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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### Access answers to ML Aggarwal Solutions for Course 9 Maths Chapter 12 – Pythagoras Theorem

**Practice 12**

**1. Lengths of sides of triangles are given below. Determine which of them are correct triangles. In instance of a correct triangle, write the length of its hypotenuse:**

**(i) 3 cm, 8 cm, 6 cm**

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

**(3) 1.4 cm, 4.8 cm, 5 cm**

**Solution:**

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

We have h^{ii}

= b^{2}+a^{2}

[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

h^{2}

= 8^{2}

= 64

hither 45 ≠ 64

Hence the given triangle is not a correct triangle.

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

b^{2}+a^{ii}

= 12^{2}+ 5^{2}

= 144+25 = 169

h^{2}

= xiii^{2}

= 169

here b^{2}+a^{2}

= 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}+a^{2}

= 1.4^{2}+ 4.8^{2}

= 1.96+23.04 = 25

h^{ii}

= 5^{ii}

= 25

here b^{2}+a^{2}

= h^{2}

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}

= PQ^{2}+QR^{2}

10^{ii}

= 6^{2}+QR^{2}

100 = 36+QR^{2}

QR^{2}

= 100-36

QR^{2}

= 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}

= AB^{ii}+BC^{two}

24^{two}

= 18^{2}+BC^{2}

576 = 324+BC^{2}

BC^{2}

= 576-324

BC^{2}

= 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,

AC^{2}

= AE^{2}+CE^{two}

AC^{2}

= 12^{2}+5^{2}

Ac^{2}

= 144+25

Air-conditioning^{2}

= 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}+a^{ii}

20^{2}

= (3x)^{ii}+(4x)^{2}

400 = 9x^{2}+16x^{2}

25x^{ii}

= 400

ten^{2}

= 400/25

x^{2}

= 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}

= BC^{ii}+AB^{ii}

BC^{2}+AB^{2}= (3x)^{2}+(4x)^{ii
}

= 9x^{2}+16x^{2}

= 25x^{2}

AC^{2}

= (5x)^{2}

= 25x^{2}

AC^{ii}

= BC^{2}+AB^{2}

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,

AB^{ii}

= CB^{two}+Ac^{two}

26^{2}

= ( ii(x+7))^{2}+(2x)^{2}

676 = 4(x^{2}+14x+49) + 4x^{2}

4x^{2}+56x+196+4x^{two}

= 676

8x^{two}+56x+196 = 676

8x^{2}+56x +196-676 = 0

8x^{2}+56x -480 = 0

x^{2}+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}

= CB^{2}+Air-conditioning^{2}

(2x+6)^{2}

= x^{ii}+(2x+4)^{2}

4x^{two}+24x+36 = 10^{2}+4x^{2}+16x+16

ten^{2}-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,

AB^{2}

= BC^{2}+Air-conditioning^{two}

AB^{two}

= Ac^{two}+AC^{2
}[∵AC = BC]

AB^{2}

= 2AC^{ii}

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,

AB^{2}

= Advertizement^{two}+BD^{two}

[Pythagoras theorem]

Advert^{2}

= AB^{2}– BD^{2}

…(i)

In ADC,

AC^{2}

= Advertizing^{2}+CD^{2}

[Pythagoras theorem]

AD^{ii}

= Air-conditioning^{2}– CD^{2}

…(ii)

Comparing (i) and (ii)

AB^{2}– BD^{2
}= Ac^{2}– CD^{two}

AB^{2}+ CD^{2}

= AC^{2}+ 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,

PQ^{ii}

= PD^{2}+QD^{2}

[Pythagoras theorem]

PD^{2
}= PQ^{2}– QD^{two}

PD^{2}

= a^{ii}– c^{ii}

…(i) [∵

**PQ = a and QD = c]**

In PDR,

PR^{2}

= PD^{2}+DR^{2}

[Pythagoras theorem]

PD^{ii}

= PR^{ii}– DR^{ii}

PD^{2}

= b^{2}– d^{2}

…(ii) [∵

**PR = b and DR = d]**

Comparing (i) and (2)

a^{2}– c^{ii}= b^{2}– d^{ii}

a^{2}– b^{2}= c^{2}– d^{2}

(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.

AB^{2}

= BD^{2}

+Advertising^{2}

[Pythagoras theorem]

AD^{2}

= AB^{2}

-BD^{2}

AD^{two}

= 12^{ii}-4^{2}

AD^{2}

= 144-xvi

Advertizement^{ii}

= 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 cm^{2}

Hence the area of triangle is 32√2 cm^{2}.

**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.

AC^{ii}

= AB^{ii}+BC^{2}

10^{two}

= x^{2}+10^{2}

100 = 2x^{2}

x^{2}

= fifty

x = √50 = √(25×two)

x = v√ii

So expanse of square = x^{ii}

= (5√2)^{2}

= 50 cm^{2}

Perimeter = 4x

= iv×5√two

= 20√2 cm

Hence surface area and perimeter of the square are fifty cm^{2}

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}
= BC^{ii}+DC^{2}
[Pythagoras theorem]**

**BD ^{two}
= 3^{two}+12^{2}
**

**BD ^{2}
= 9+144**

**BD ^{2}
= 153**

**ABD is a right triangle.**

**AD ^{ii}
= AB^{2}+BD^{2}
[Pythagoras theorem]**

**xiii ^{2}
= AB^{2}+153**

**169 = AB ^{2}+153**

AB^{2}

= 169-153

AB^{ii}

= 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.**

**BD ^{ii}
= BC^{2}+DC^{2}
[Pythagoras theorem]**

**BD ^{2}
= eight^{2}+6^{2}
**

**BD ^{2}
= 64+36**

**BD ^{2}
= 100**

**Taking square root on both sides,**

**BD = 10 cm**

**ABD is a right triangle.**

**BD ^{2}
= AB^{2}+Advertising^{ii}
[Pythagoras theorem]**

**10 ^{2}
= 2AB^{2}
[∵AB = Advertizing]**

**100 = 2AB ^{ii}
**

**AB ^{2}
= 100/two**

**AB ^{2}
= 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.**

**fifteen.**

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

**(b) In effigy (two) given below, ∠PSR = xc°, PQ = 10 cm, QS = half-dozen cm and RQ = ix 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.**

**Solution:**

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

**ABC is a right triangle.
**

**AC ^{two}
= AB^{2}+BC^{two}
[Pythagoras theorem]**

**13 ^{2}
= 12^{two}+BC^{2}
**

**BC ^{ii}
= xiii^{2}-12^{2}
**

**BC ^{2}
= 169-144**

**BC ^{2}
= 25**

**Taking square root on both sides,**

**BC = v cm**

**CDE is a right triangle.
**

**CE ^{ii}
= CD^{2}+DE^{ii}
[Pythagoras theorem]**

**10 ^{2}
= CD^{two}+6^{2}
**

**100 = CD ^{2}+36**

**CD ^{2}
= 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.**

**PQ ^{2}
= PS^{2}+QS^{ii}
[Pythagoras theorem]**

**10 ^{2}
= PS^{2}+vi^{2}
**

**100 = PS ^{2}+36**

**PS ^{2}
= 100-36**

**PS ^{2}
= 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**

**PR ^{2}
= PS^{2}+RS^{2}
[Pythagoras theorem]**

**PR ^{2}
= 8^{2}+fifteen^{2}
**

**PR ^{2}
= 64+225**

**PR ^{2}
= 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.**

**Ac ^{2}
= AD^{two}+CD^{two}
[Pythagoras theorem]**

**six ^{2}
= AD^{2}+CD^{2}
…..(i)**

**ABD is a right triangle.**

**AB ^{ii}
= Advertising^{two}+BD^{2}
[Pythagoras theorem]**

**16 ^{2}
= AD^{2}+(BC+CD)^{two}
**

**16 ^{2}
= AD^{2}+(12+CD)^{2}
**

**256 = Advert ^{two}+144+24CD+CD^{2}**

**256-144 = Ad ^{2}+CD^{2}+24CD**

**AD ^{2}+CD^{2}**

= 112-24CD

**half-dozen ^{2}
= 112-24CD [from (i)]**

**36 = 112-24CD**

**24CD = 112-36**

**24CD = 76**

**CD = 76/24 = nineteen/vi**

**16.**

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

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

**(b) In the figure (two) given below, ABC is a right triangle correct angled at C. If D is mid-indicate of BC, prove that AB ^{two}
= 4AD² – 3AC².**

**Solution:**

**(a)Given BC = v cm,B =90°, AB = 5AE,
**

**CD = 2AE and Air-conditioning = ED**

**ABC is a correct triangle.**

**AC ^{2}
= AB^{2}+BC^{2
}…(i) [Pythagoras theorem]**

**BED is a correct triangle.**

**ED ^{2}
= Be^{two}+BD^{2
}
[Pythagoras theorem]**

**Air-conditioning ^{2}
= BE^{ii}+BD^{two
}…(two) [∵AC = ED]**

**Comparing (i) and (ii)**

**AB ^{2}+BC^{2}**

= Be^{2}+BD^{2}

**(5AE) ^{ii}+5^{2}**

= (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}**

**25x ^{ii}+25 = 16x^{2}+25+20x+4x^{2}**

**25x ^{2}
= 20x^{2}+20x**

**5x ^{2}
= 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.**

**AC ^{2}
= AB^{2}+BC^{2}
[Pythagoras theorem]**

**Air-conditioning ^{2}
= 20^{2}+5^{ii}
**

**Air-conditioning ^{2}
= 400+25**

**AC ^{ii}
= 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.**

**AB ^{2}
= AC^{2}+BC^{ii}
…(i) [Pythagoras theorem]**

**ADC is a right triangle.**

**Ad ^{2}
= AC^{two}+DC^{ii}
…(ii) [Pythagoras theorem]**

**AC ^{two}
= Advertizement^{2}-DC^{2}
**

**AC ^{2}
= AD^{2}– (½ BC)^{2}
[∵DC = ½ BC]**

**Air-conditioning ^{2}
= AD^{2}– ¼ BC^{ii}
**

**4AC ^{2}
= 4AD^{two}– BC^{2}
**

**AC ^{ii}+3AC^{two}**

= 4AD^{2}– BC^{ii}

**Air conditioning ^{2}+BC^{ii}**

= 4AD^{2}-3AC^{2}

**AB ^{ii}
= 4AD^{2}-3AC^{2}
[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 cm^{2}

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

AD = threescore×two/10

Advertizement = 60/five

Ad = 12cm

ADC is a right triangle.

AC^{2}

= AD^{2}+DC^{2}

x^{2}

= 12^{ii}+v^{2}

10^{2}

= 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.

BC^{2}

= OC^{2}+OB^{2}

[Pythagoras theorem]

BC^{2
}= 15^{2}+twenty^{two}

BC^{two
}= 225+400

BC^{2
}= 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.

**19.**

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

**(b) In effigy (2) given below, AB || DC, ∠ A = ninety°, DC = 7 cm, AB = 17 cm and Ac = 25 cm. Calculate BC.**

**(c) In effigy (iii) given below, ABCD is a square of side seven cm. if**

**AE = FC = CG = HA = 3 cm,**

**(i) prove that EFGH is a rectangle.**

**(ii) find the surface area and perimeter of EFGH.**

**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.**

**Ad ^{2}
= AM^{ii}+DM^{2}
[Pythagoras theorem]**

**thirteen ^{ii}
= 5^{2}+DM^{two}
**

**DM ^{2}
= xiii^{2}-5^{2}
**

**DM ^{2}
= 169-25**

**DM ^{2}
= 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}
= AD^{2}+DC^{ii
}[Pythagoras theorem]**

**25 ^{two}
= Advertising^{2}+7^{2}
**

**Advertising ^{ii}
= 25^{2}-7^{2}
**

**AD ^{2}
= 625-49**

**AD ^{2}
= 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.**

**BC ^{2}
= BM^{2}+CM^{2}
**

**BC ^{2}
= 10^{2}+24^{ii}
**

**BC ^{2}
= 100+576**

**BC ^{2}
= 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.**

**HE ^{2}
= AE^{2}+AH^{ii}
[Pythagoras theorem]**

**HE ^{ii}
= 3^{2}+3^{2}
**

**HE ^{two}
= ix+9 = eighteen**

**HE = √(9×2) = 3√ii cm**

**Similarly GF = iii√2 cm**

**EBF is a right triangle.**

EF^{2}

= BE^{ii}+BF^{2}

[Pythagoras theorem]

**EF ^{2}
= 4^{2}+iv^{2}
**

**EF ^{2}
= 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**

**EG ^{2}
= EF^{2}+GF^{ii}
**

**EG ^{2}
= (iv√2)^{2}+(3√2)^{2}
**

**EG ^{2}
= 32+eighteen = fifty**

**EG = √50 = 5√2 cm …(i)**

**Join HF.**

**Also HF ^{2}
= EH^{2}+HG^{2}
**

= (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 cm ^{2}
**

**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 cm ^{2}
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}

= AD^{2}+BD^{ii}

BD = ½ BC

AB^{two}

= AD^{ii}+( ½ BC)^{two}

AB^{two}

= Advertizement^{2}+( ½ AB)^{two}

[∵BC = AB]

AB^{2}

= AD^{2}+ ¼ AB^{2}

AB^{ii}

= (4AD^{ii}+ AB^{ii})/4

4AB^{two}

= 4AD^{2}+ AB^{two}

4AD^{2
}= 4AB^{2}– AB^{2}

4AD^{two
}= 3AB^{ii}

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)4AD ^{ii}
= 4AC^{two}+BC^{2}
**

**(ii)4BE ^{2}
= 4BC^{2}+AC^{2}
**

**(three)4(Advertisement ^{2}+Be^{ii}) = 5AB^{2}**

**Solution:**

Proof:

(i)C = 90˚

So ACD is a right triangle.

Advert^{2}

= Air conditioning^{two}+CD^{2}

[Pythagoras theorem]

Multiply both sides by 4, we get

4AD^{2}

= 4AC^{2}+4CD^{2}

4AD^{2}

= 4AC^{two}+4BD^{2}

[∵D is the midpoint of BC, CD = BD = ½ BC]

4AD^{2}

= 4AC^{2}+(2BD)^{2}

4AD^{ii}

= 4AC^{2}+BC^{2}….(i) [∵BC = 2BD]

Hence proved.

(ii)BCE is a right triangle.

BE^{ii}

= BC^{ii}+CE^{ii}

[Pythagoras theorem]

Multply both sides by four , we get

4BE^{2}

= 4BC^{two}+4CE^{2}

4BE^{2}

= 4BC^{two}+(2CE)^{2}

4BE^{2}

= 4BC^{2}+Air conditioning^{2}

….(ii) [∵E is the midpoint of AC, AE = CE = ½ AC]

Hence proved.

(3)Adding (i) and (ii)

4AD^{2}+4BE^{2}

= 4AC^{2}+BC^{2}+4BC^{2}+Air conditioning^{2}

4AD^{2}+4BE^{2}

= 5AC^{2}+5BC^{2}

4(Advertising^{2}+Be^{2}

) = 5(Air conditioning^{2}+BC^{two})

4(AD^{2}+BE^{2}

) = 5(AB^{2}) [∵ABC is a right triangle, AB^{2}

= AC^{2}+BC^{2}]

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.

AB^{2}

= AP^{2}+BP^{2}

[Pythagoras theorem]

AB^{2}

= AP^{2}+(BD-PD)^{ii}

AB^{two}

= AP^{2}+BD^{two}+PD^{2}-2BD×PD

AB^{ii}

= (AP^{2}+PD^{ii})+BD^{two}-2BD×PD

AB^{2
}= AD^{two}+ (½ BC)^{2}-two×( ½ BC)×PD [∵AP^{2}+PD^{2}

= Advertizing^{2}

and BD = ½ BC]

AB^{ii
}= Advert^{two}+ ¼ BC^{2}– BC×PD ….(i)

APC is a right triangle.

Ac^{ii}

= AP^{ii}+PC^{2}

[Pythagoras theorem]

AC^{ii}

= AP^{ii}+(PD^{2}+DC^{ii})

Air conditioning^{ii}

= AP^{two}+PD^{2}+DC^{2}+2×PD×DC

Air conditioning^{2}

= (AP^{two}+PD^{two})+ (½ BC)^{2}+2×PD×( ½ BC) [DC = ½ BC]

AC^{2}

= (Advertisement)^{2}+ ¼ BC^{2}+PD× BC …(two) [In APD, AP^{two}+PD^{ii}

= AD^{2}]

Adding (i) and (ii), we get

AB^{2}+AC^{two}

= 2AD^{2}+ ½ BC^{2}

…..(iii)

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

Similarly nosotros get,

BC^{2}+CA^{ii}

= 2CF^{two}+ ½ AB^{2}

….(iv)

AB^{ii}+BC^{ii}

= 2BE^{2}+ ½ Ac^{two}

….(v)

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

2(AB^{two}+BC^{2}+CA^{ii}) = 2(Advertising^{two}+BE^{2}+CF^{2})+ ½ (BC^{2}+AB^{two}+Air-conditioning^{two})

2(AB^{2}+BC^{2}+CA^{two}) = two(AB^{ii}+BC^{2}+CA^{2}) – ½ (AB^{2}+BC^{two}+CA^{2})

two(Advertisement^{2}+BE^{2}+CF^{2}) = (iii/2)× (AB^{2}+BC^{two}+CA^{two})

four(AD^{2}+BE^{2}+CF^{2}) = 3(AB^{two}+BC^{2}+CA^{2})

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 that****AB² + 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.

AB^{ii}

= OB^{2}+OA^{2}

…(i) [Pythagoras theorem]

COD is a correct triangle.

CD^{ii}

= OC^{2}+OD^{2}

…(ii) [Pythagoras theorem]

Calculation (i) and (ii), we get

AB^{2}+ CD^{2}

= OB^{two}+OA^{2}+ OC^{2}+OD^{2}

AB^{2}+ CD^{2}

= (OA^{2}+OD^{ii})+ (OC^{2}+OB^{2}) …(iii)

AOD is a correct triangle.

Advertising^{2}

= OA^{2}+OD^{two}

…(iv) [Pythagoras theorem]

BOC is a right triangle.

BC^{ii}

= OC^{ii}+OB^{2}

…(5) [Pythagoras theorem]

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

AB^{2}+ CD^{two}

= AD^{2}+BC^{2}

Hence proved.

**24. In a quadrilateral ABCD, B = 90° = D. Evidence that 2 Air-conditioning² – BC ^{ii}
= AB² + Advertizement² + DC².**

**Solution:**

Given B = D = 90˚

Then ABC and ADC are correct triangles.

In ABC,

AC^{2}

= AB^{2}+BC^{two}

…(i) [Pythagoras theorem]

In ADC,

AC^{ii}

= Ad^{2}+DC^{2}

…(2) [Pythagoras theorem]

Calculation (i) and (ii)

2AC^{2}

= AB^{ii}+BC^{2}+ Advertizing^{2}+DC^{2}

2AC^{2}

-BC^{two}

= AB^{2}+AD^{ii}+DC^{2}

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,

DC^{2}

= CA^{2}+Advertizement^{ii}

[Pythagoras theorem]

DC^{2}

= CA^{2}+(AB+BD)^{2}

DC^{2}

= CA^{two}+AB^{2}+BD^{ii}+2AB×BD

DC^{two}

-BD^{2
}= CA^{two}+AB^{2}+2AB×BD

DC^{ii}

-BD^{2}

= AB^{ii}+AB^{2}+2AB×BD [∵CA = AB]

DC^{ii}

-BD^{2}

= 2AB^{2}+2AB×BD

DC^{2}

-BD^{two}

= 2AB(AB+BD)

DC^{2}

-BD^{2}

= 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.

AD^{ii}

= AP^{2}+PD^{ii}

[Pythagoras theorem]

AD^{2}

= AP^{2}+(PC+CD)^{two}

[PD = PC+CD]

AD^{2}

= AP^{2}+PC^{2}+CD^{2}

+2PC×CD ….(i)

APC is a right triangle.

AC^{2}

= AP^{2}+PC^{ii}

…(ii) [Pythagoras theorem]

Substitute (two) in (i)

Advertizement^{ii}

= Air-conditioning^{2}

+CD^{2}+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]

AD^{2}

= AC^{2}

+CD^{2}+ii× ½ BC ×CD

Advertising^{ii}

= Air conditioning^{2}

+CD^{2}+BC×CD

Advert^{2}

= Air conditioning^{2}

+CD(CD+BC)

AD^{ii}

= AC^{2}

+CD×BD [CD+BC = BD]

AD^{2}

= AC^{2}

+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}

= Advert^{2}+DC^{2}

[Pythagoras theorem]

17^{two}

= xv^{ii}+DC^{2}

289 = 225+DC^{2}

DC^{2}

= 289-225

DC^{2}

= 64

Taking square root on both sides,

DC = viii cm

ADB is a right triangle.

AB^{2}

= Advertizing^{2}+BD^{ii}

[Pythagoras theorem]

25^{ii}

= fifteen^{ii}+BD^{two}

625 = 225+ BD^{ii}

BD^{2}

= 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}

= Advertisement^{ii}+DC^{ii}

[Pythagoras theorem]

AC^{two}

= 6^{2}+8^{2}

Ac^{2}

= 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.

BC^{ii}

= AC^{2}+AB^{2}

[Pythagoras theorem]

26^{2}

= x^{2}+AB^{2}

AB^{ii}

= 26^{2}-10^{ii}

AB^{2}

= 676-100

AB^{2}

= 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 cm^{ii}

Surface area of ADC = ½ ×Advert×DC

= ½ ×6×8

= 24 cm^{2}

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

= 120-24

= 96 cm^{2}

Hence the area of shaded region is 96 cm^{ii}.

(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.

AC^{2}

= AB^{2}+BC^{two}

[Pythagoras theorem]

15^{2}

= 9^{two}+BC^{2}

225 = 81+BC^{2}

BC^{2}

= 225-81

BC^{2}

= 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.

AE^{2}

= AD^{two}+DE^{2}

[Pythagoras theorem]

(15/2)^{2}

= (9/2)^{2}+DE^{2}

DE^{2}

= (xv/2)^{2}

– (9/2)^{2}

DE^{2}

= 225/4 -81/4

DE^{2}

= 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 cm^{2}

Hence the area of the ADE is xiii.5 cm^{2}.

**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,

AB^{2}

= Ad^{2}+BD^{2}

[Pythagoras theorem]

Advertizing^{two}

= AB^{2}-BD^{ii}

…(i)

In ADC,

AC^{2}

= Ad^{2}+CD^{two}

[Pythagoras theorem]

Advertising^{2}

= Air conditioning^{2}-CD^{2}

…(two)

Equating (i) and (2)

AB^{2}-BD^{2}

= Air conditioning^{ii}-CD^{ii}

AB^{two}-AC^{2}

= BD^{ii}– 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}

= AC^{2}+CQ^{2}

[Pythagoras theorem]

Multiplying both sides by 9, we get

9AQ^{2}

= 9AC^{2}+9CQ^{ii}

9AQ^{two
}= 9AC^{2}+(3CQ)^{two}

…(i)

Given BQ: CQ = 1:2

CQ/BC = CQ/(BQ+CQ)

CQ/BC = two/3

3CQ = 2BC ….(ii)

Substitute (2) in (i)

9AQ^{2
}= 9AC^{2}+(2BC)^{two}

9AQ^{2
}= 9AC^{2}+4BC^{ii}

…(iii)

Hence proved.

(ii) ) In BPC,

BP^{ii}

= BC^{2}+CP^{2}

[Pythagoras theorem]

Multiplying both sides by 9, we get

9BP^{two}

= 9BC^{2}+9CP^{2}

9BP^{2}

= 9BC^{2}+(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}

= 9BC^{2}+(2AC)^{2}

9BP^{2}

= 9BC^{2}+4AC^{ii}

..(vi)

Hence proved.

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

9AQ^{two}+9BP^{two}

= 9AC^{2}+4BC^{2}+9BC^{2}+4AC^{2}

9(AQ^{2}+BP)^{2}

= 13AC^{2}+13BC^{2}

9(AQ^{2}+BP)^{2}

= xiii(AC^{2}+BC^{ii})…(vii)

In ABC,

AB^{2}

= AC^{2}+BC^{2}

…..(eight)

Substitute (eight) in (viii), we get

nine(AQ^{2}+BP)^{2}

= 13AB^{2}

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,

PR^{2}

= RQ^{two}+PQ^{ii}

[Pythagoras theorem]

PR^{2}

= (3x)^{ii}+PQ^{2}

PR^{2}

= 9x^{2}+PQ^{ii}

Multiply above equation past three

3PR^{2}

= 27x^{2}+3PQ^{ii}

….(i)

Similarly in PTS,

PT^{ii}

= TQ^{2}+PQ^{2}

[Pythagoras theorem]

PT^{2}

= (2x)^{2}+PQ^{2}

PT^{2}

= 4x^{2}+PQ^{2}

Multiply above equation past 8

8PT^{2}

= 32x^{2}+8PQ^{2}

….(ii)

Similarly in PSQ,

PS^{two}

= SQ^{2}+PQ^{two}

[Pythagoras theorem]

PS^{2}

= x^{two}+PQ^{2}

Multiply higher up equation past five

5PS^{2}

= 5x^{two}+5PQ^{2}

…(3)

Add (i) and (3), nosotros become

3PR^{2}

+5PS^{ii}

= 27x^{ii}+3PQ^{2}+5x^{2}+5PQ^{2}

3PR^{2}

+5PS^{2}

= 32x^{2}+8PQ^{2}

3PR^{2}

+5PS^{2}

= 8PT^{2}

[From (ii)]

8PT^{two}

= 3PR^{ii}

+5PS^{2}

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 conditioning^{2}

= AB^{2}+BC^{ii}

….(i) [Pythagoras theorem]

Given

**AD² = AB² + BC² + CD²**

AD² = AC^{2}+CD^{2}

[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-conditioning^{2}+AB^{2}

[Pythagoras theorem]

BC^{2}

= b^{2}+c^{2}

BC = √( b^{2}+c^{2}) …(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 ×√( b^{2}+c^{2}) ×Advertizement

Advertising = bc /(√( b^{2}+c^{ii})

Hence Advertizing is bc /(√( b^{2}+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 cm ^{2}]**

**108 = (one/12)x ^{ii}
**

**x ^{2}
= 108×12**

**x ^{2}
= 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 ,

BC^{2}

= Be^{2}+EC^{ii}

[Pythagoras theorem]

BC^{2}

= Be^{two}+(AC-AE)^{two}

BC^{ii}

= BE^{2}+AC^{two}+AE^{ii}-2 Air-conditioning×AE

BC^{ii}

= Be^{2}+AE^{2}+Air-conditioning^{2}-2 AC×AE …(i)

In ABC,

AB^{2}

= BE^{2}+AE

^{2}

..(2)

Substitute (ii) in (i)

BC^{2}

= AB^{2}+Air-conditioning^{2}-2 AC×AE

BC^{2}

= AC^{2}+Air conditioning^{2}-2 Ac×AE [∵AB = AC]

BC^{two}

= 2AC^{2}-2 AC×AE

BC^{2}

= 2AC(Air-conditioning-AE)

BC^{2}

= 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.

## If Ef Bisects Cd Cg 5x 1 Gd 7x 13

Source: https://byjus.com/ml-aggarwal-solutions/class-9-maths-chapter-12/

Originally posted 2022-08-07 07:16:12.