The area of rectangle gets reduced …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Maunisha Mallan Mallan 6 years, 6 months ago
- 1 answers
Test
Related Questions
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 0 answers
Posted by Hari Anand 6 months, 2 weeks ago
- 0 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Lakshay Kumar 1 year, 1 month ago
- 0 answers
Posted by Kanika . 1 month, 1 week ago
- 1 answers
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 6 years, 6 months ago
Let length of given rectangle be x and breadth be y
{tex}\therefore{/tex} area of rectangle = xy
According to the first condition
(x - 5)(y + 3) = x y - 9
or, xy + 3x - 5 y - 15 = xy - 9
or, xy + 3x - 5y - xy = 15 - 9
or, 3x - 5y = 6........(i)
According to the second condition,
(x +3)(y +2) = xy + 67
or, xy + 2x + 3y + 6 = xy + 67
or, xy + 2x + 3y - xy = 67 - 6
or, 2x + 3y = 61 ...(ii)
Multiplying eqn. (i) by 3 and eqn. (ii) by 5 and then adding,
9x-15y = 18
10x + 15y = 305
or, 19x = 323
{tex}\therefore{/tex} {tex}x = \frac { 323 } { 19 } = 17{/tex}
Substituting this value of x in eqn. (i),
3(17)-5y = 6
51 - 5y = 6
or, 5y = 51 - 6
{tex}\therefore{/tex} y = 9
0Thank You