Question

in a right angled isosceles triangle.if its sides are increased by 75%,then the percentage change its areas?

Answer 1

Solution :-

In an isosceles triangle, there are two sides with equal length.

Let the two sides with equal length of the original triangle before the increase be a and b and their length is 10 cm each.

Area of an isosceles triangle = 1/2*Base*Height

⇒ 1/2*10*10

⇒ 100/2

⇒ 50 sq cm

Now, length of each of these two sides after increase = 17.5 cm

Increased area = 1/2*17.5*17.5

⇒ 306.25/2

= 153.125 sq cm

Percentage increase in area = (Change in area*100)/Original area


⇒ [(153.125 - 50)*100]/50

⇒ (103.125*100)/50

⇒ 10312.5/50

⇒ 206.25 %

So, if the sides are increased by 75 %, then the area will increase by 206. 25 %.

Answer.

Answer 2

In an isosceles ∆, two sides are of equal length.

Let the two sides of equal length of the original isosceles ∆ = x

Original Area of an isosceles ∆ = ½ Base × height


=( ½) × x × x


Original Area of an isosceles ∆ = x²/2 cm²

New side after 75% increase = 75%of x + x

= (75/100)x +x= 3x/4+x=( 3x+4x) /4= 7x/4

New side after 75% increase =7x/4


Increased area =( ½)× 7x/4 × 7x/4= 49x²/32


Change in area = increased area - original area = (49x²/32 - x²/2)

Percentage increase in area = (Change in area×100)/Original area


= [( (49x²/32 - x²/2))×100]/x²/2

=( (49x²-16x²)/32) × 100 / (x²/2)

= 33x²/32 × 2/x²

= (33 ×100)/16


= 3300/16

= 825/4

206.25 %

Hence, if the sides are increased by 75 %, then the area will increase by 206. 25 %.

==========°==========================°============================

Hope this will help you....