taking square root on both sides,

AC = √425 = √(25×17)

Ac = v√17 cm

Hence EA = iv cm, CD = viii cm, AB = 20 cm and AC = 5√17 cm.

(b)Given D is the midpoint of BC.

DC = ½ BC

ABC is a correct triangle.

AB²
= AC²+BCⁱⁱ
…(i) [Pythagoras theorem]

ADC is a right triangle.

Ad²
= AC^two+DCⁱⁱ
…(ii) [Pythagoras theorem]

AC^two
= Advertizement²-DC²

AC²
= AD²– (½ BC)²
[∵DC = ½ BC]

Air-conditioning²
= AD²– ¼ BCⁱⁱ

4AC²
= 4AD^two– BC²

ACⁱⁱ+3AC^two
= 4AD²– BCⁱⁱ

Air conditioning²+BCⁱⁱ
= 4AD²-3AC²

ABⁱⁱ
= 4AD²-3AC²
[from (i)]

Hence proved.

17.
In ∆ ABC, AB = AC = x, BC = x cm and the area of ∆ ABC is lx cm². Find ten.

Solution:

Given AB = AC = x

So ABC is an isosceles triangle.

Advertising BC

The altitude to the base of an isosceles triangle bisects the base of operations.

BD = DC = x/2 = five cm

Given area = 60 cm²

½ ×base of operations ×height = ½ ×ten×AD = threescore

AD = threescore×two/10

Advertizement = 60/five

Ad = 12cm

ADC is a right triangle.

AC²
= AD²+DC²

x²
= 12ⁱⁱ+v²

10²
= 144+25

x^two
= 169

Taking foursquare root on both sides

x = 13 cm

Hence the value of 10 is 13 cm.

18.
In a rhombus, If diagonals are 30 cm and 40 cm, detect its perimeter.

Solution:

Let ABCD be the rhombus.

Given AC = 30cm

BD = xl cm

Diagonals of a rhombus are perpendicular bisectors of each other.

OB = ½ BD = ½ ×xl = 20 cm

OC = ½ AC = ½ ×30 = 15 cm

OCB is a correct triangle.

BC²
= OC²+OB²
[Pythagoras theorem]

BC²
= 15²+twenty^two

BC^two
= 225+400

BC²
= 625

Taking square root on both sides

BC = 25 cm

And then side of a rhombus, a = 25 cm.

Perimeter = 4a = 4×25 = 100 cm

Hence the perimeter of the rhombus is 100 cm.

Solution:

(i)
Given AB || DC, BC = Advertising = 13 cm.

AB = 22 cm and DC = 12cm

Here DC = 12

MN = 12 cm

AM = BN

AB = AM+MN+BN

22 = AM+12+AM [∵AM = BN]

2AM = 22-12 = 10

AM = 10/two

AM = v cm

AMD is a right triangle.

Ad²
= AMⁱⁱ+DM²
[Pythagoras theorem]

thirteenⁱⁱ
= 5²+DM^two

DM²
= xiii²-5²

DM²
= 169-25

DM²
= 144

Taking square root on both sides,

DM = 12 cm

Hence the height of the trapezium is 12 cm.

(b) Given AB || DC, A = 90°, DC = 7 cm,

AB = 17 cm and Air conditioning = 25 cm

ADC is a right triangle.

AC^two
= AD²+DCⁱⁱ
[Pythagoras theorem]

25^two
= Advertising²+7²

Advertisingⁱⁱ
= 25²-7²

AD²
= 625-49

AD²
= 576

Taking foursquare root on both sides

Advertizement = 24 cm

CM = 24 cm [∵ABCD]

DC = 7 cm

AM = 7 cm

BM = AB-AM

BM = 17-7 = 10 cm

BMC is a right triangle.

BC²
= BM²+CM²

BC²
= 10²+24ⁱⁱ

BC²
= 100+576

BC²
= 676

Taking foursquare root on both sides

BC = 26 cm

Hence length of BC is 26 cm.

(c) (i)Proof:

Given ABCD is a square of side 7 cm.

So AB = BC = CD = Ad = 7 cm

Likewise given AE = FC = CG = HA = 3 cm

BE = AB-AE = 7-3 = four cm

BF = BC-FC = 7-iii = 4 cm

GD = CD-CG = 7-iii = iv cm

DH = AD-HA = vii-3 = four cm

A = ninety˚ [Each angle of a foursquare equals 90˚]

AHE is a right triangle.

HE²
= AE²+AHⁱⁱ
[Pythagoras theorem]

HEⁱⁱ
= 3²+3²

HE^two
= ix+9 = eighteen

HE = √(9×2) = 3√ii cm

Similarly GF = iii√2 cm

EBF is a right triangle.

EF²
= BEⁱⁱ+BF²
[Pythagoras theorem]

EF²
= 4²+iv²

EF²
= 16+xvi = 32

Taking square root on both sides

EF = √(16×2) = 4√ii cm

Similarly HG = 4√ii cm

Now bring together EG

In EFG

EG²
= EF²+GFⁱⁱ

EG²
= (iv√2)²+(3√2)²

EG²
= 32+eighteen = fifty

EG = √50 = 5√2 cm …(i)

Join HF.

Also HF²
= EH²+HG²

= (3√2)²+(4√2)^two

= eighteen+32 = 50

HF = √50 = v√2 cm …(ii)

From (i) and (ii)

EG
= HF

Diagonals of the quadrilateral are congruent. So EFGH is a rectangle.

Hence proved.

(2)Surface area of rectangle EFGH = length × breadth

= HE ×EF

= 3√ii×4√2

= 24 cm²

Perimeter of rectangle EFGH = 2(length+breadth)

= 2×(4√2+three√2)

= two×seven√2

= fourteen√ii cm

Hence surface area of the rectangle is 24 cm²
and perimeter is 14√two cm.

20.
AD is perpendicular to the side BC of an equilateral Δ ABC. Prove that 4AD² = 3AB².

Solution:

Given AD BC

D = ninety˚

Proof:

Since ABC is an equilateral triangle,

AB = Air conditioning = BC

ABD is a right triangle.

According to Pythagoras theorem,

AB^two
= AD²+BDⁱⁱ

BD = ½ BC

AB^two
= ADⁱⁱ+( ½ BC)^two

AB^two
= Advertizement²+( ½ AB)^two
[∵BC = AB]

AB²
= AD²+ ¼ AB²

ABⁱⁱ
= (4ADⁱⁱ+ ABⁱⁱ)/4

4AB^two
= 4AD²+ AB^two

4AD²
= 4AB²– AB²

4AD^two
= 3ABⁱⁱ

Hence proved.

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

Prove that :

(i)4ADⁱⁱ
= 4AC^two+BC²

(ii)4BE²
= 4BC²+AC²

(three)4(Advertisement²+Beⁱⁱ) = 5AB²

Solution:

Proof:

(i)C = 90˚

So ACD is a right triangle.

Advert²
= Air conditioning^two+CD²
[Pythagoras theorem]

Multiply both sides by 4, we get

4AD²
= 4AC²+4CD²

4AD²
= 4AC^two+4BD²
[∵D is the midpoint of BC, CD = BD = ½ BC]

4AD²
= 4AC²+(2BD)²

4ADⁱⁱ
= 4AC²+BC²….(i) [∵BC = 2BD]

Hence proved.

(ii)BCE is a right triangle.

BEⁱⁱ
= BCⁱⁱ+CEⁱⁱ
[Pythagoras theorem]

Multply both sides by four , we get

4BE²
= 4BC^two+4CE²

4BE²
= 4BC^two+(2CE)²

4BE²
= 4BC²+Air conditioning²
….(ii) [∵E is the midpoint of AC, AE = CE = ½ AC]

Hence proved.

(3)Adding (i) and (ii)

4AD²+4BE²
= 4AC²+BC²+4BC²+Air conditioning²

4AD²+4BE²
= 5AC²+5BC²

4(Advertising²+Be²
) = 5(Air conditioning²+BC^two)

4(AD²+BE²
) = 5(AB²) [∵ABC is a right triangle, AB²
= AC²+BC²]

Hence proved.

22.
If AD, Be and CF are medians of ABC, prove that 3(AB² + BC² + CA²) = 4(Advertizement² + Be² + CF²).

Solution:

Construction:

Draw APBC

Proof:

APB is a right triangle.

AB²
= AP²+BP²
[Pythagoras theorem]

AB²
= AP²+(BD-PD)ⁱⁱ

AB^two
= AP²+BD^two+PD²-2BD×PD

ABⁱⁱ
= (AP²+PDⁱⁱ)+BD^two-2BD×PD

AB²
= AD^two+ (½ BC)²-two×( ½ BC)×PD [∵AP²+PD²
= Advertizing²
and BD = ½ BC]

ABⁱⁱ
= Advert^two+ ¼ BC²– BC×PD ….(i)

APC is a right triangle.

Acⁱⁱ
= APⁱⁱ+PC²
[Pythagoras theorem]

ACⁱⁱ
= APⁱⁱ+(PD²+DCⁱⁱ)

Air conditioningⁱⁱ
= AP^two+PD²+DC²+2×PD×DC

Air conditioning²
= (AP^two+PD^two)+ (½ BC)²+2×PD×( ½ BC) [DC = ½ BC]

AC²
= (Advertisement)²+ ¼ BC²+PD× BC …(two) [In APD, AP^two+PDⁱⁱ
= AD²]

Adding (i) and (ii), we get

AB²+AC^two
= 2AD²+ ½ BC²
…..(iii)

Draw perpendicular from B and C to AC and AB respectively.

Similarly nosotros get,

BC²+CAⁱⁱ
= 2CF^two+ ½ AB²
….(iv)

ABⁱⁱ+BCⁱⁱ
= 2BE²+ ½ Ac^two
….(v)

Adding (iii), (iv) and (v), we get

2(AB^two+BC²+CAⁱⁱ) = 2(Advertising^two+BE²+CF²)+ ½ (BC²+AB^two+Air-conditioning^two)

2(AB²+BC²+CA^two) = two(ABⁱⁱ+BC²+CA²) – ½ (AB²+BC^two+CA²)

two(Advertisement²+BE²+CF²) = (iii/2)× (AB²+BC^two+CA^two)

four(AD²+BE²+CF²) = 3(AB^two+BC²+CA²)

Hence proved.

23.(a)
In fig. (i) given beneath, the diagonals Air conditioning and BD of a quadrilateral ABCD intersect at O, at right angles. Show thatAB² + CD² = AD² + BC².

Solution:

Given diagonals of quadrilateral ABCD, Air conditioning and BD intersect at O at right angles.

Proof:

AOB is a correct triangle.

ABⁱⁱ
= OB²+OA²
…(i) [Pythagoras theorem]

COD is a correct triangle.

CDⁱⁱ
= OC²+OD²
…(ii) [Pythagoras theorem]

Calculation (i) and (ii), we get

AB²+ CD²
= OB^two+OA²+ OC²+OD²

AB²+ CD²
= (OA²+ODⁱⁱ)+ (OC²+OB²) …(iii)

AOD is a correct triangle.

Advertising²
= OA²+OD^two
…(iv) [Pythagoras theorem]

BOC is a right triangle.

BCⁱⁱ
= OCⁱⁱ+OB²
…(5) [Pythagoras theorem]

Substitute (iv) and (v) in (iii), we get

AB²+ CD^two
= AD²+BC²

Hence proved.

24. In a quadrilateral ABCD, B = 90° = D. Evidence that 2 Air-conditioning² – BCⁱⁱ
= AB² + Advertizement² + DC².

Solution:

Given B = D = 90˚

Then ABC and ADC are correct triangles.

In ABC,

AC²
= AB²+BC^two
…(i) [Pythagoras theorem]

In ADC,

ACⁱⁱ
= Ad²+DC²
…(2) [Pythagoras theorem]

Calculation (i) and (ii)

2AC²
= ABⁱⁱ+BC²+ Advertizing²+DC²

2AC²
-BC^two
= AB²+ADⁱⁱ+DC²

Hence proved.

25.
In a ∆ ABC, A = 90°, CA = AB and D is a bespeak on AB produced. Testify that :DC² – BD² = 2AB×AD.

Solution:

Given A = 90˚

CA = AB

Proof:

In ACD,

DC²
= CA²+Advertizementⁱⁱ
[Pythagoras theorem]

DC²
= CA²+(AB+BD)²

DC²
= CA^two+AB²+BDⁱⁱ+2AB×BD

DC^two
-BD²
= CA^two+AB²+2AB×BD

DCⁱⁱ
-BD²
= ABⁱⁱ+AB²+2AB×BD [∵CA = AB]

DCⁱⁱ
-BD²
= 2AB²+2AB×BD

DC²
-BD^two
= 2AB(AB+BD)

DC²
-BD²
= 2AB×Advertizement [A-B-D]

Hence proved.

26. In an isosceles triangle ABC, AB = Ac and D is a point on BC produced.

Evidence that AD² = AC²+BD.CD.

Solution:

Given ABC is an isosceles triangle.

AB = Ac

Construction: Describe AP BC

Proof:

APD is a right triangle.

ADⁱⁱ
= AP²+PDⁱⁱ
[Pythagoras theorem]

AD²
= AP²+(PC+CD)^two
[PD = PC+CD]

AD²
= AP²+PC²+CD²
+2PC×CD ….(i)

APC is a right triangle.

AC²
= AP²+PCⁱⁱ
…(ii) [Pythagoras theorem]

Substitute (two) in (i)

Advertizementⁱⁱ
= Air-conditioning²
+CD²+2PC×CD ….(iii)

Since ABC is an isosceles triangle,

PC = ½ BC [The distance to the base of an isosceles triangle bisects the base of operations]

AD²
= AC²
+CD²+ii× ½ BC ×CD

Advertisingⁱⁱ
= Air conditioning²
+CD²+BC×CD

Advert²
= Air conditioning²
+CD(CD+BC)

ADⁱⁱ
= AC²
+CD×BD [CD+BC = BD]

AD²
= AC²
+BD×CD

Hence proved.

Chapter test

1.
a) In fig. (i) given beneath, Advertisement ⊥ BC, AB = 25 cm, Air conditioning = 17 cm and Advertizing = 15 cm. Detect the length of BC.

(b) In figure (2) given below, BAC = 90°, ADC = 90°, Advertizing = six cm, CD = viii cm and BC = 26 cm.

Observe :(i) AC

(ii) AB

(3) surface area of the shaded region.

(c) In figure (three) given beneath, triangle ABC is right angled at B. Given that AB = ix cm, AC = 15 cm and D, East are mid-points of the sides AB and Air-conditioning respectively, calculate

(i) the length of BC

(2) the area of ∆ ADE.

Solution:

(a) Given
AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm

ADC is a right triangle.

AC^two
= Advert²+DC²
[Pythagoras theorem]

17^two
= xvⁱⁱ+DC²

289 = 225+DC²

DC²
= 289-225

DC²
= 64

Taking square root on both sides,

DC = viii cm

ADB is a right triangle.

AB²
= Advertizing²+BDⁱⁱ
[Pythagoras theorem]

25ⁱⁱ
= fifteenⁱⁱ+BD^two

625 = 225+ BDⁱⁱ

BD²
= 625-225 = 400

Taking square root on both sides,

BD = 20 cm

BC = BD+DC

= twenty+8

= 28 cm

Hence the length of BC is 28 cm.

(b)
Given BAC = 90°, ADC = 90°

Advert = 6 cm, CD = 8 cm and BC = 26 cm.

(i) ADC is a correct triangle.

Air-conditioning^two
= Advertisementⁱⁱ+DCⁱⁱ
[Pythagoras theorem]

AC^two
= 6²+8²

Ac²
= 36+64

AC^two
= 100

Taking square root on both sides,

Air-conditioning = 10 cm

Hence length of AC is 10 cm.

(ii) ABC is a right triangle.

BCⁱⁱ
= AC²+AB²
[Pythagoras theorem]

26²
= x²+AB²

ABⁱⁱ
= 26²-10ⁱⁱ

AB²
= 676-100

AB²
= 576

Taking square root on both sides,

AB = 24 cm

Hence length of AB is 24 cm.

(iii)Area of ABC = ½ ×AB×Ac

= ½ ×24×10

= 120 cmⁱⁱ

Surface area of ADC = ½ ×Advert×DC

= ½ ×6×8

= 24 cm²

Area of shaded region = area of ABC- expanse of ADC

= 120-24

= 96 cm²

Hence the area of shaded region is 96 cmⁱⁱ.

(c) Given B = 90˚.

AB = 9 cm, AC = 15 cm .

D, East are mid-points of the sides AB and AC respectively.

(i)ABC is a correct triangle.

AC²
= AB²+BC^two
[Pythagoras theorem]

15²
= 9^two+BC²

225 = 81+BC²

BC²
= 225-81

BC²
= 144

Taking square root on both sides,

BC = 12 cm

Hence the length of BC is 12 cm.

(ii) AD = ½ AB [D is the midpoint of AB]

Advertizement = ½ ×9 = 9/2

AE = ½ AC [E is the midpoint of Air conditioning]

AE = ½ ×15 = 15/2

ADE is a right triangle.

AE²
= AD^two+DE²
[Pythagoras theorem]

(15/2)²
= (9/2)²+DE²

DE²
= (xv/2)²
– (9/2)²

DE²
= 225/4 -81/4

DE²
= 144/iv

Taking square root on both sides,

DE = 12/2 = 6 cm.

Area of ADE = ½ ×DE×Ad

= ½ ×half dozen×9/two

= 13.five cm²

Hence the area of the ADE is xiii.5 cm².

two.
If in ∆ ABC, AB > AC and AD BC, testify that AB² – Air-conditioning² = BD² – CD²

Solution:

Given Advertizement BC, AB>Air-conditioning

So ADB and ADC are right triangles.

Proof:

In ADB,

AB²
= Ad²+BD²
[Pythagoras theorem]

Advertizing^two
= AB²-BDⁱⁱ
…(i)

In ADC,

AC²
= Ad²+CD^two
[Pythagoras theorem]

Advertising²
= Air conditioning²-CD²
…(two)

Equating (i) and (2)

AB²-BD²
= Air conditioningⁱⁱ-CDⁱⁱ

AB^two-AC²
= BDⁱⁱ– CD^two

Hence proved.

3.
In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio ii:1.

Prove that

(i) 9AQ² = 9AC² + 4BC²

(2) 9BP² = 9BC² + 4AC²

(iii) 9(AQ² + BP²) = 13AB².

Solution:

Construction:

Join AQ and BP.

Given C = xc˚

Proof:

(i) In ACQ,

AQ^two
= AC²+CQ²
[Pythagoras theorem]

Multiplying both sides by 9, we get

9AQ²
= 9AC²+9CQⁱⁱ

9AQ^two
= 9AC²+(3CQ)^two
…(i)

Given BQ: CQ = 1:2

CQ/BC = CQ/(BQ+CQ)

CQ/BC = two/3

3CQ = 2BC ….(ii)

Substitute (2) in (i)

9AQ²
= 9AC²+(2BC)^two

9AQ²
= 9AC²+4BCⁱⁱ
…(iii)

Hence proved.

(ii) ) In BPC,

BPⁱⁱ
= BC²+CP²
[Pythagoras theorem]

Multiplying both sides by 9, we get

9BP^two
= 9BC²+9CP²

9BP²
= 9BC²+(3CP)²
…(4)

Given AP: PC = 1:2

CP/AC = CP/AP+PC

CP/Air conditioning = two/three

3CP = 2AC ….(5)

Substitute (5) in (iv)

9BP^two
= 9BC²+(2AC)²

9BP²
= 9BC²+4ACⁱⁱ
..(vi)

Hence proved.

(3)Adding (three) and (vi), we become

9AQ^two+9BP^two
= 9AC²+4BC²+9BC²+4AC²

9(AQ²+BP)²
= 13AC²+13BC²

9(AQ²+BP)²
= xiii(AC²+BCⁱⁱ)…(vii)

In ABC,

AB²
= AC²+BC²
…..(eight)

Substitute (eight) in (viii), we get

nine(AQ²+BP)²
= 13AB²

Hence proved.

iv.
In the given figure, ∆PQR is correct angled at Q and points S and T trisect side QR. Prove that 8PT² = 3PR² + 5PS².

Solution:

Given Q = 90˚

S and T are points on RQ such that these points trisect it.

So RT = TS = SQ

To prove : 8PT² = 3PR² + 5PS².

Proof:

Let RT = TS = SQ = 10

In PRQ,

PR²
= RQ^two+PQⁱⁱ
[Pythagoras theorem]

PR²
= (3x)ⁱⁱ+PQ²

PR²
= 9x²+PQⁱⁱ

Multiply above equation past three

3PR²
= 27x²+3PQⁱⁱ
….(i)

Similarly in PTS,

PTⁱⁱ
= TQ²+PQ²
[Pythagoras theorem]

PT²
= (2x)²+PQ²

PT²
= 4x²+PQ²

Multiply above equation past 8

8PT²
= 32x²+8PQ²
….(ii)

Similarly in PSQ,

PS^two
= SQ²+PQ^two
[Pythagoras theorem]

PS²
= x^two+PQ²

Multiply higher up equation past five

5PS²
= 5x^two+5PQ²
…(3)

Add (i) and (3), nosotros become

3PR²
+5PSⁱⁱ
= 27xⁱⁱ+3PQ²+5x²+5PQ²

3PR²
+5PS²
= 32x²+8PQ²

3PR²
+5PS²
= 8PT²
[From (ii)]

8PT^two
= 3PRⁱⁱ
+5PS²

Hence proved.

5. In a quadrilateral ABCD, B = xc°. If Advert² = AB² + BC² + CD², prove that ACD = xc°.

Solution:

Given : B = 90˚ in quadrilateral ABCD

Advertising² = AB² + BC² + CD²

To prove: ACD = ninety˚

Proof:

In ABC,

Air conditioning²
= AB²+BCⁱⁱ
….(i) [Pythagoras theorem]

Given
AD² = AB² + BC² + CD²

AD² = AC²+CD²
[from (i)]

In ACD, ACD = xc˚ [Converse of Pythagoras theorem]

Hence proved.

6. In the given figure, find the length of Ad in terms of b and c.

Solution:

Given : A = ninety˚

AB = c

Ac = b

ADB = 90˚

In ABC,

BC^two
= Air-conditioning²+AB²
[Pythagoras theorem]

BC²
= b²+c²

BC = √( b²+c²) …(i)

Expanse of ABC = ½ ×AB×Ac

= ½ ×bc …(ii)

Also, Expanse of ABC = ½ ×BC×Ad

= ½ ×√( b^two+c^two) ×AD …(three)

Equating (ii) and (iii)

½ ×bc = ½ ×√( b²+c²) ×Advertizement

Advertising = bc /(√( b²+cⁱⁱ)

Hence Advertizing is bc /(√( b²+c^two).

7.
ABCD is a square, F is mid-point of AB and Be is one-third of BC. If area of ∆FBE is 108 cm², notice the length of AC.

Solution:

Let x exist each side of the square ABCD.

FB = ½ AB [∵ F is the midpoint of AB]

FB = ½ x …(i)

Exist = (1/iii) BC

Exist = (1/iii) x …(two)

Ac = √2 ×side [Diagonal of a square]

AC = √2x

Area of FBE = ½ FB×Be

108 = ½ × ½ ten ×(1/3)x [given area of FBE = 108 cm²]

108 = (one/12)xⁱⁱ

x²
= 108×12

x²
= 1296

Taking square root on both sides.

x = 36

Ac = √two×36 = 36√2

Hence length of AC is 36√2 cm.

8.
In a triangle ABC, AB = Air-conditioning and D is a point on side Air conditioning such that BC² = Air conditioning × CD, Testify that BD = BC.

Solution:

Given : In ABC, AB = AC

D is a point on side AC such that BC² = Air-conditioning × CD

To bear witness : BD = BC

Construction: Draw BEAC

Proof:

In BCE ,

BC²
= Be²+ECⁱⁱ
[Pythagoras theorem]

BC²
= Be^two+(AC-AE)^two

BCⁱⁱ
= BE²+AC^two+AEⁱⁱ-2 Air-conditioning×AE

BCⁱⁱ
= Be²+AE²+Air-conditioning²-2 AC×AE …(i)

In ABC,

AB²
= BE²+AE
²
..(2)

Substitute (ii) in (i)

BC²
= AB²+Air-conditioning²-2 AC×AE

BC²
= AC²+Air conditioning²-2 Ac×AE [∵AB = AC]

BC^two
= 2AC²-2 AC×AE

BC²
= 2AC(Air-conditioning-AE)

BC²
= 2AC×EC

Given
BC² = Ac × CD

2AC×EC =
AC × CD

2EC = CD ..(ii)

East is the midpoint of CD.

EC = DE …(three)

In BED and BEC,

EC = DE [From (3)]

Be = Exist [common side]

BED = BEC

BED BEC [By SAS congruency dominion]

BD = BD [c.p.c.t]

Hence proved.