Question

1)Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40square meters.
2)The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm
then find its base and altitude.
​

Answer 1

Answer:

Step-by-step explanation:Perimeter of rectangle = 28mts

1)

2 ( l + b ) = 28

2 (l + b ) = 28

l + b = 28/2

l + b = 14

l = 14 - b ---------(1)

Now area of rectangle = 40

l × b = 40

from eq 1

( 14-b ) × b = 40

Therefore, length of rectangle = 10 , breadth of rectangle =4..

Base of a triangle is 4 cm longer than its altitude.

2)

Area of triangle =48 cm  

2

 

Let base of triangle =b

Altitude =b−4

Area =  

2

1

​  

×b×h=48

2

1

​  

×b×(b−4)=48

b  

2

−4b−96=0

(b−12)(b+8)=0

But, base cannot be negative

So, base =12 cm

Altitude=12−4=8 cm.