Question

The sides of a triangle are 30cm, 70cm, 80cm. If an altitude is dropped upon the side of length 80cm. Then find the length of the larger segment cut off this side.

Answer 1

by using Helens formula you get equation 1 and remains solution as FOLLOWS
i Hope it is correct

Answer 2

Answer:

The length of the larger segment is 65 cm.

Solution:

Note: Refer the diagram in the image attached below.

The sides of the triangle A= 30 cm, B= 70 cm and C = 80 cm,

The perimeter of the triangle,

$(30+70+80) c m=180 \mathrm{cm}$

Semi-perimeter,  

$S=(180 / 2) \mathrm{cm}=90 \mathrm{cm}$

According to the Heron’s formula,  

Area $S1 = \sqrt{S(S-A)(S-B)(S-C)}=\sqrt{90(90-30)(90-70)(90-80)} \mathrm{cm}$

$S 1=\sqrt{1080000} \mathrm {cm}$ …. (i)

Let us assume that the height of the triangle is h,  

So the area of the triangle S2= $\frac{1}{2}(80 \times h)=40 h$ …………….. (ii)

Equating,

  $S 2=S 1$

$(40 h)^{2}=1080000$

$h^{2}=  675$

$h=\sqrt{675}$

Let$x \text { and }(80-x)$are the two segments created by the altitude on the side of length 80.

According to Pythagoras Theorem, $A^{2}+B^{2}=C ^{2}$

$x^{2}+h^{2}=300$

$x^{2}= 900-675$

$x^{2}=225$

On taking square roots on both the sides,

$x=15$

So, the length of the larger segment is $(80-x) c m=(80-15) c m=65 \mathrm{cm}$