Q.7: Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Q.8: Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
Q.9: Is it possible to design a rectangular park of perimeter 80 and area 400 sq.m.? If so find its length and breadth.
Answers
Answer:
Question 1 :-
The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Given :-
Sum of areas of two squares = 468 m²
Difference of their perimeter = 24 m
To Find :-
Sides of two squares.
Solution :-
Let the side of the first square be x and second square be y.
Hence,
Area of first square = (x)²
Area of second square = (y)²
According to the question,
\dashrightarrow\:\:\sf{x^2+y^2=468}\:\:...(1)⇢x
2
+y
2
=468...(1)
Difference of perimeter of square = 24 m.
So,
\dashrightarrow\:\:\sf{4x-4y= 24\:m}\:\:...(2)⇢4x−4y=24m...(2)
From equation (2) we get,
➝ \sf{4x-4y= 24}4x−4y=24
➝ \sf{4(x-y)= 24}4(x−y)=24
➝ \sf{x-y= \dfrac{24}{2} }x−y=
2
24
➝ \sf{x-y= 6 }x−y=6
➝ \sf{x= 6 +y}\:\:...(3)x=6+y...(3)
Putting the value of x in equation (1)
➔ \sf{x^2+y^2=468}x
2
+y
2
=468
➔ \sf{(6+y)^2+y^2=468}(6+y)
2
+y
2
=468
➔ \sf{(6)^2 + (y)^2 + 2 \times 6 \times y + (y)^2 = 468}(6)
2
+(y)
2
+2×6×y+(y)
2
=468
➔ \sf{36 + y^2 + 12y + y^2 = 468}36+y
2
+12y+y
2
=468
➔ \sf{2y^2 + 12y - 468 +36 = 0}2y
2
+12y−468+36=0
➔ \sf{2y^2 + 12y -432 = 0}2y
2
+12y−432=0
➔ \sf{2( y^2+ 6y - 216) = 0}2(y
2
+6y−216)=0
➔ \sf{y^2 + 6y-216 = 0}y
2
+6y−216=0
➔ \sf{y^2 + 18y -12y -216 = 0}y
2
+18y−12y−216=0
➔ \sf{y(y+18) - 12(y+18) = 0}y(y+18)−12(y+18)=0
➔ \sf{(y+18) (y-12) = 0}(y+18)(y−12)=0
∴ y = 12
Putting y = 12 in equation (3),
➙ \sf{x = 6+y}x=6+y
➙ \sf{x=y + 6 }x=y+6
➙ \sf{12 + 6 = 18 }12+6=18
Therefore,
Side of first square = x = 18 m
Side of second square = y = 12 m.
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