LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

PERIMETER & AREA OF QUADRILATERAL - PROBLEM & SOLUTION

**Example.1) The length and breadth of a rectangular
grassy plot are in the ration 7 : 4. A path 4 m wide running all around outside
it has an area of 416 m². Find the dimensions of the grassy plot.**

**Ans.)
Let the length & breadth of the grassy plot be 7a and 4a meters
respectively. Then the area of the plot is = (7a X 4a) m² = (28 a²) m²**

**Now,
as per the given condition, a path of 4 m wide running all around outside area
and it has area of 416 m²**

**As you can find from the above
picture, in length there is = 4 m + 4 m = 8 m more length to be added in given
length, and in breadth also there is = 4 m + 4 m = 8 m more breadth is to be
added in the given breadth size.**

**Now, the length along with running
area should be = (7a + 8) m**

**And, the breadth along with running
area should be = (4a + 8) m**

**So, area of the plot including the
path = (7a + 8) (4a + 8) m²**

**Now, area of the path => [{(7a + 8)(4a + 8)} – 28a² ]m² = 416 m²**

** Or, (28a² + 32a + 56a + 64) – 28a² = 416**

** Or, 88 a = 416 – 64**

** Or, 88a = 352**

** Or, a = 352/88 = 4**

**So, now the length of the plot is 7a
= 7 X 4 = 28 m, and width of the given plot is = 4a = 4 X 4 = 16 m (Ans.)**

**Example.2) A rectangular lawn 64 m by 48 m has two roads,
each 5 m wide , running in the middle of it, one parallel to length and other parallel
breadth. Find the cost of graveling them at $ 2.80 per m²**

**From the above picture, area of road
ABCD = (64 X 5) m² = 320 m²**

**Area of road EFGH = (48 X 5) m² = 240
m²**

**Clearly, area JKLM is common to both
of the above roads.**

**Area JKLM = (5 X 5) m² = 25 m²**

**Area of the road to graveled = {(320
+ 240) – 25} m² = 535 m²**

**Cost of graveling of the path = $
(2.80 X 535) = $ 1498 (Ans.)**

**Example.3) If the
length and breadth of a rectangular room are each increased by 1 m, then the
area of the floor increased by 21 m². if the length is increased by 1 m and
breadth is decreased by 1 m, then the area is decreased by 7 m². Find the
perimeter of the floor.**

**Ans.) Let, the original length and
breadth of the room be x and y meters respectively. Then, original area of the
floor = (xy) m²**

**When both length and breadth are
increased by 1 m**

**New length = (x + 1) m **

**New breadth = (y + 1) m **

**New area = (x + 1)(y + 1) m²**

**New area => [{(x + 1) (y + 1)} –
xy] = 21**

** => xy + y + x + 1 – xy =
21**

** => y + x = 21 - 1 **

** => y + x = 20 ……………………….(i)**

**When length is increased by 1 m and
breadth is decreased by 1 m**

**New length = (x + 1) m **

**New breadth = (y - 1) m **

**New area = (x + 1)(y - 1) m²**

**New area => [xy - {(x + 1) (y -
1)}] = 7**

** => xy – (xy + y - x – 1) = 21**

** => x - y + 1 = 7 **

** => x - y = 6 ……………………….(ii)**

**Adding (i) & (ii), and we get –**

** x + y = 20 **

** x – y = 6**

** -----------------**

** 2x = 26**

** Or, x = 13**

**Now, we will substitute the value of
x in (i), and we get - **

** y + x = 20**

** or, y + 13 = 20**

** or, y = 7**

**so, the perimeter of the floor –**

** [2 (length + Breadth)] = 2 (13 +
7) = 2 X 20 **

**
**

** = 40 m (Ans.)**