0

Guest

Courses New

please solve.........reply soon sir.............

please solve.........reply soon sir.............

Question Image
Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
Hello Student,
Please find answer to your question below

241-723_pTIUp.png

In place of y, we have x & vice – versa.
Area of triangle ABC
\frac{1}{2}bh…..........(1)
Area of rectangle:
xy…...........(2)
Area of triangle CDE
\frac{1}{2}y(h-x)…............(3)
Area of bottom two triangle:
\frac{1}{2}x(b-y)….............(4)
Now we are going to compare the area’s
(1)=(2)+(3)+(4)
\frac{1}{2}bh = xy + \frac{1}{2}y(h-x) + \frac{1}{2}x(b-y)
bh = hy + bx
y = \frac{b}{h}(h-x)
Perimeter of rectangle:
2(x+y)
2(x+b-\frac{bx}{h})
2(b+x(1-\frac{b}{h}))
Area of rectangle:
xy
\frac{bx}{h}(h-x)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Differential Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More