Ann Theressa   11 years AGO

Let the sides of squares be x and (x+4)

x^2 + (x+4)^2 = 400
2x^2 + 8x + 16 = 0
x^2 + 4x - 192 = 0
x = 12,-16
take x = 12(positive value)

So the dimension of 2 squares are 12cm and 16cm

Like? Yes (11) | No (4)  

Sumit yadav   11 years AGO

Let the side of the smaller square be x cm.
Side of the larger square = Side of the smaller square + 4 cm
∴ Side of the larger square = (x + 4) cm
Given, Area of the smaller square + Area of the larger square = 400 cm2

Side of the smaller square = 12 cm
Side of the larger square = (x + 4) cm = (12 + 4) cm = 16 cm

Like? Yes (10) | No (6)  

SAIF UDDIN   11 years AGO

12*12=144
16*16=256
Now, 144+256=400

Like? Yes (5) | No (1)  

M VINOD KUMAR   11 years AGO

Let the side of one square is x cm then another one is x+4 cm
sum of areas (x+4)^2 + x^2 = 400
x^2 + 8x + 16 + x^2 =400
x^2 + 4x -192 = 0
solving x= 12 and -16
we took x=12cm , x+4 = 16cm

Like? Yes (6) | No (3)  

NAGENDRA P   11 years AGO

A^2+(A+4)^2=200...
SOLVING A=12CM..AND 16CM

Like? Yes (5) | No (2)  

manish chauhan   11 years AGO

12*12=144
16*16=256
16-12=4
144+256=400

Like? Yes (5) | No (3)  

prateek   11 years AGO

here 400=a^2+b^2
so 20^2=a^2+b^2
so we have to find pythagorean triplets matching the eqn.we know that 3,4,5 is a pythagorean triplet.so we get (5*4)^2=(4*4)^2+(3*4)^2.
thus we get sides as 12 and 16

Like? Yes (4) | No (2)  

shefali jain   11 years AGO

let the side of a square is=a
then the side of other square is=a+4
then
(a+4)^2+a^2=400
by solving this equation the value of
a=12cm (which is the side of first square)
a+4=16cm(which is the side of other square)

Like? Yes (3) | No (1)  

Meghana   11 years AGO

area of square=a^2
12^2+16^2=144+256=400&16-12=4

Like? Yes (3) | No (1)  

ANUP KUMAR   11 years AGO

let x = dimension of 1st square
Given, x+4 = dimension of other square
and x^2 + (x+4)^2 = 400
by solving,
x = 12
and x+4 = 16
option (7)

Like? Yes (6) | No (5)  

Dr.Dipin Singh   11 years AGO

x^2+(x+4)^2=400
solving, we get, x=12 cms
so sides of squares are 12 and 16 cms.

Like? Yes (8) | No (7)  

Karan   11 years AGO

if sides of squares are x and x+4 cms, then
x^2+(x+4)^2=400
x^2+4x-192=0
(x+16)*(x-12)=0
x=12 cms
so sides are 12 and 16 cms.

Like? Yes (8) | No (7)  

Varunraj Reddy   11 years AGO

Let the side of a square=x cm
Then the side of another square=x+4 cm
Sum of the areas of two squares=400 sq cm
x^2+(x+4)^2=400
x^2+x^2+16+8x-400=0
2x^2+8x-384=0
x^2+4x-192=0
x^2+16x-12x-192=0
x(x+16)-12(x+16)=0
x=12,-16
Option (7)

Like? Yes (5) | No (4)  

vishnuvardhan   11 years AGO

x^2+(x+4)^2=400
2x^2+8x-384=0
x=12,-16(as side is +ve)
x=12,x+4=16
12,16

Like? Yes (4) | No (3)  

sruthi   11 years AGO

12, 16 is answer

Like? Yes (6) | No (6)  

MD SOHAIL AKHTER   11 years AGO

let the side of the square be x cm
side of other square is (x+4) cm
A/Q
x^2+(x+4)^2=400
on solving we get
side of two squares are=12cm and 16cm

so correct option is (7)

Like? Yes (4) | No (5)  

hareesh   11 years AGO

x^2+(x+4)^2=400
x^2+4x-192=0
x^2+16x-12x-192=0
(x+16)(x-12)=0
x=-16,x=12

Like? Yes (4) | No (5)  

Devendra Marghade   11 years AGO

If side of squares are x & x+4
x^2+(x+4)^2=400
x^2+4x-192=0
(x+16)(x-12)=0, so x=12
So sides are 12 & 16

Like? Yes (2) | No (4)  

Jigyasa Kumar   11 years AGO

Answer : 12, 16
on solving (a-b)^2, we get ab = 192
and a,b = 12 and 16

Like? Yes (2) | No (4)  

abhilasha   11 years AGO

on solving (a-b)^2, we get ab = 192
and a,b = 12 and 16

Like? Yes (2) | No (4)  

SIBNATH HALDER   11 years AGO

X^2+(X+4)^2=400
X^2+4X-192=0
X=16
X=12

Like? Yes (2) | No (4)  

rajani   11 years AGO

Let side of square be xcm
Side of another square =(x+4)cm
And the sum of area of two squares=400 sqcm
--> x^2+(x+4)^2=400
2x^2+8x+16=400
x^2+4x-192=0
x^2+16x-12x-192=0
(x+16)(x-12)=0
x=12 or -16
So,x+4=12+4=16cm
the sides of both squares are 12cm and 16cm

Like? Yes (2) | No (4)  

P.Someshwar Reddy   11 years AGO

Answer is option(7)

Let the side of the square be s1
and the side of the other square be s2

Given the side of square exceeds the side of the other square by 4.c.m
=> s2=s1+4

Given sum of areas is 400.sq.c.m.
=>(s1^2)+(s2^2)=400
=>(s1^2)+((s1+4)^2))=400
=>(s1^2)+(s1^2+8s1+16)=400
=>2s1^2+8s1+16=400
=>2s1^2+8s1-384=0
dividing the equation by 2
=>s1^2+4s1-192=0
=>s1^2+16s1-12s1-192=0
=>s1(s1+16)-12(s1+16)=0
=>(s1+16)(s1-12)=0
=> s1= -16 or 12
since length can not be negative
the side of square is 12.cm.
=>s1=12.c.m.
=>s2=12+4=16.c.m.

therefore the two lengths are 12cm and 16cm

Like? Yes (3) | No (5)  

Deb Das   11 years AGO

x^2+(x+4)^2=400
x^2+x^2+8x+16=400
x^2+4x-192=0

(x+16)(x-12)=0

So, x=12
and the dimensions are 12 and (12+4)=16 cm

Like? Yes (2) | No (4)  

sanghavi    11 years AGO

let l1 be the side of one square
and side of another square is l2=l1+4
sum of area (l1)^2+(l1+4)^2=400
2(l1^2+4l1+8)=400
l1^2+4l+4+4=200
(l1+2)^2=196
l1+2=sqrt(196)
l1+2=14
l1=12
then l2=12+4=16

Like? Yes (2) | No (4)  

Arvind Yadav   11 years AGO

here it is given that one square exceeds the side to another by 4 cm .
so suppose one sqare side is x cm then another side is x+4.
and
we know that area of sqare =side^2
so sum of area of both sqares =x^2+(x+4)^2
and it is given that sum of area of squares =400
so x^2+(x+4)^2=400
so from option best ans is 12 and 16 which satisfy the above condition.

Like? Yes (2) | No (4)  

bithika das   11 years AGO

side of square1=a,side of another square =a+4
ares of two square=(a)^2+(a+4)^2=400
solving ,
we get a=12 and 16

Like? Yes (2) | No (4)  

subramani   11 years AGO

Area of Square= A^2
So...Two Squares area A1^2+ A2^2=400
A^2=400 then
a=20
To find square A1^2, Less Exceed side size 4 cm.. so
A1= 20-4=16cm

so A1^2=16^2=256
put the A1^2 in Total Area=A1^2+A2^2=400
16^2+A2^2=400
A2^2=400-256=144
A2^2=144 so
A2=Sqrt(144)
A2=12cm

so Sides of Two Squares are 12 & 16 Cm
Option 7 is correct

Like? Yes (2) | No (4)  

ABHOY KUMAR   11 years AGO

let the sides of two squares be s1,s2
s1^2+s2^2=400 ...(i)
s1=s2+4 ....(ii)
solving (i) & (ii)
s1=16 cm
s2=12cm

Like? Yes (2) | No (4)  

AMIT SINGH   11 years AGO

sides of square are x and x+4 cm
then x^2 +(x+4)^2 =400
=>x=12
so, dimension of square are 12cm and 16cm.

Like? Yes (2) | No (4)  

Pranjali Das   11 years AGO

let one side be x and the other be y
so according to the qs
x-y=4 and x^2 + y^2= 400
so solving the equations y=12 and x=16
so the answer is option 7

Like? Yes (2) | No (4)  

Dhruva Rathore   11 years AGO

X is 12 after solving so when the second side exceeds by 4 . Which implies other side is x+4 = 16

Like? Yes (2) | No (4)  

SREEVIDHYA P P   11 years AGO

12cm &16cm
Area of a square of side 12cm is 144sq.cm.
Area of a square of side 16cm is 256sq cm.
Sum of the area of squares is 144+256=400sq cm.
Ans. Is option 7...12cm &16cm

Like? Yes (2) | No (4)  

Ravi Raj   11 years AGO

12^2+16^2 = 400

Like? Yes (2) | No (4)  

SHUBHAM KUMAR   11 years AGO

x*x+(x+4)*(x+4)=400
on solving x=12

Like? Yes (2) | No (4)  

Leelanand Kumar Yalla   11 years AGO

x^2+(x+4)^2=400; by solving this we will get x=12 and x+4= 16cms......

Like? Yes (2) | No (4)  

Naveen Ganesh   11 years AGO

a^2 + (a+4)^2 = 400

Like? Yes (2) | No (4)  

Rownaz Haque   11 years AGO

Let the side of the square be x.
Side of other square=x+4
Sum of area of two squares= 400
so,
x sq + (x +4) sq = 400
2x sq + 16+ 8x = 400

putting the values from options we see that option 7 satisfies the equation....so ans is option 7.

Like? Yes (1) | No (4)  

ASHISH MITTAL   11 years AGO

12^2+16^2=144+256=400

Like? Yes (1) | No (4)  

Arun Kumar Tiwari   11 years AGO

x^2 + (x+4)^2 = 400
=>x^2 + x^2 + 16 + 8x = 400
=>2x^2 +16+8x=400
=>x^2 +4x+8=200
=>(x+16)(x-12)=0
=>x=-16 or 12

since the length of the square can not be negative
Therefore x=12

so, lenght of first square is
12
and second square is x+4=16

Thus answer is 12 and 16

Like? Yes (1) | No (4)  

pravesh singh khatri   11 years AGO

since
area of square1 + area of square2= 400,
a1^2 + a2^2 = 400,
since a2= a1+4
a1^2 + (a1+4)^2=400,
a1=12,
a2=16

Like? Yes (1) | No (4)  

yeshwanth   11 years AGO

let x be one side of the small square,
big squares side = x+4
area sum =x*x+(x+4)*(x+4)=400
x2+x2+4x+4x+16=400
2x2+8x+16=400
x2+4x+8=200


solving this x=

Like? Yes (1) | No (4)  

srilakshmi   11 years AGO

let the side of one square is xcm,then then the second one will be (x+4)cm.
some of area of those 2 squares is 400sqrcm
so
x^2+(x+4)^2 = 400
x^2+x^2+16+8x = 400
x^2+4x+8 = 200
x^2+4x+4 = 196
(x+2)^2 = 196
x+2 = 14
x = 12
so
x+4 = 16
12cm and 16cm is answer

Like? Yes (1) | No (5)  

harshal agrawal   11 years AGO

a-b = 4
400-2ab = 16
ab=192
hence
a=16 b=12

Like? Yes (1) | No (5)  

DILEEP KUMAR   11 years AGO

a^2 + (a+4)^2 = 400
a=12,16.

Like? Yes (1) | No (5)  

Gokul Krishnan   11 years AGO

x^2+(x+4)^2=400

x=12,16

Like? Yes (1) | No (5)