Question

HlO MATHS ARYABHATTA,S☺

Q2.The area of a rectangle is equal to 96 square meters. Find the length and width of the rectangle if
its perimeter is equal to 40 meters.


Q3.The height of a triangle is 3 feet longer than its corresponding base. The area of the triangle is

equal to 54 square feet. Find the base and the height of the triangle.


Q4.The product of the first and the third of three consecutive positive numbers is equal to 1 subtracted from the square of the second of these numbers. Find the three numbers.

Answer 1

Heya mate, Here is ur answer

1) Perimeter of a rectangle= 2(l+b)

40=2(l+b)

40/2=l+b

20=l+b

20-l=b --------(1)

Area of rectangle= l×b

96 = l×b

from eq (1) , b=20-l

96 = l (20-l)

96 = 20l - l^2

l^2 -20l +96 = 0

By middle term splitting

l^2 - (12+8) l +96 =0

l^2 -12 l -8 l +96=0

l(l-12)+8(l-12)=0

(l+8)(l-12) =0

l+8= 0

l=-8

b=20-(-8)

b=20+8

b=28

l-12=0

l=12

b=20-12

b=8

Since dimensions can never be negative

So, the possible dimensions are :

l=12 and b=8

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2) Let the corresponding base be x.

Then, A/Q,

Height =3+x

Area of a triangle=1/2 ×base × height

$54 = \frac{1}{2} \times x \times 3+x$

$54 = \frac{{x}^{2} +3x}{2}$

${54 \times 2}= {x}^{2} +3x$

$\frac{108}= {x}^{2} +3x$

$0 = {x}^{2} +3x-108$

0=x^2 + 12x -9x +108

0= x(x+12) -9(x+12)

0=(x-9)(x+12)

x-9=0

x=9

x+12=0

x=-12

Since sides of triangle can never be negative,

So, base = 9 feet

Height = 3+ 9

=12 feets

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3) Let three consecutive positive numbers be x, x+1 , and x+2

A/Q

x(x+2) = (x+1)^2 -1

${x}^{2} + 2x = {x}^{2} + 1 + 2x - 1$

${x}^{2} + 2x = {x}^{2} + 2x$

$0 = 0$

This problem has infinity solutions, one can be -

So, the numbers are ---

0 , 0+1 , 0+2

0 , 1 , 2

Let us check

0(2) =1^2-1

0=1-1

0=0

Hence verified

=============================

Warm regards

@Laughterqueen

Be Brainly✌✌✌

Answer 2

(2)

Let the length of the rectangle be 'x' and breadth be 'y' m.

Given, Area of rectangle = 96 m^2.

Given, Perimeter = 40 m.

We know that Area of rectangle = l * b.

⇒ 96 = xy

⇒ x = 96/y  ---- (i)

We know that perimeter of rectangle = 2(l + b)

⇒ 40 = 2x + 2y.

⇒ 40 = 2(x + y)

⇒ 20 = x + y

⇒ 20 = (96/y) + y

⇒ 20 = (96 + y^2)

⇒ 96 + y^2 - 20y  = 0

⇒ y^2 - 20y + 96 = 0

⇒ y^2 - 12y - 8y + 96 = 0

⇒ y(y - 12) - 8(y - 12) = 0

⇒ (y - 8)(y - 12) = 0

⇒ y = 8,12

When y = 12:

⇒ x = 8.

When y = 8:

⇒ x = 12.

Therefore, the length and breadth of the rectangle are 12 m and 8 m.

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(3)

Let the corresponding base be 'x'.

Then the height of the triangle = x + 3.

Given Area of triangle is 54 sq.feet.

We know that Area of triangle = (1/2) * b * h.

⇒ 54 = (1/2) * x * (x + 3)

⇒ 108 = x^2 + 3x

⇒ x^2 + 3x - 108 = 0

⇒ x^2 + 12x - 9x - 108 = 0

⇒ x(x + 12) - 9(x + 12) = 0

⇒ (x - 9)(x + 12) = 0

⇒ x = 9,-12 {x cannot be negative}

⇒ x = 9.

Then:

⇒ x + 3 = 12.

Therefore,

The base of triangle is 9 feet and height of the triangle = 12 feet

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(4)

Let the three consecutive numbers be x,x + 1 and x + 2.

According to the given condition,

⇒ (x)(x + 2) = (x + 1)^2 - 1

⇒ x^2 + 2x = x^2 + 1 + 2x - 1

⇒ x^2 + 2x = x^2 + 2x

This problem has infinite number of solutions.

Hope it helps!