Question

A rectangular plot is to be designed whose breadth is 3m less than length,it's area is to be 4 square metre more than the area of a park that has already been made in the shape isosceles triangle with it's base as the breadth of the rectangular park and of altitude 12 metre find it's length and breadth ?​

Answer 1

Answer:

MATHS

A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find length and breadth of rectangular park.

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ANSWER

Here, Length of Rectangular Park =l

Breadth of Rectangular Park =l−3

Altitude of Triangular Park =12

Breadth of Triangular Park =l−3

Now,

A

Rectangle

=l(l−3)

A

Triangle

=

2

1

×12(l−3)

A

Rectangle

−A

Triangle

=4

⇒l(l−3)−

2

1

×12(l−3)=4

⇒l

2

−9l+14=0

⇒(l−2)(l−7)=0⇒l=2,7

Since, breadth is l−3 , l



=2

∴l=7 m and breadth =7−3=4 m

Answer 2

Given:

• Breadth of rectangular plot is 3m less than length of rectangular plot

• Area of rectangular field is 4m² more than area of triangle.

• Altitude = 12m

Find:

• Length and breadth of Rectangular plot.

Solution:

Here,

Let, Length be x m

and Breadth be (x - 3) m

we, know that

$\underline{ \boxed{ \sf \leadsto Area \: of \: triangle = \frac{1}{2} \times b \times h }}$

where,

• Base b = (x - 3) m
• Attitude h = 12 m

So,

$\sf \rightharpoonup Area \: of \: triangle = \frac{1}{2} \times b \times h$

$\sf \rightharpoonup Area \: of \: triangle = \frac{1}{2} \times (x - 3) \times (12)$

$\sf \rightharpoonup Area \: of \: triangle = 6(x - 3)$

$\sf \rightharpoonup Area \: of \: triangle = 6(x - 3) {m}^{2}$

________________

we, know that

$\underline{ \boxed{ \sf \leadsto Area \: of \: rectangle =l \times b}}$

where,

• Length l = x m
• Breadth b = (x - 3) m

So,

$\sf \longmapsto Area \: of \: rectangle = l \times b$

$\sf \longmapsto Area \: of \: rectangle = (x) \times (x - 3)$

$\sf \longmapsto Area \: of \: rectangle = ({x}^{2} - 3x) {m}^{2}$

__________________

Now,

$\bold{ACCORDING \: TO \: QUESTION}$

➥ Area of Rectangle = 4 + Area of triangle

where,

• Area of Rectangle = (x² - 3x) m²
• Area of Triangle = 6(x - 3) m²

So,

$\sf \leadsto ( {x}^{2} - 3x) = 4 + 6(x - 3)$

$\sf \leadsto {x}^{2} - 3x= 4 + 6x - 18$

$\sf \leadsto {x}^{2} - 3x - 6x= 4 - 18$

$\sf \leadsto {x}^{2} - 9x= - 14$

$\sf \leadsto {x}^{2} - 9x + 14 = 0$

Now, solve by middle split term

$\sf \leadsto {x}^{2} - 7x - 2x + 14 = 0$

$\sf \leadsto x(x - 7) - 2(x - 7) = 0$

$\sf \leadsto (x - 7) (x - 2) = 0$

So,

$\sf \leadsto x = 7,2$

_______________

Now,

Case 1:

Length = x = 7m

Breadth = x - 3 = 7 - 3 = 4m

_____________

Case 2:

Length = x = 2m

Breadth = x - 3 = 2 - 3 = -1m

Here, Case 2 can not be possible.

_______________

So, Length of Rectangular plot = 7m

and Breadth of Rectangular plot = 4m