Chapter – 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9

Let a be the side of the square.

Therefore, the diagonal = a√2

But, a√2 = 20√2

So, a = 20

Therefore, the length for fencing the wall surrounding the square

= 4 × side

= 4 × 20 m = 80 m

The length of the rectangular land = 2.5 Dm = 25 meters

The width of the rectangular land = 1.7 Dm = 17 meters

Length of the outer rectangle (including path)

= 25 + 2 × 5 = 35 meters

Width of the outer rectangle (including path)

= 17 + 2 × 5 = 27 meters

Perimeter (fencing length) of the outer rectangle

= 2 × (Length + Width)

= 2 × (35 + 27)

= 2 × 62 = 124 meters

Cost of fencing = ₹18 × 124 = ₹2232

(i) Perimeter = 2 × (18 + 12) cm

= 2 × 30 cm

= 60 cm

(ii) Perimeter = 4 × 9 cm

= 36 cm

(iii) Perimeter = 8 cm + 9 cm + 15 cm + 7 cm

= 39 cm

(iv) Perimeter = 12 cm + 12 cm + 21 cm + 21 cm

= 66 cm

(v) Perimeter = 5 cm + 13 cm + 12 cm

= 30 cm

(vi) Perimeter = 14 cm + 14 cm + 17 cm

= 45 cm

(i) We have:

AC = 13 cm, BC = 5 cm

In the right-angled triangle ABC, we use the Pythagoras theorem:

AB² = AC² − BC²

= (13)² − (5)²

= 169 − 25

= 144

So, AB = √144 = 12 cm

Therefore, Area of triangle ABC

= ½ × BC × AB

= ½ × 5 × 12 sq. cm

= 30 sq. cm

(ii) Area of triangle ABC = (√3 / 4) × (6)²

= (√3 / 4) × 36

= 9√3 sq. cm

(iii) Area of triangle ABC = 1\over2 × 8 × √[(6)² − (8/2)²]

= 1\over2 × 8 × √(36 − 16)

= 4 × √20

= 4 × √(2 × 2 × 5)

= 4 × 2 × √5

= 8√5 sq. cm

(iv) Let a = 16 cm, b = 14 cm, c = 10 cm

Semi-perimeter (s) = \text{a + b + c} \over 2

= \text{16 + 14 + 10} \over 2

= 40 \over 2 = 20 cm

Area of triangle BCD = √[s(s − a)(s − b)(s − c)]

= √[20 × (20 − 16) × (20 − 14) × (20 − 10)]

= √(20 × 4 × 6 × 10)

= √(10 × 2 × 2 × 2 × 2 × 3 × 10)

= 2 × 2 × 10 × √3

= 40√3 sq. cm

Area of quadrilateral ABCD = 2 × area of triangle BCD

= 2 × 40√3

= 80√3 sq. cm

Let the depth of water be x.

Therefore,

AC = x cm, AB = 15 cm

So, BC = (x + 2) cm

We use the Pythagoras theorem:

BC² = AB² + AC²

(x + 2)² = 15² + x²

x² + 4x + 4 = 225 + x²

Now cancel x² from both sides:

4x + 4 = 225

4x = 221

x =221 \over 4

x = 55.25 cm

Since triangle ABC is an isosceles right-angled triangle,

therefore, AB = BC

Let AB = BC = x

We know that:

AB² + BC² = AC²

x² + x² = (12√2)²

x² + x² = 144 × 2

2x² = 288

x² = 144

x = √144 = 12

So, AB = BC = 12 cm

Now, area of triangle ABC

= 1\over2 × BC × AB

= 1\over2 × 12 × 12 = 72 sq. cm

Let a = 65 m, b = 70 m, c = 75 m

Step 1: Calculate semi-perimeter (s)

s = (a + b + c) / 2

= (65 + 70 + 75) / 2

= 210 / 2 = 105 m

Step 2: Use Heron’s formula to find the area of triangle ABC

Area of triangle ABC = √[s(s − a)(s − b)(s − c)]

= √[105 × (105 − 65) × (105 − 70) × (105 − 75)]

= √[105 × 40 × 35 × 30]

= √[35 × 3 × 10 × 2 × 2 × 35 × 3 × 10]

= 35 × 3 × 2 × 10 = 2100 sq. m

Step 3: Use area formula to find the length of perpendicular (CD)

We know:

Area = (1/2) × base × height

2100 = (1/2) × 75 × CD

75 × CD = 4200

CD = 4200 / 75

CD = 56 m

We have.

Ratio of the two triangles = 3 : 4

Ratio of area of the two triangles = 4 : 3

\therefore \frac{\text { Area of } 1 \text { st. triangle }}{\text { Area of } 2 \text { nd. triangle }}=\frac{4}{3}

or, \frac{\text { Base of } 1 \text { st. triangle }}{\text { Base of } 2 \text { nd. triangle }} \times \frac{\text { Height } 1 \text { st triangle }}{\text { Height } 2 \text { nd triangle }}=\frac{4}{3}

or, \frac{\text { Base of } 1 \text { st. triangle }}{\text { Base of } 2 \text { nd. triangle }} \times \frac{3}{4}=\frac{4}{3}

\therefore \quad \frac{\text { Base of } 1 \text { st. triangle }}{\text { Base of } 2 \text { nd triangle }} \times \frac{16}{9}

\therefore Ratio of two bases = 16 : 9

(i) Area of Kamal’s garden = 20 m × 20 m

= 400 sq. m

(ii) Area of Kamal’s verandah = 10 m × 5 m

= 50 sq. m

Therefore, cost of repairing the floor of Kamal’s verandah

= ₹30 × 50

= ₹1500

(iii) Length of the reading room = 6 m = 600 cm

Breadth of the reading room = 5 m = 500 cm

Size of one tile = 25 cm × 25 cm

No. of tiles =\frac{600 \times 500}{25 \times 25} = 480

(i)

Area of rectangle ABCD = AB × AD

= 12 m × 8 m

= 96 sq. m

Now,

AE = 12 m − 3 m = 9 m

AG = 8 m − 3 m = 5 m

Area of rectangle AEFG = AE × AG

= 9 m × 5 m = 45 sq. m

Area of the coloured part = 96 − 45 = 51 sq. m

(ii)

AB = 14 m, BC = 26 m

Area of rectangle ABCD = AB × BC

= 14 m × 26 m

= 364 sq. m

DE = (26 − 3) / 2 = 23 / 2 = 11.5 m

DG = (14 − 3) / 2 = 11 / 2 = 5.5 m

Area of one small square (DEFG) = DE × DG

= (23 / 2) × (11 / 2)

= 253 / 4 = 63.25 sq. m

Area of 4 small squares = 4 × 63.25 = 253 sq. m

Area of the coloured part = 364 − 253 = 111 sq. m

(iii)

AB = 16 m, AD = 9 m

EF = 16 m + 2 × 4 m = 16 m + 8 m = 24 m

HE = 9 m + 2 × 4 m = 9 m + 8 m = 17 m

Area of rectangle ABCD = AB × AD

= 16 m × 9 m

= 144 sq. m

Area of rectangle EFGH = EF × HE

= 24 m × 17 m

= 408 sq. m

Area of coloured part = 408 − 144 = 264 sq. m

(iv)

AB = 28 m, BC = 20 m

EF = 28 − 2 × 3 = 28 − 6 = 22 m

FG = 20 − 2 × 3 = 20 − 6 = 14 m

Area of rectangle ABCD = AB × BC

= 28 m × 20 m

= 560 sq. m

Area of rectangle EFGH = EF × FG

= 22 m × 14 m

= 308 sq. m

Area of the coloured part = 560 − 308

= 252 sq. m

(v)

Because BC = 120 cm, CD = 90 cm

Area of rectangle ABCD = BC × CD

= 120 cm × 90 cm

= 10,800 sq. cm

II = BC = 120 cm, IJ = 3 cm

Area of rectangle IJKL = BC × IJ

= 120 cm × 3 cm

= 360 sq. cm

EH = 3 cm,

FE = BJ = (90 − 3) / 2 = 87 / 2 cm

IL = BC = 120 cm, IJ = 3 cm

EH = 3 cm,

FE = BJ = 87 / 2 cm

Area of rectangle EFGH = EH × FE

= 3 cm × (87 / 2) cm = 261 / 2 sq. cm

Area of 4 small rectangles = 4 × (261 / 2)

= 522 sq. cm

Area of coloured part = 360 + 522

= 882 sq. cm

Let length = 4x

Let breadth = 3x, where x is the common ratio.

Therefore, Perimeter = 2(4x + 3x)

= 2 × 7x

= 14x

According to the problem,

14x = 336

x = 336 ÷ 14 = 24

Length = 4 × 24 = 96 m

Breadth = 3 × 24 = 72 m

Therefore, Area of the rectangular field = Length × Breadth

= 96 m × 72 m

= 6912 sq. m

The cost of farming a square land of Samar at the rate of ₹3.50 per square meter is ₹1400.

Therefore, area of the square land = 1400 ÷ 3.50

= 400 sq. meters

Length of side of the square land = √Area

= √400

= 20 meters

Length of fencing around its four sides = 4 × side

= 4 × 20

= 80 meters

∴ Total cost of fencing = ₹8.50 × 80

= (850 ÷ 100) × 80

= ₹680

Let the length of the rectangular land be x

and the breadth be y

Area of the rectangular land = 500 sq. m

So, x × y = 500 ………….. (i)

If the length is decreased by 3 m and the breadth is increased by 2 m, the land becomes a square.

or, x − 3 = y + 2

Solving,

x = y + 2 + 3

x = y + 5 ……………. (ii)

Putting the value of x in equation (i):

(y + 5) × y = 500

or, y² + 5y − 500 = 0

or, y² + 25y − 20y − 500 = 0

or, y(y + 25) − 20(y + 25) = 0

or, (y + 25)(y − 20) = 0

So, either

y + 25 = 0

y = −25

or

y − 20 = 0

y = 20

Since distance cannot be negative,

y = 20

From equation (ii):

x = y + 5 = 20 + 5 = 25

Therefore,

Length of the rectangular field = 25 m

Breadth = 20 m

Side of a square land = 300m

Area \text{ " " " " " " " " "} =(3.00)^2 \text { Sq.m. } \\

=90000 \text { Sq.m. } \\

\text { Wide of the wall }=3 \mathrm{dcm} . \quad=0.3 \mathrm{~m} \text {. } \\

\text { Length of the square with wide }=(300+2 \times 0.3)^2 \mathrm{~m} \\

=(300+0.6) \mathrm{m} \\

=300.6 \mathrm{~m} \\

=(300.6)^2 \\

=90360.36 \mathrm{Sq} \cdot \mathrm{m} \\

\text { Area of } 4 \text { walls }=(90360.36-90000) \text { Sq.m. } \\

=360.36 \text { Sq.m. } \\

Cost of the wall 100 Sq.m at the rate of Rs. 5,000

\text { " " " } 1 \text { " } 1 \text { " } 360.36 \text { sq.m:" } "

=\frac{5,0\cancel{00}}{1\cancel{00}}=\text { Rs. } 50 \\

\text { " " " "360.36sq.m. " "" " " } =\text { Rs. } 360.36 \times 50 \\

=\text { Rs. } \frac{36036}{100} \times 50 \\

=\text { Rs. } 18018 \text { (Ans.) }

Length of rectangular garden = 14m

Breadth ” ” ” ” ” ” ” = 12m

Let the width of the park = x

\text { Length of rectangular garden } =14-2 × x \\

= 14 – 2x

Breadth ” ” ” ” ” ” = 12 – 2 × x

= 12 – 2x

Cost of constructing of wide path of 20 sq.m = Rs. 1380

\text{ " " " " " " " " " " " " "} 1 \ sqm. =Rs\frac{1380}{20} \\

=\text { Rs. } 69

According to the condition of the problem

(14-2 x)(12-2 x)=69 \\

\text { or, } 168-28 x-24 x+4 x^2=69 \\

\text { or, } 4 x^2-52 x+168-69=0 \\

\text { or, } 4 x^2-52 x+69=0 \\

\text { or, } 4 x^2-46 x-6 x+69=0 \\

\text { or, } 2 x(2 x-23)-3(2 x-23)=0 \\

\text { or, }(2 x-23)(2 x-3)=0

either,

2 x-23=0 \quad \text { or, } \quad 2 x-3=0

or, 2 x=23 \quad \text { or, } 2 x=3

\therefore x=\frac{23}{2}=11.5 \quad \therefore x=\frac{3}{2}=1.5

\therefore \quad x = 11.5 is not possible because wide can not be greater or equal or nearest to recatangular length

\therefore x = 1.5

\therefore \quad Wide of the path = 1.5m (Ans.)

Length of rectangular garden = 40cm.

Area \text{ " " " " " " " " " " " " "} = 1200 Sq.cm.

Breadth \text{ " " " " " " " " " " " " "} = \frac{\cancel{1200}}{\cancel{40}} cm

= 30cm.

\therefore \text { Lenth of diagonal of rectangle } =\text { side of the square. } \\

=\sqrt{(40)^2+(30)^2} \\

=\sqrt{1600+900} \\

=\sqrt{2500} \\

=50 \mathrm{~cm} .

\therefore \text { Area of the square field } =(\text { Side })^2 \\

=(50)^2 \\

=2500 \text { Sq.cm. } \quad \text { (Ans.) }

Length = 4m, Breadth = 6cm, height = 4m :

\text { Area of one door } =1.5 \times 1 \mathrm{~m} =1.5 \mathrm{Sq} . \mathrm{m} . \\

\text { Area of three doors } =3 \times 1.5 \mathrm{Sq} . \mathrm{m} =4.5 \mathrm{Sq} . \mathrm{m} . \\

\text { Area of one window } =1.2 \mathrm{~m} \times 1 \mathrm{~m} =1.2 \mathrm{Sq} . \mathrm{m} \\

” ” ” ” \quad 4 \text { Windows } =1.2 \times 4 \mathrm{Sq} . \mathrm{m} . =4.8 \mathrm{Sq} . \mathrm{m} .

Area of the walls with doors and windows

=2 \text { (Length }+ \text { Breadth) } \times \text { height. } \\

=2(4 \mathrm{~m}+6 \mathrm{~m}) \times 4 \mathrm{~m} \\

=2 \times 10 \mathrm{~m} \times 4 \mathrm{~m} . \\

=80 \mathrm{Sq} . \mathrm{m} .

Area of the walls without doors and windows.

=\{80-(4.5+4.8)\} \text { Sq.m. } \\

=(80-9.3) \mathrm{Sq} \cdot \mathrm{m} \\

=70.7 \mathrm{Sq} \cdot \mathrm{m} .

\text { Cost of colouring } =\text { Rs. } 70 \times 70.7 \\

=\text { Rs. } 4949 \quad \text { (Ans.) }

Length of room = 4m

Area of the floor = 12 \mathrm{Sq} \cdot \mathrm{m}

or, Length × Breadth = 12

\text { or, Breadth }=\frac{12}{\text { Length }} \\

=\frac{12}{4}=3 \mathrm{~m} \text {. } \\

\therefore Area of four walls of a room = 42 sq.cm

or, 2 (Length + Breadth) \times height =42

or, 2(4+3) \times height =42

or, 2 \times 7 \times height =42

or, 14 \times height =42

or, height =\frac{\cancel{42}}{\cancel{14}}

\therefore height =3 \mathrm{~m}.

\therefore Height of room =3 \mathrm{~m} (Ans)

Let the length of the rectangular picture be x

and ” breadth” \text{ " " " " " " " " " " " " "} y

\therefore \quad Area = 84 Sq.cm.

or, x \times y=84

or, xy = 84…………… (i)

According to the condition of the problem,

x – y = 5

\text { or, } \quad x=5+y………. (ii)

Putting the value of x in equation ……………… (i)

(5 + y) y = 84

\text { or, } 5 y+y^2=84 \\

\text { or, } y^2+5 y-84=0 \\

\text { or, } y^2+12 y-7 y-84=0 \\

\text { or, } y(y+12)-7(y+12)=0 \\

\text { or, }(y+12)(y-7)=0

Either,

y + 12 = 0

or, y = -12

or, y – 7 = 0

\therefore \quad y = 7

\because Length can’t be negative

\therefore y=7

From (ii),

x =5 + y

= 5 + 7 = 12

\therefore Length of the rectangular picture = 12m.

\text { Breadth " " " " " } = 7m.

\therefore Perimeter ” ” ” ” = 2(12+7)m

=2 \times 19 \mathrm{~m}

=38 \mathrm{~m} \text {. (Ans.) }

Let the side of the square garden be x There is a 2.5m wide path around the square.

Side of square garden with path

=(x+2 \times 25) \quad =(x+5)

Accodring to the condition of the problem,

(x+5)^2-(x)^2=165

or, (x)^2+2 \cdot x .5+(5)^2-x^2=165

or, \cancel{x^2}+10 x+25- \cancel{x^2}=165

or, 10 \mathrm{x}=165-25

or, 10 \mathrm{x}=140

\therefore \quad \mathrm{x}=14

\therefore Side of the square garden

\text { Side of the square garden } =14 \mathrm{~m} \\

\text { Area" } =(\text { side })^2 \\

=(14)^2 \\

=196 \text { Sq.m. }

\text { Length of diagonal } =\text { side } \sqrt{2} \mathrm{~m} \\

=14 \sqrt{2} \mathrm{~m} (Ans.)

Length of the diagonal of the square land = 20 \sqrt{2} \mathrm{~m}

\therefore \quadSide of the square land = 20m

Length of wall for walling outside round the square field = perimeter of the square.

= 4 × side

= 4 × 20m = 80m

\therefore \quad Area of the square land = (\text { side })^2

=(20)^2 \\

=400 \text { Sq. } \cdot \mathrm{m}

\therefore Cost for planting grass at the rate of Rs. 20 per Sq.m

=\text { Rs. } 20 \times 400 \\

=\text { Rs. } 8000 \quad \text { (Ans.) }

Length of the rectangular garden = 12m.

\text { Breadth" " " " " " " " " " " " } = 7m

\text { Length of diagonal } =\sqrt{(\text { Length })^2+(\text { breadth })^2} \\

=\sqrt{(12)^2+(7)^2}

=\sqrt{193 \mathrm{~m}} \\

\text { Perimeter of the triangle } =12 \mathrm{~m}+7 \mathrm{~m}+\sqrt{193 \mathrm{~m}} \\

= (19+\sqrt{193}) \mathrm{m} \text { (Ans.) }

Let length of rectangle be 9x

\text { Breadth" " " " " " " " " " " " } 5 x

\therefore Perimeter of rectangle = 140m

or, 2 (Length + Breadth )= 140

or, 2(9x + 5x) = 140

or, x =\frac{140}{4 \times \sqrt{4}}

\therefore \quad x = 5

\therefore \quad Length of rectangle = 9 \mathrm{x}=9 \times 5=45 \mathrm{~m}

Breadth of rectangle =5 \mathrm{x}=5 \times 5=25 \mathrm{~m}

Area of rectangle = length \times breadth

=45 \mathrm{~m} \times 25 \mathrm{~m} \\

=1125 \text { Sq.m }

\text { Dimension of one tile } =25 \mathrm{~cm} \times 20 \mathrm{~cm} \\

=\frac{25}{100} \times \frac{20}{100} \text { Sq.m. } \\

=\frac{1}{20}

\text { No. of tiles } =1125 \div \frac{1}{20} \\

=1125 \times 20=22500

Rate of each 100 tiles = Rs. 500

\text { " " " " } 1tile = Rs. \frac{500}{100} \\

\text { " " " } 22500 Rs. \frac{500}{100} \times 22500 \\

=\mathrm{Rs}: 112500 \text { (Ans.) }

Length of Rectangle = 18m

Let breadth of Rectangle be x

Total cost = Rs. 2160

Area of Rectangle = 18 \times x=18 \times Sq.m.

If the breadth of the floor would be 4 meter less.

\therefore Breadth of Rectangle = x – 4

Area of Rectangle =18(x-4) \mathrm{Sq} . \mathrm{m}

Cost of carpeting of 18x Sq. m. = Rs .2160

\text {" " " " " " " " " " " " }18(x – 4) Sq.m = Rs. 1620

Cost of carpeting of 18x – 18(x – 4)=Rs. (2160-1620)

\text {" " " " " " " " " " " " }18x – 18x + 72 = Rs. 540

\text {" " " " " " " " " " " " } 72 \mathrm{Sq} . \mathrm{m} .=Rs. 540

If Rs. 540, cost of carpeting of 72 Sq.m

” 1 \text {" " " " " " " " " " " " } \frac{72}{540} \mathrm{Sq} \cdot \mathrm{m}

” 1 \text {" " " " " " " " " " " " } \frac{72 \times 2160}{540} \mathrm{Sq} . \mathrm{m} \\

=288 \mathrm{Sq} . \mathrm{m}

\therefore \quad Area of Rectangle =288 Sq.m

or, 18x = 288

or, x =\frac{288}{18}

\therefore \quad x = 16

\therefore \quad \text { Breadth of Rectangle }=16 \mathrm{~m} \\

\text { Perimeter " " " " " " "} =2(\text { Length }+ \text { Breadth }) \\

=2(18 \mathrm{~m}+16 \mathrm{~m}) \\

=2 \times 34 \mathrm{~m}=68 \mathrm{~m}.\\

Area of Rectangle = 288 Sq.m. (Ans.)

Let the length of the rectangular land be x

and Breadth ” ” ” ” ” ” y

According to the condition of the problem,

\sqrt{x^2+y^2}=15

or, x^2+y^2=225

And x – y = 3

or, x = 3 + y

Putting the value of x in equation (i)

(3+y)^2+y^2=225

or, (3)^2+2 \cdot 3 \cdot y+y^2+(y)^2=225

or, 9+6 y+y^2+y^2-225=0

or, 2 y^2+6 y-216=0

or, y^2+3 y-108=0

or, y^2+12 y-9 y-108=0

or, y(y+12)-9(y+12)=0

or, (y+12)(y-9)=0

either, y + 12 = 0

y = -12

or, y – 9 = 0

y = 9

\therefore Length can not be negative.

\therefore y = 9

From (ii),

x = 3 + y

= 3 + 9 = 12

\therefore Length =12 \mathrm{~m} . \quad Breadth =9 \mathrm{~m}.

\therefore Perimeter of the rectangular land

=2 \text { (Length }+ \text { Breadth) } \\

=2(12 \mathrm{~m}+9 \mathrm{~m}) \\

=2 \times 21 \mathrm{~m} \\

=42 \mathrm{~m} .

Area of the rectangular land

=\text { length } \times \text { breadth } \\

=12 \mathrm{~m} \times 9 \mathrm{~m}=108 \text { Sq.m (Ans.) }

First we find the H.C.F of 385 and 60

\therefore \quad H.C.F of 385 and 60 = 5

Side of the square tile = 5m

Area of rectangular courtyard = 385 \mathrm{~m} \times 60 \mathrm{~m}

Area of one square tile = (\text { Side })^2

=(5)^2 \\

=25 \mathrm{Sq} \cdot \mathrm{m} .

\therefore \quad \text { No. of tiles } =\frac{385 \times 60}{25} \\

=924 \text { (Ans.) }

Length of diagonal of square 12 \sqrt{2} \mathrm{~m}

\therefore \quad \text { Side } \sqrt{2}=12 \sqrt{2} \mathrm{~m}.

\therefore \quad \text { Side of a square } =12 \mathrm{~m} \\

\therefore \quad \text { Area of square } =(\text { Side })^2 \\

=(12)^2 \\

=144 \mathrm{~m}^2

\therefore (b) is correct option

\quad Let A_1=a^2

and A_2=(a \sqrt{2})^2=2 a^2

\therefore A_1: A_2=1: 2

\therefore (a) is correct option

Area of Rectangle = 6 \mathrm{~m} \times 4 \mathrm{~m}

=24 \mathrm{~m}^2

\text { Area of square }=2 \mathrm{dm} \times 2 \mathrm{dm} \\

=\frac{2}{10} \times \frac{2}{10} \mathrm{~m}^2 \\

\text { No. of tiles }=\frac{24}{\frac{2}{10} \times \frac{2}{10}}=\frac{24 \times 10 \times 10}{2 \times 2}=600

No. of tiles =\frac{24}{\frac{2}{10} \times \frac{2}{10}}=\frac{24 \times 10 \times 10}{2 \times 2}=600

\therefore \quad(c) is correct option

\therefore (b) is correct option

Given,

Length of diagonal = 10

\sqrt{(\text { Length })^2+(\text { Breadth })^2}=10

or, (\text { Length }+ \text { Breadth })^2-2.length \times breadth =100

or, (\text { Length }+ \text { Breadth })^2-2 \times 62.5=100

or, (\text { Length }+ \text { Breadth) }^2-125=100

or, (\text { Length }+ \text { Breadth })^2=100+125=225

or, (\text { Length }+ \text { Breadth })^2=(15)^2

\therefore Length + breadth = 15cm.

\therefore (b) is correct option

Let the side of square be a

\therefore Area of square =\mathrm{a}^2

If the length of square is increased by 10%

Side of square = a + 10% of a

= a+\frac{10}{100} \times a \\

= a+\frac{a}{10}=\frac{11 a}{10} \\

\text { Area of square } =\left(\frac{11 a}{10}\right)^2 \\

=\frac{121 a^2}{100} \\

\text { Increased area of square } =\frac{121 a^2}{100}-a^2=\frac{121 a^2-100 a^2}{100}=\frac{21 a^2}{100}

Percentage of the area of square will be increased

=\frac{21 a^2}{\frac{100}{a^2}} \times 100 \\

=21 \% \text { (Ans.) }

Let’ l ‘ be the length of rectangle.

and ‘b'” ” breadth ” “

Area of rectangle = l \times b

If length is increased by 10% and breadth is decreased by 10%

\therefore \text { Increased length } =l+10 \% l \\

=l+\frac{10}{100} \times l \\

=l+\frac{l}{10}=\frac{11 l}{10}

Decreased beadth =\mathrm{b}-10 \% of \mathrm{b}

=b-\frac{10}{100} \times b \\

=b+\frac{b}{10}=\frac{9 b}{10} \\

\text { Area }=\frac{11 l}{10} \times \frac{9 b}{10}=\frac{99 l b}{100}

\text { Decreased area } =m-\frac{99 l b}{100} \\

=\frac{l b}{100}

\text { Percentage of area will be decreased } =\frac{l b}{100} \\

= lb \times 100 \% \\

=1 \% \text { (Ans.) }

From the figure,

\mathrm{AC} =5 \mathrm{~cm} \\

\mathrm{OE} =2 \mathrm{~cm}, \mathrm{OF}=2 \mathrm{~cm} \\

\therefore \quad \mathrm{EF}=\mathrm{BC} =2 \mathrm{~cm}+2 \mathrm{~cm} \\

= 4 \mathrm{~cm}

Let breadth of rectangle be b Then,

\sqrt{b^2+(4)^2}=5 \\

\text { or, } b^2+16=25 \\

\text { or, } b^2=25-16 \\

\text { or, } b^2=9 \\

\text { or, } b^2=3^2 \\

\therefore \quad b=3 \\

\therefore \quad \text { Length of breadth }=3 \mathrm{~cm}

\therefore \mathrm{EG}=\mathrm{FG}=2 \sqrt{2} \mathrm{~cm}.

\mathrm{EF} =(2 \sqrt{2}+2 \sqrt{2}) \mathrm{cm} \\

=4 \sqrt{2} \mathrm{~cm} \\

\therefore \quad \mathrm{EF} =\mathrm{BC}=4 \sqrt{2} \mathrm{~cm}

\text { Length of diagonal }=\operatorname{Side} \sqrt{2}

=4 \sqrt{2} \times \sqrt{2}

=4 \times 2 \mathrm{~cm}=8 \mathrm{~cm} \text {. (Ans.) }

Let ‘ l ‘ be the length and ‘ b ‘ be the breadth of the rectangle

Perimeter =34 \mathrm{~cm}.

or, 2(l+b)=34

or, l+\mathrm{b}=17

\therefore \quad l=17-b

\therefore \quad Area =60 Sq.cm.

l \times \mathrm{b}=60

or, (17-b) \times b=60

or, \quad 17 b-b^2=60

or, b^2-17 b+60=0

or, b^2-12 b-5 b+60=0

or, b(b-12)-5(b-12)=0

or, (b-12)(b-5)=0

either,

b – 12 = 0 \quad \text { or, } b-5=0 \\

b = 12 \quad b = 5

\therefore Breadth is always less than length

\therefore \quad Breadth = 5cm

\therefore \quad l=17-5=12 \\

\therefore \quad \text { Length }=12 \mathrm{~cm}

\text { Length of each diagonal } =\sqrt{l^2+\mathrm{b}^2} \\

=\sqrt{(12)^2+(5)^2} \mathrm{~cm} . \\

=\sqrt{144+25} \mathrm{~cm} . \\

=\sqrt{169} \mathrm{~cm} . \\

=13 \mathrm{~cm} . \quad \text { (Ans.) }

\text { Area of } \triangle A B C=\frac{\sqrt{3}}{4} \times(10)^2 \\

=\frac{\sqrt{3}}{4} \times 100 \\

=25\sqrt{3} \text { sq.cm } \\ (Ans.)

Area of \triangle \mathrm{ABC} =\frac{1}{2} \times 8 \times \sqrt{(10)^2 - (\frac{8}{2} )^2}

=4 \times \sqrt{100-16} \\

=4 \times \sqrt{84} \text { Sq. units. } \\

=4 \times \sqrt{4 \times 21} \text { Sq. units. } \\

=4 \times 2 \sqrt{21} \text { Sq. units. } \\

=8 \sqrt{21 \mathrm{Sq} \text {. units. }} \text { (Ans.) }

Area of Trapazium \mathrm{ABCD}=\frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{CD}

=\frac{1}{2}(5+4) \times 3 \\

=\frac{1}{2} \times 9 \times 3 \\

=\frac{27}{2} \text { Sq. Units } \\

=13.5 \text { Sq. units. }

\text { Area of Trapazium }=\frac{1}{2}(\mathrm{AD}+\mathrm{CD}) \times \mathrm{AD}

=\frac{1}{2}(15+40) \times 9 \\

=\frac{1}{2} \times 55 \times 9 \\

=\frac{495}{2} \\

=247.5 \text { Sq.cm (Ans.) }

\mathrm{AC}=42 \mathrm{~cm}, \mathrm{CD}=38 \mathrm{~cm}

\because \angle \mathrm{ADC}=90^{\circ}

We know that,

A D^2 =A C^2-C D^2 \\

=(42)^2-(38)^2 \\

=(42+38(42-38) \\

=80 \times 4=320

\therefore \quad A D =\sqrt{320} \\

=\sqrt{4 \times 4 \times 2 \times 2 \times 5} \\

=4 \times 2 \sqrt{5}=8 \sqrt{5} \mathrm{~cm}

\therefore \quad \text { Area of } \mathrm{ABCD} =\mathrm{CD} \times \mathrm{AD} \\

=38 \times 8 \sqrt{5} \text { Sq.cm. } \\

=304 \sqrt{5} \text { Sq.cm. (Ans.) }

\therefore The perimeter of an equilateral triangle = 48cm.

\therefore \quad Side of the equilateral triangle

=\frac{48}{3} \mathrm{~cm} . \\

=16 \mathrm{~cm} .

\therefore \quad Area of the equilateral triangle

=\frac{\sqrt{3}}{4} \times(16)^2 \\

=\frac{\sqrt{3}}{4} \times 256 \\

=44 \sqrt{3} \mathrm{~cm} . \quad \text { (Ans.) }

The height of an equilateral triangle \mathrm{ABC}=5 \sqrt{3} \mathrm{~cm}. We know that,

\therefore \frac{\sqrt{3}}{2} \times \text { side }=5 \sqrt{3}

or, \frac{\text { Side }}{2}=5

\therefore \quad Side = 10

\therefore \quad Side of an equilateral triangle = 10cm.

\therefore Area of an equilateral triangle.

=\frac{\sqrt{3}}{4} \times(10)^2 \\

=\frac{\sqrt{3}}{4} \times 100 \text { Sq.cm } \\

=25 \sqrt{3} \text { Sq.cm. }(Ans.)

Each equal side of an isosceles triangle ABC = 10cm.

Length of base = 4cm.

\text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 4 \times \sqrt{(10)^{2}-(_{2}^{4})^{2}} \\

=2 \times \sqrt{100-4} \\

=2 \times \sqrt{96} \\

=2 \times \sqrt{4 \times 4 \times 6} \\

=2 \times 4 \sqrt{6} \\

=8 \sqrt{8} \text { Sq.cm. (Ans.) }

Each equal side of an isosceles triangle ABC = 10cm.

Length of base = 12cm.

\therefore \quad \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times \sqrt{(10)^{2}-(\frac{12}{2})^{2}} \\

=6 \times \sqrt{100-36} \\

=6 \times \sqrt{64} \\

=6 \times 8= 48 \text { Sq.cm. (Ans.) }

Let the length of base be x

\therefore Length of each side =\frac{5}{6} \times x=\frac{5 x}{6}

\therefore \quad Perimeter =544 \mathrm{~cm} . \therefore \quad Perimeter =544 \mathrm{~cm}.

or, x+\frac{5 x}{6}+\frac{5 x}{6}=544

or, \frac{6 x+5 x+5 x}{6}=544

or, \frac{16 x}{6}=544

\text { or, } \frac{x}{6}=34 \\

\therefore x = 204

\therefore Length of the base \quad = 204 \mathrm{~cm} .

\text { Length of each side } =\frac{5}{6} \times 204 \\

= 170cm

\text { Area of } \mathrm{ABC}=\frac{1}{2} \times 204 \times \sqrt{(170)^{2}- \frac{204}{2} } \\

=102 \times \sqrt{28900-10404} \\

=102 \times \sqrt{18496} \\

=102 \times 136 \\

=13872 \text { Sq.cm. (Ans.) }

Let the length of equal side be x

We have,

A B^{2}+B C^{2}=A C^{2}

or, x^{2}+x^{2}=(12 \sqrt{2})^{2}

or, 2 x^{2}=144 \times 2

or, x^{2}=144

or, x=\sqrt{144}

\therefore x = 12

\therefore Length of equal side be 12cm.

\therefore \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times 12 \text { Sq.cm. } \\

=72 \text { Sq.cm (Ans.) }

\text { Let } A C=6 \mathrm{~cm}, \quad B D=8 \mathrm{~cm} . \\

\therefore \quad \mathrm{AO}=\mathrm{OC}=3 \mathrm{~cm} \\

\mathrm{BO}=\mathrm{OD}=4 \mathrm{~cm} . \\

\therefore \quad \angle C O D=90^{\circ} \\

\triangle \mathrm{In} \triangle \mathrm{COD} \\

\mathrm{CD}^{2}=O C^{2}+\mathrm{OD}^{2} \\

=(3)^{2}+(4)^{2} \\

=9+16=25 \\

\therefore \quad \mathrm{or}, \quad \mathrm{CD}=\sqrt{25}=5 \mathrm{~cm} . \\

\therefore \quad A B=B C=C D=D A=5 \mathrm{~cm} .

Hence ABCD is rhombus

Let a = 2x

b = 3x

c = 4x

\therefore \quad Perimeter =216m.

or, a + b + c = 216m

or, 2x + 3x + 4x = 216

or, 9x = 216

or, x =\frac{216}{9}

\therefore \quad x =24 \\

\therefore \quad a =2 x=2 \times 24=48 \mathrm{~m} \\

b = 3 \mathrm{x}=3 \times 24=72 \mathrm{~m} \\

c = 4 \mathrm{x}=4 \times 24=96 \mathrm{~m} \\

\therefore \quad s =\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} \\

=\frac{48+72+96}{2}=\frac{216}{2}=108 \mathrm{~m}.

\text { Area of the park } =\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\

=\sqrt{108(108-48)(108-72)(108-96)} \\

=\sqrt{108 \times 60 \times 36 \times 12} \\

=\sqrt{36 \times 3 \times 12 \times 5 \times 36 \times 12} \\

=36 \times 12 \sqrt{5 \times 3} \\

=432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m} \\

\text { Area of the park } =432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m}

Area of the park =432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m}

or, \quad \frac{1}{2} \times \mathrm{BC} \times \mathrm{AD}=432 \sqrt{15}

or, \frac{1}{2} \times 96 \times \mathrm{AD}=432 \sqrt{15}

\text { or, } \mathrm{AD}=\frac{432 \sqrt{15}}{48}

\text { or, } \mathrm{AD}=9 \sqrt{15 \mathrm{~m}} \text { (Ans.) }

\text { Let } a=26 \mathrm{~m} \\

\mathrm{~b}=28 \mathrm{~m} \\

\mathrm{c}=30 \mathrm{~m} \\

\mathrm{~s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} \\

=\frac{26+28+30}{2}=\frac{84}{2}=42 \mathrm{~m} . \\

Area of the triangular field =\sqrt{s(s-a)(s-b)(s-c)} =\sqrt{42(42-26)(42-28)(42-30)} \\

=\sqrt{42 \times 16 \times 14 \times 12} \\

=\sqrt{14 \times 3 \times 4 \times 4 \times 4 \times 14 \times 2 \times 2 \times 3} \\

=14 \times 4 \times 3 \times 3 \\=336 \text { Sq.m }

Cost of planting grass in the triangular field =336 \times 5

= Rs. 1680 (Ans.)

Perimeter of the triangular field =26 \mathrm{~m}+28 \mathrm{~m}+30 \mathrm{~m}=84 \mathrm{~m}

But leaving a space 5m for constructing entrance gate,

\therefore \quad \text { Length for fencing around three sides }=84 \mathrm{~m}-5 \mathrm{~m}

= 79m

\text { Total cost }=\text { Rs. } 79 \times 18=\text { Rs. } 1422 \text { (Ans.) }

Let ‘a’ be the side of an equilateral triangle.

Area of the triangle PQR =\frac{\sqrt{3}}{4} \times(\mathrm{a})^{2}

=\frac{\sqrt{3}}{4} \times a^{2}

According to the condition of the problem

\frac{\sqrt{3}}{4} \times \mathrm{a}^{2}=\frac{1}{2} \mathrm{a} \times 10+\frac{1}{2} \times a \times 12+\frac{1}{2} \times a \times 8

or, \frac{\sqrt{3}}{4} \times a^{2}=5 a+6 a+4 a

or, \frac{\sqrt{3}}{4} \times a^{2}=15 a

or, \frac{\sqrt{3}}{4} \times a=15 \quad[\because a \neq 0]

or, \sqrt{3} \mathrm{a}=60

or, a=\frac{60}{\sqrt{3}}

\text { or, }=\frac{20 \times \sqrt{3} \times \sqrt{3}}{\sqrt{3}} \\

\therefore \quad a=20 \sqrt{3}

\text { Area of } \triangle P Q R =\frac{\sqrt{3}}{4} \times(20 \sqrt{3})^{2} \\

=\frac{\sqrt{3}}{4} \times 20 \times 20 \times 3 \\

=300 \sqrt{3} \text { Sq.m (Ans.) }

In \triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}=20 \mathrm{~m} \ and \ \angle \mathrm{A}=45^{\circ},

CD is drawn perpendicular to AB

Now, if we take AB as base of the triangle,

Then its altitude is CD.

According to construction, \angle \mathrm{ACD}=45^{\circ}, \therefore \mathrm{AD}=\mathrm{CD}

In the right angled triangle ADC,

\mathrm{CD}^{2}+\mathrm{AD}^{2}=\mathrm{AC}{ }^{2}=(20)^{2} \mathrm{Sqm}=400 \mathrm{Sq} \cdot \mathrm{m} \\

\therefore \quad 2 \mathrm{CD}^{2}=400 \mathrm{Sq} \cdot \mathrm{m}[\because \mathrm{AD}=\mathrm{CD}]

\therefore \quad 2 \mathrm{CD}^{2}=200 \text { Sq.m } \therefore \mathrm{CD}=\sqrt{200} \\

\text { Area of the triangle }=\frac{1}{2} \times \mathrm{AB} \times \mathrm{CD} \\

=\frac{1}{2} \times 20 \times 10 \sqrt{2} \\

=100 \sqrt{2} \text { Sq } \mathrm{m} \text {. } \\

In \triangle A B C \\

AB = AC = 20cm

We draw perpendicular CD on AB

Then, \angle A D C=90^{\circ},

\angle \mathrm{CAD}=30^{\circ} (Given)

and, \angle \mathrm{ACD}=60^{\circ}

\therefore In any right angled triangle,

the three angles are 90^{\circ}, 60^{\circ} and \ 30^{\circ},

Then,\mathrm{CD}=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \times 20 \mathrm{~cm}=10 \mathrm{~cm}

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

=\frac{1}{2} 10 \times 10 \text { Sq.cm } \\

=100 \text { Sq.cm (Ans.) }

Let the equal sides be ‘ a ‘

And length of the base be ‘ b ‘

\therefore \quad Perimeter =\sqrt{2}+1

or, a+a+b=\sqrt{2}+1

or, 2 a+b=\sqrt{2}+1…………. (i)

We have,

A C^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}

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

or, b^{2}=2 a^{2}

or, b=\sqrt{2 a^{2}}

or, b=\sqrt{2} \mathrm{a}…………. (ii)

From (i),

2 a+b=\sqrt{2}+1

or, 2 a+\sqrt{2 a}=\sqrt{2}+1 ………….by (ii)

or, \sqrt{2} a(\sqrt{2}+a)=\sqrt{2}+1

or, \sqrt{2} \mathrm{a}=1

\therefore a=\frac{1}{\sqrt{2}}

From (ii),

b=\sqrt{2 a} \\

= \sqrt{2} \times \frac{1}{\sqrt{2}} \quad[\because a=1 \sqrt{2}] \\

= 1

\therefore \quad Length of the hypotenuse = 1cm.

\text { Area of the triangle } =\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB} \\

=\frac{1}{2} \times \mathrm{a} \times \mathrm{a} \\

=\frac{1}{2} \times \mathrm{a}^{2} \\

=\frac{1}{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2} \\

=\frac{1}{2} \times \frac{1}{2} \\

=\frac{1}{4}=0.25 \text { Sq.cm }(Ans.)

Total distance = speed × time

=18 \times \frac{10}{60} \mathrm{Km} . \\

= 3km.

Perimeter of equilateral trianglar park = 3 km .

\text { length of each side }=\frac{3}{3} \mathrm{Km} .=1 \mathrm{~km} \text {. }

Distance between mid point of a side and its opposite vertex = AD

=\frac{\sqrt{3}}{2} \times \text { side }=\frac{\sqrt{3}}{2} \times 1 \mathrm{~km}=\frac{\sqrt{3}}{2} \mathrm{~km}

Time taken to go from opposite vertex to mid-point of a side

=\frac{\text { Distance }}{\text { Speed }}

=\frac{\sqrt{3}}{2 \times 18} hours

=\frac{\sqrt{3}}{2 \times 18} \times 60 minutes

=\frac{5 \sqrt{3}}{3} minutes(Ans.)

Let ‘ x ‘ be the side of an equilateral triangle.

\text { Area of the equilateral triangle } =\frac{\sqrt{3}}{4} \times(x)^{2} \\

=\frac{\sqrt{3}}{4} x^{2}

If length of each side of an equilateral triangle is increased by 1 meter,

Then, side of an equilateral triangle (x + 1)

\therefore \text { Area of the equilateral triangle }=\frac{\sqrt{3}}{4} \times(x+1)^{2}

According to the condition of problem,

\frac{\sqrt{3}}{4} \times(x+1)^{2}-\frac{\sqrt{3}}{4} x^{2}=\sqrt{3}

or, \frac{\sqrt{3}}{4}\{(x+1)^{2}-x^{2}\}=\sqrt{3}

or, \frac{1}{4}(x+1+x)(x+1-x)=1

or, \frac{1}{4}(2 x+1)=1

or, 2x + 1 = 4

or, 2x = 4 – 1

or, 2x = 3

or, \mathrm{x}=\frac{3}{2} \quad \therefore \mathrm{x}=1.5

\therefore \quad Length of side equilateral triangle = 1.5m.

Let ‘a’ be the side of a square

Then, diagonal of square = 60cm.

\therefore \quad \mathrm{a} \sqrt{2}=60 \mathrm{~cm}

or, a=\frac{60}{\sqrt{2}} \mathrm{~cm}

By the problem,

Area of equilateral triangle : area of square =\frac{\sqrt{3}}{2}

\text { or, } \frac{\text { Area of equilateral triangle }}{\text { Area of square }}=\frac{\sqrt{3}}{2} \\

\text { or, } \frac{\text { Area of equilateral triangle }}{(\frac{60}{\sqrt{2}})^{2}}=\frac{\sqrt{3}}{2} \\

\text { or, } \frac{\text { Area of equilateral triangle }}{\frac{3600}{2}}=\frac{\sqrt{3}}{2}

or, Area of equilateral triangle =900 \sqrt{3}

or, \frac{\sqrt{3}}{4}(\text { Side })^{2}=900 \sqrt{3}

or, (\text { side })^{2}=3600 \\

or, \text { Side }=\sqrt{3600} \\

=60 \mathrm{~m} \text {. } \\

Perimeter of an equilateral triangle

=3 \times \text { side }=3 \times 60 \mathrm{~m}=180 \mathrm{~m}(Ans.)

Length of hypotenuse = 13cm.

Perimeter of right angled triangle = 30cm.

\therefore Sum of remaining sides other than

hypotenuse = (30 – 13)cm = 17cm.

Let ‘ p ‘ be the perpendicular and ‘ b ‘ be the base of a right angled triangle.

\therefore \mathrm{p}+\mathrm{b}=17 ……….. (i)

Also,

p^{2}+b^{2}=h^{2}

or, (p+b)^{2}-2 p b=(13)^{2}

or, (17)^{2}-2 \mathrm{pb}=(13)^{2}

or, 289-2 \mathrm{pb}=169

or, -2 \mathrm{pb}=169-289

\text { or, }+2 \mathrm{pb}=f 120 \\

\text { or, } \mathrm{pb}=\frac{120}{2} \\

\therefore \mathrm{pb}=60

Again,

p^{2}+b^{2}=h^{2}

or, (p-b)^{2}+2 p b=h^{2}

or, (p-b)^{2}+2 \times 60=(13)^{2}

or, (\mathrm{p}-\mathrm{b})^{2}+120=169

or, (p-b)^{2}=169-120

or, (p-b)^{2}=49

or, (p-b)=\sqrt{49}

\therefore p – b = 7 ……….. (ii)

\therefore \quad p + b = 17 ......... (i) \\ \underline{p-b =7 ........... (ii)} \\ 2 \mathrm{p}=24(by \ adding)

or, p=\frac{24}{2}

\therefore \mathrm{p}=12

Putting the value of p in equation (i)

p + b = 17

or, 12 + b = 17

\text { or, } b=17-12 \quad \therefore b=5

\text { Area of right angled triangle }

=\frac{1}{2} \times \mathrm{b} \times \mathrm{p} \\

=\frac{1}{2} \times 5 \times 12 \\

=30 \text { Sq. cm. (Ans.) }

Let \mathrm{AB}=12 \mathrm{~cm},

\mathrm{BC}=5 \mathrm{~cm}

We know that,

A C^{2}=A B^{2}+B C^{2}

or, \mathrm{AC}^{2}=(12)^{2}+(5)^{2}

or, \mathrm{AC}^{2}=144+25

or, \mathrm{AC}^{2}=169

or, \mathrm{AC}^{2}=\sqrt{169}

\therefore \quad \mathrm{AC}=13

\therefore Area of \triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}

=\frac{1}{2} \times 5 \times 12

= 30 Sq. \mathrm{cm}

\therefore \quad \text{Area of} \triangle \mathrm{ABC}=30 Sq.cm.

or, \frac{1}{2} \times \mathrm{BD} \times \mathrm{AC}=30

or, \frac{1}{2} \times \mathrm{BD} \times 13=30

or, 13 \mathrm{BD}=60

or, \mathrm{BD}=\frac{60}{13}

or, \mathrm{BD}=4.615 \mathrm{~cm} (Approx)

\therefore \quadLength of the perpendicular =4.615 \mathrm{~cm} (Approx)(Ans.)

Let \mathrm{AB}=4 \mathrm{~cm}, \mathrm{BC}=3 \mathrm{~cm}

\mathrm{AC}=5 \mathrm{~cm}

Let \mathrm{BD}=\mathrm{DE}=\mathrm{EF}=\mathrm{BF}=\mathrm{a}

\text { Area of } \triangle A B C= \text { Area of } \triangle A E F+ \text { Area of Square } BDEF + \text { Area of } \triangle D E C

or \frac{1}{2} \times 3 \times 4=\frac{1}{2} \times(4-a) \times a+a^{2}+\frac{1}{2} \times(3-a) \times a

or, 12=(4-a) a+2 a^{2}+(3-a) a

or, 12=4 a-a^{2}+2 a^{2}+3 a-a^{2}

or, 12=7 a

or, a=\frac{12}{7}

\therefore \quad \text { Length of square } =\frac{12}{7} \mathrm{~cm} \\

=1 \frac{5}{7} \mathrm{~cm}(Ans.)

Height of an equilateral triangle

=\frac{\sqrt{3}}{2} \times(\text { side }) \\

=\frac{\sqrt{3}}{\cancel2} \times {\cancel 4} \\

=2 \sqrt{3} \mathrm{~cm}.

\therefore \quad (d) is correct option (Ans.)

Let AB = a, BC = a

or, A C^2=a^2+a^2

or, \mathrm{AC}=\sqrt{2 \mathrm{a}^2}

or, \mathrm{AC}=\sqrt{2} \mathrm{a}

\therefore \quad \text{Perimeter of triangle} =a+a+\sqrt{2} a

=2 a+\sqrt{2} a

=(2+\sqrt{2}) a unit.

\therefore \quad (b) is correct option (Ans.)

We have,

\frac{2 a}{\operatorname{sh}}=\frac{2 \times \frac{\sqrt{3}}{4} \times(\text { side })^{2}}{3 \times \operatorname{side} \times \frac{\sqrt{3}}{2} \times \text { side }}=\frac{\frac{2}{4}}{\frac{3}{2}} \\

= \frac{2}{4} \times \frac{2}{3}=\frac{1}{3}

\therefore (c) is correct option (Ans.)

\text { Area of triangle } =\frac{1}{2} \times 6 \times \sqrt{(5)^2-(\frac{6}{3})^2} \\

=\frac{1}{2} \times 6 \times \sqrt{25-9} \\

=3 \times \sqrt{16} \\

=3 \times 4=12 \text { Sq.cm. }

\therefore (b) is correct option (Ans.)

Area of triangle ABC = 40 Sq.cm

or, \mathrm{AD}: \mathrm{DC}=3: 2

or, \triangle \mathrm{ABD}: \triangle \mathrm{BDC}

or, \frac{\triangle A B D}{\triangle B D C}=\frac{3}{2}

or, 2 \triangle \mathrm{ABD}=3 \triangle \mathrm{BDC}

or, \mathrm{ABD}=\frac{3}{2} \triangle \mathrm{BDC}

Also \triangle \mathrm{ABC}=\triangle \mathrm{ABD}+\triangle \mathrm{BDC}

or, \triangle \mathrm{ABC}={ }_2^3 \triangle \mathrm{BDC}+\triangle \mathrm{BDC}

or, \triangle \mathrm{ABC}=\frac{5}{2} \triangle \mathrm{BDC}

or, \triangle B D C={ }_5^2 \triangle A B C

= \frac{2}{\cancel{5}} \times \cancel{40} \mathrm{Sq} \cdot \mathrm{cm}

=16 \mathrm{Sqcm}.

\therefore (a) is correct option (Ans.)

Let, s – a = 8

s – b = 7

s – c = 5

adding these,

3s – (a + b + c) = 8 + 7 + 5

\text { or, } 3 s-2 s=20 \quad\left[\therefore s=\frac{a+b+c}{2}\right]

\therefore \mathrm{s}=20

Area of Triangle =\sqrt{s(s-a)(s-b)(s-c)} \\

=\sqrt{20 \times 8 \times 7 \times 5} \\

=\sqrt{5 \times 4 \times 4 \times 2 \times 7 \times 5} \\

=5 \times 4 \sqrt{2 \times 7} \\

=20 \sqrt{14} \mathrm{Sq} \cdot \mathrm{cm}

\therefore \quad (c) is correct option (Ans.)

By question,

\quad \frac{\sqrt{3}}{4}(\text { side })^2=\frac{\sqrt{3}}{2} \times(\text { side }) \\

\text { or, } \frac{\text { side }}{2}=1 \\

\text { or, side }=2 \\

\text { Length of side of triangle }=2 \text { units }

\therefore Length of side of triangle = 2 units (Ans.)

Let ‘ a ‘ be the side of an equilateral triangle,

\therefore \quad \text { Area of the triangle }=\frac{\sqrt{3}}{4} \mathrm{a}^2

If length of each side of an equilateral triangle is doubled

\therefore \quad \text { Area of the triangle } =\frac{\sqrt{3}}{4}(2 a)^2 \\

=\frac{\sqrt{3}}{4} \times 4 a^2

Increase area of the triangle

=\frac{\sqrt{3}}{4} \times 4 a^2-\frac{\sqrt{3}}{4} a^2 \\

=\frac{\sqrt{3}}{4} \times 3 \mathrm{a}^2 \\

\text { Perimeter of increased in area }=\frac{\frac{\sqrt{3}}{4} \times 3 a^2}{\frac{\sqrt{3}}{4} \times a^2} \times 100 \% \\

=300 \% \text { (Ans.) } \\

Let ‘ a ‘ be the side of an equilateral triangle,

\therefore \text { Area of the triangle }=\frac{\sqrt{3}}{4} \mathrm{a}^2

If length of each side of an equilateral triangle is trippled

\therefore \quad \text { Area of the triangle } =\frac{\sqrt{3}}{4}(3 a)^2 \\

=\frac{\sqrt{3}}{4} \times 9 a^2

Increased area of the triangle

=\frac{\sqrt{3}}{4} \times 9 a^2-\frac{\sqrt{3}}{4} a^2 \\

=\frac{\sqrt{3}}{4} \times 8 a^2 \\

=\sqrt{3} \times 2 a^2

\text { Perimeter of increased in area } =\frac{\sqrt{3} \times 2 \mathrm{a}^2}{\sqrt{3}} \times 100 \% \\

=8 \times 100 \% \\

=800 \% \text { (Ans.) }

We have, (\text { hypotenuse} )^2=(\text { base })^2+(\text { Perpendicular })^2

(x+2)^2=(x-2)^2+(x)^2

or, (x+2)^2-(x-2)^2=(x)^2

or, 4 \cdot x \cdot 2=x^2

\text { or, } 8 x=x^2 \\

\text { or, } 8=x \quad [\therefore x \neq 0] \\

\therefore \quad x=8

\therefore \quad \text { Length of hypotenuse } =(x+2) \mathrm{cm} \\

=(8+2) \mathrm{cm} \\

=10 \mathrm{~cm} \text {. (Ans.) }

Area of triangle : Area of square

=\frac{\sqrt{3}}{4} \times(\text { side })^2:\left(\frac{\sqrt{3}}{2} \times(\text { side })\right)^2 \\

=\frac{\sqrt{3}}{4} \times(\text { side })^2: \frac{\sqrt{3} \times \sqrt{3}}{4} \times(\text { side })^2 \\

= 1: \sqrt{3} \text { (Ans.) }

\text { Length of base }=5 \mathrm{~cm} \\

\text { Height }=4 \mathrm{~cm}

\text { Area of the parallelogram } =\text { base } \times \text { height } \\

=5 \mathrm{~cm} \times 4 \mathrm{~cm} \\

=20 \text { sq.cm. (Ans.) }

Let ‘ x ‘ be the height of a parallelogram

\therefore Base =2 \times height =2 \mathrm{x}

\therefore \text{Area of parallelogram} =98 sq. \mathrm{cm}.

or, Base × height = 98

or, 2x × x = 98

or, 2x^2 = 98

or, x^2-\frac{98}{2}

or, x^2=49

or, \mathrm{x}=\sqrt{49}

\therefore \quad \mathrm{x}=7

Length of base = 2x

=2 \times 7=14 \mathrm{~cm} \text {. }

Height of the parallelogram = x =7cm (Ans.)

Let, a =13 \mathrm{~cm} \\

b =14 \mathrm{~cm} \\

c =15 \mathrm{~cm}

s=\frac{a+b+c}{2}

=\frac{13 m+14 m+15 m}{2}=\frac{42}{2} m=21 m

\text { Area of } \triangle \mathrm{ABD} =\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\

=\sqrt{21(21-13)(21-14)(21-15)}

=\sqrt{21 \times 8 \times 7 \times 6} \\

=\sqrt{7 \times 3 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3} \\

=2 \times 2 \times 3 \times 7 \\

=84 \mathrm{Sq} . \mathrm{m}

\text { Area of parallelogram } \mathrm{ABCD} =2 \times \text { Area of } \triangle \mathrm{ABD} \\

=2 \times 84 \mathrm{Sq} \cdot \mathrm{m}=168 \mathrm{Sq} \cdot \mathrm{m}

Let, \mathrm{a}=25 \mathrm{~cm}

b=20 \mathrm{~cm} \\

c=15 \mathrm{~cm}

s=\frac{a+b+c}{2} \\

=\frac{25+20+15}{2}=\frac{60}{2}=30 \mathrm{~cm}

=\frac{25+20+15}{2}=\frac{60}{2}=30 \mathrm{~cm}

\text { Area of } \triangle A B D =\sqrt{s(s-a)(s-b)(s-c)} \\

=\sqrt{30(30-25)(30-20)(30-15)} \\

=\sqrt{30 \times 5 \times 10 \times 15} \\

=\sqrt{15 \times 2 \times 2 \times 5 \times 15} \\

=2 \times 5 \times 15 \\

=150 \text { Sq.cm }

Area of parallelogram = 2 \times Area of \triangle \mathrm{ABD}

=2 \times 150 \text { Sq.cm. } \\

=300 \text { Sq. } \\

\therefore \quad Base \times height =300 Sq. \mathrm{cm}.

or, 25 \times height =300

or, height ^2=\frac{300}{25}

\therefore height =12 \mathrm{~cm}(Ans.)

Let, A B=15 \mathrm{~cm}

\mathrm{BC}=12 \mathrm{~cm} \\

\mathrm{AE}=7.5 \mathrm{~cm} .

Area of the \triangle A B C

=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AE} \\

=\frac{1}{2} \times 12 \times 7.5 \mathrm{Sq} . \mathrm{cm} . \\

=45 \text { Sq.cm. }

Let the distance between the longer two sides be ‘ x ‘

\therefore \text { Area of } \triangle \mathrm{ACD}=\text { Area of } \triangle \mathrm{ABC}

or, \frac{1}{2} \times B C \times distance between the longer two sides = 45

or, \frac{1}{2} \times 15 \times x=45

or, \frac{15 x}{2}=45

or, \frac{x}{2}=3

\therefore \quad x = 6

\therefore \quad The distance between the longer two sides = 6cm (Ans.)

\text { Let, } \mathrm{AC} =15 \mathrm{~m} \\

\mathrm{BD} =20 \mathrm{~m}

\mathrm{AO}=\mathrm{OC}=\frac{15}{2} \mathrm{~m} \\

\mathrm{BO}=\mathrm{OD}=\frac{20}{2} \mathrm{~m}=10 \mathrm{~m}

Area of Rhombus ABCD

=\frac{1}{2} \mathrm{AC} \times \mathrm{BD} \\

=\frac{1}{2} 15 \times 20 \mathrm{~m} \\

=150 \text { Sq.m (Ans.) }

In \triangle \mathrm{OAB}, \quad \angle \mathrm{AOB}=90^{\circ}

A B^2=A O^2+O B^2

or, A B^2=\left(\frac{15}{2}\right)^2+(10)^2

or, A B^2=\frac{225}{4}+100^2

or, \mathrm{AB}^2=\frac{225+400}{4}

or, A B^2=\frac{625}{4}

\therefore \quad \mathrm{AB}=\sqrt{\frac{625}{4}}=\frac{25}{2} \mathrm{~m}

\therefore \quad \text{Perimeter of} \mathrm{ABCD}=4 \times \mathrm{AB}

=4 \times \frac{25}{2} \\

=50 \mathrm{~m} \text { (Ans.) }

\therefore \quad Area of Rhombus ABCD =150 Sq.m

\therefore \quad Base \times height =150

or, \frac{25}{2} \times height ^2=150

or, \frac{\text { height }}{2}=6

\therefore height =12 \mathrm{~m} (Ans.)

\text { Side of Rhombus } =\frac{\text { Perimeter }}{4} \\

=\frac{440}{4} \\

=110 \mathrm{~m}

\therefore Base \mathrm{CD} \text{of Rhombus} \mathrm{ABCD}=110 \mathrm{~m}

\therefore Distance between two parallel sides = Altitude AE

\therefore \quad \text { Area of the Rhombus } =\text { Base } \times \text { Altitude } \quad=22 \mathrm{~m} \\

=\mathrm{CD} \times \mathrm{AE} \\

=110 \times 22 \text { Sq.m } \\

=2420 \text { Sq.m (Ans.) }

\text { Perimeter of Rhombus }=20 \mathrm{~cm} \\

\text { Side of a rhombus } =\frac{20}{4} \mathrm{~cm} \\

=5 \mathrm{~cm} .

Length of its one diagonal = 6cm.

Let B D=6 \mathrm{~cm}.

\mathrm{OB}=\frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm} . \\

AB = 5cm

In \triangle \mathrm{OAB}

\mathrm{OA}^2 =\mathrm{AB}^2-\mathrm{OB}^2 \\

\mathrm{OA}^2 =(5)^2-(3)^2 \\

\text { or, } \mathrm{OA}^2 =25-9

or, \mathrm{OA}^2=16

or, \mathrm{OA}=\sqrt{16}

\therefore \mathrm{OA} =4 \mathrm{~cm} \\

\mathrm{AC} =2 \times \mathrm{OA} \\

=2 \times 4 \mathrm{~cm}=8 \mathrm{~cm}

\text { Area of rhombus } \mathrm{ABCD} =\frac{1}{2} \times \mathrm{BD} \times \mathrm{AC} \\

=\frac{1}{2} \times 6 \times 8 \text { Sq.cm. } \\

=24 \text { Sq.cm. (Ans.) }

\text { Let } \mathrm{AE}=20 \mathrm{dm} \\

\mathrm{AD}=3 \mathrm{x} \\

\mathrm{BC}=4 \mathrm{x} \\

\text { trapazium }=1400 \text { Sq.cm }

Area of trapazium = 1400 Sq.cm

\text { or } \quad \frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AE}=1400 \\

\text { or } \quad \frac{1}{2}(3 \mathrm{x}+4 \mathrm{x}) \times 20=1400

7x = 1400

x=\frac{140}{7} \\

\therefore \quad x = 20

\therefore \quad A D=3 x=3 \times 20 \mathrm{dcm} .=60 \mathrm{dcm} \\

\quad \mathrm{BC}=4 \mathrm{x}=4 \mathrm{x} \times 20 \mathrm{dcm} .=80 \mathrm{dcm} \text {. (Ans.) }

If we draw diagonals we get equal six equilater triangles.

Area of \triangle A O B

=\frac{\sqrt{3}}{4}(\text { Side })^2 \\

=\frac{\sqrt{3}}{4}(8)^2 \\

=\frac{\sqrt{3}}{4} 64 \\

=16 \sqrt{3} \mathrm{Sq} \cdot \mathrm{cm} .

\text { Area of Hexagon } \mathrm{ABCDEF} =6 \times 16 \sqrt{3} \\

=96 \sqrt{3} \text { Sq. } \mathrm{cm} \text {. }(Ans.)

\text { Let } A B=5 \mathrm{~m}, B C=12 \mathrm{~m} \text {, } \\

C D=14 \mathrm{~m}, \mathrm{DA}=15 \mathrm{~m} \text {, } \\

\therefore \angle A B C=90^{\circ} \\

\text { or, } A C^2=\sqrt{\mathrm{AB}^2+B C^2} \\

=\sqrt{(5)^2+(12)^2} \text {. } \\

=\sqrt{25+144} \\

=\sqrt{109} \\

=13 \mathrm{~m} \\

\therefore \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times 5 \\

=30 \mathrm{Sq} . \mathrm{cm}

\text { Let } a=13 \mathrm{~m}, b=14 \mathrm{~m}, \mathrm{c}=15 \mathrm{~m}

s=\frac{a+b+c}{2} \\

s=\frac{13+14+15}{2}=\frac{42}{2}=21 \mathrm{~m}

Area of \triangle A C D=\sqrt{s(s-a)(s-b)(s-c)}

=\sqrt{21(21-13)(21-14)(21-15)} \\

=\sqrt{21 \times 8 \times 7 \times 6} \\

=\sqrt{3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3} \\

=2 \times 2 \times 3 \times 7 \mathrm{Sq} \cdot \mathrm{m} \\

=84 \text { Sq.cm. }

\text { Area of Quadrilateral } A B C D=(3+84) \text { Sq.m. } \\

=114 \text { Sq.m (Ans.) } \\

\text { Let, } \mathrm{BD} =11 \mathrm{~cm}, \\

\mathrm{AE} =5 \mathrm{~cm} \\

\mathrm{CF} =11 \mathrm{~cm}

\text { Area of } \triangle \mathrm{ABD}=\frac{1}{2} \times \mathrm{BD} \times \mathrm{AE} \\

=\frac{1}{2} \times 11 \times 5 \\

=\frac{\jmath \rho}{2} \text { Sq.cm } \\

\text { Area of } \triangle B C D=\frac{1}{2} \times B D \times C F \\

= \frac{1}{2} \times 11 \times 11 \\

=\frac{121}{2} \text { Sq.cm } \\

Area of Trapezium ABCD

=\text { Area of } \triangle A B D+\text { Area of } \triangle B C D

=(\frac{55}{2}+\frac{121}{2}) \text { Sq.cm } \\

=\frac{176}{2} \\

=88 \mathrm{Sq} . \mathrm{cm} \text { (Ans.) }

\text { Let } P Q=\frac{4}{9}+PE= \frac{4}{9} \times 9cm \\

= 4cm

\therefore \quad \mathrm{QE}=(9-4) \mathrm{cm} .=5 \mathrm{~cm} \\

\text { Now, } \because A D \| B C \text { and } E P \perp B C \\

\therefore \mathrm{EQ} \perp \mathrm{AD} \\

\therefore \text { Area of pentagon } A B C D E \\

=\text { Area of } \triangle \mathrm{ADE}+\text { Area of trapezium } \mathrm{ABCD} \\

=\frac{1}{2} \times \mathrm{AD} \times \mathrm{QE}+\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC} ) \times \mathrm{PQ} \text { Sq.unit. } \\

=\frac{1}{2} \times 13 \times 5+\frac{1}{2} \times(13+7) \times 4 \text { Sq.unit. } \\

=(32.5+40) \cdot \mathrm{Sq} \cdot \mathrm{cm} \\

=72.5 \text { Sq.cm } \\

Let ‘ a ‘ be the side of a square and also a side of a rhombus. The side of a rhombus.

The length of diagonal = 40 \sqrt{2} \mathrm{~cm}. or, a \sqrt{2}=40 \sqrt{2}\\

\therefore \quad a=40, \therefore \mathrm{CD}=40 \mathrm{~cm} \text {. }\\

Let \mathrm{AC}=3 \mathrm{x}, \mathrm{BD}=4 \mathrm{x}\\

\mathrm{OC}=\frac{3 \mathrm{x}}{2}, \mathrm{OD}=\frac{4 \mathrm{x}}{2} 2 \mathrm{x}\\

In \triangle \mathrm{COD}\\

O C^2+O D^2=C D^2\\

(\frac{3 x}{2})^2+(2 x)^2=(40)^2\\

or, \frac{9 x^2}{4}+4 x^2=1600\\

or, \frac{9 x^2+16 x^2}{4}=1600\\

or, \frac{25 x^2}{4}{ }^2=1600\\

or, 25 \mathrm{x}^2=1600 \times 4\\

or, x^2=\frac{1660 \times 4}{25}{ }^2\\

or, x^2=64 \times 4\\

=8 \times 2=16\\

\therefore \quad x =16\\

\mathrm{AC}=3 x=3 \times 16 \mathrm{~cm}=48 \mathrm{~cm} \\

\mathrm{BD}=4 \mathrm{x}=4 \times 16 \mathrm{~cm}=64 \mathrm{~cm} .\\

\text { Area of Rhombus }\mathrm{ABCD} =\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD} \\

=\frac{1}{2} \times 48 \times 64 \\

Let \mathrm{AB}=\mathrm{CD}=10 \mathrm{~cm}

A D=5 \mathrm{~cm} \\

B C=17 \mathrm{~cm} \\

A D \| B C \text { and } A B=C D \\

A D=E F=5 \mathrm{~cm}

\therefore A D \| B C and A B=C D

\therefore A D =E F=5 \mathrm{~cm} . \\

\therefore \mathrm{BF} =\mathrm{CE}=\frac{17-5}{2}=\frac{12}{2} \mathrm{~cm}=6 \mathrm{~cm}

In \triangle \mathrm{ABF}

\mathrm{AF} =\sqrt{\mathrm{AB}^2-\mathrm{BF}^2} \\

=\sqrt{(10)^2-(6)^2} \\

=\sqrt{100-36} \\

=\sqrt{64} \\

=8 \\

\therefore \quad A F =8 \mathrm{~cm} . \quad \therefore \quad \mathrm{DE}=\mathrm{AF}=8 \mathrm{~cm}.

And BE = BF + EF = (6 + 5)cm =11cm (Ans.)

Area of Trapazium ABCD

=\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AF} \\

=\frac{1}{2} \times(5+17) \times 8 \\

=\frac{1}{2} \times 22 \times 8 \\

=88 \text { Sq. } \mathrm{cm}

In \triangle \mathrm{BED}

\mathrm{BD}=\sqrt{\mathrm{DE}^2+\mathrm{BE}^2}

or, \mathrm{BD}=\sqrt{(8)^2+(11)^2}

or, \mathrm{BD}=\sqrt{64+121}

or, \mathrm{BD}=\sqrt{64+121}

\therefore \mathrm{BD}=\sqrt{185} cm.

\therefore Length of diagonal =\sqrt{185 cm}. (Ans.)

Let \mathrm{AD}=9 cm

\mathrm{BC} =19 cm \\

\mathrm{AB} =8 cm \\

\mathrm{CD} =6 cm \\

\mathrm{BC} =\mathrm{BF}+\mathrm{FE}+\mathrm{CE} \\

=\mathrm{BF}+\mathrm{CE}+9 \\

\mathrm{BF} =19-9-\mathrm{CE} \\

=10-\mathrm{CE}

In \triangle \mathrm{ABF},

A B^2=A F^2+B F^2

or, (8)^2=A F^2+(10-C E)^2

In \triangle \mathrm{CDE}

\mathrm{CD}^2=\mathrm{DE}^2+\mathrm{CE}^2

or, (6)^2=\mathrm{AF}^2+\mathrm{CE}^2[\because \mathrm{DE}=\mathrm{AF}]

or, (6)^2=(8)^2-(10-C E)^2+C E^2

or, 36=64-\{(10)^2-2.10 . \mathrm{CE}+(\mathrm{CE})^2\}-(\mathrm{CE})^2

or, 36=64-(10)^2+20 \mathrm{CE}-\mathrm{CE}^2+\mathrm{CE}^2

or, 36=64-100+20 \mathrm{CE}

or, 20 \mathrm{CE}=36+36

or, 20 \mathrm{CE}=72

or, \mathrm{CE}=\frac{72}{20}

\text { or, } \mathrm{CE} =\frac{36}{10}=3.6 cm \\

\therefore \mathrm{DE} =\sqrt{\mathrm{CD}^2-\mathrm{CE}^2} \\

=\sqrt{(6)^2-(3.6)^2} \\

=\sqrt{(6+3.6)(6-3.6))} \\

=\sqrt{9.6 \times 2.4} \\

=\sqrt{4.8 \times 4.8} \\

= 4.8 cm

Area of Trapazium AECE

=\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC}) \times \mathrm{DE} \\

=\frac{1}{2} \times(9+19) \times 4.8 \\

=\frac{1}{2} \times 28 \times 4.8 \\

=67.2 \text { Sq.cm. (Ans.) }

Area of parallelogram = 192 Sq.cm.

or, Base × height = 192

or, Base \times \frac{1}{3} base =192

or, (\text { Base })^2=476

or, Base =\sqrt{476}=24 cm.

\therefore \quad \text { Height } =\frac{1}{3} \times \text { base } \\

=\frac{1}{3} \times 24=8 cm .

\therefore (b) is correct option

\because \quad A B=B C=6 cm \\

\therefore \angle A B C = 60

\therefore \quad C is an equilateral triangle

\therefore \text { Area of } \triangle A B C=\frac{\sqrt{3}}{4} \times(6)^2 \\

=\frac{\sqrt{3}}{4} \times 36 \\

=9 \sqrt{3} \text { Sq.cm. } \\

\text { Area of rhombus } \mathrm{ABCD} =2 \times \triangle \mathrm{ABC} \\

=2 \times 9 \sqrt{3} \mathrm{Sq} \cdot \mathrm{cm} \\

=18 \sqrt{3} \mathrm{Sq} . \mathrm{sm}

\therefore \quad(b) is correct option

Let 1st diagonal =x

2nd diagonal = 3x

Area of Rhombus = 96 Sq.cm.

\frac{1}{2} \times x \times 3 x=96

or, \frac{3 x^2}{2}=96

or, x^2=64

or, x=\sqrt{64}=8

Length of long diagonal = 3 \mathrm{x} =3 \times 8 cm=24 cm.

\therefore \quad (d) is correct option

\therefore (b) is correct option

We have,

DE = 6 cm,

AB = ?, CD = 23 cm,

Area of trapazium =162 \mathrm{Sq} . \mathrm{cm}.

or, \frac{1}{2} \times(\mathrm{AB}+\mathrm{CD}) \times \mathrm{DE}=162

or, \frac{1}{2} \times(\mathrm{AB}+23) \times 6=162

or, \mathrm{AB}+23=54

or, \mathrm{AB}=54-23

\therefore \quad AB = 31cm.[/katex]

\therefore (b) is correct option

Area of parallelogram ABCD = 96 sq. cm.

\mathrm{cm} \therefore Area of \bigtriangleup ABD = \frac{96}{2} Sq.cm.

=48 \text { Sq. } \mathrm{cm}

or, \frac{1}{2} \times \mathrm{AE} \times \mathrm{BD}=48

or, \frac{1}{2} \times \mathrm{AE} \times 12=48

\therefore \quad \mathrm{AE}=8 cm \quad (Ans.)

\text { Let } AB \& = CD=3 cm . \\

AD = BC = 5 cm

AF = 2 cm

Area of \triangle \mathrm{ABD} =\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE} \\

=\frac{1}{2} \times 3 \times \mathrm{DE} \\

=\frac{3}{2} \mathrm{DE}

Area of parallelogram ABCD

=2 \times \frac{3}{2} \times \mathrm{DE}=3 \times \mathrm{DE}

\therefore \text{Area of the parallelogram} \mathrm{ABCD} =\mathrm{BC} \times \mathrm{AF} \\

=5 cm \times 2 cm \\

=10 \mathrm{Sq} . \mathrm{cm} . \\

\therefore \quad 3 \times \mathrm{DE} =10 \\

\text { or, } \quad \mathrm{DE} =\frac{10}{3} \\

\therefore \quad \mathrm{DE} =3 \frac{1}{3} cm \text { (Ans.) }

\text { Area of rhombus } =\text { Base } \times \text { Height } \\

=5 cm \times 14 cm . \\

=70 \mathrm{Sq} . \mathrm{cm} .

\text { (Ans.) }

Let \mathrm{AB}=62, cm

\angle \mathrm{ABE}=\angle \mathrm{EAB}=45^{\circ}

AE = Height of trapezium

We have,

\mathrm{AE}^{2}+\mathrm{BE}^{2}=\mathrm{AB}^{2}

or, \mathrm{AE}^{2}+\mathrm{AE}^{2}=(62)^{2} \quad[\because \mathrm{BE}=\mathrm{AE}]

or, 2 \mathrm{AE}^{2}=(62)^{2}

or, \mathrm{AE}^{22}=\frac{(62)^{2}}{2}

\text { or, } \mathrm{AE} =\frac{(62)^{2}}{4} \times 2 \\

\text { or, } \mathrm{AE} =\sqrt{\frac{(62)^{2}}{4} \times 2} \\

\text { or, } \mathrm{AE} =\frac{62}{2} \times \sqrt{2} \\

=31 \sqrt{2} cm

\therefore Distance between two parallel sides = 31 \sqrt{2} cm (Ans.)

\mathrm{AB}=4 cm, \mathrm{BC}=6 cm

We draw perpendicular

AE on BC

Then, \angle B A E=60^{\circ} ,

\angle \mathrm{AEB}=90^{\circ}, \angle \mathrm{ABE}=30^{\circ}

In a right angled triangle, If the angles are 90^{\circ}, 60^{\circ} \ 30^{\circ}, then,

\mathrm{AE} =\frac{1}{2} \mathrm{AB} \\

=\frac{1}{2} \times 4 cm=2 cm

\therefore \text { Area of parallelogram } =\text { Base } \times \text { altitude } \\

=\mathrm{BC} \times \mathrm{AE} \\

=6 \times 2 \mathrm{Sq} \cdot \mathrm{cm}. \\

=12 \mathrm{Sq} . \mathrm{cm} . \text { (Ans.) }