Question

The sum and difference of two sides of a triangle is 52 cm and 28 cm. If the third side of the
angle is 36 cm. Then the area of the triangle is​

Answer 1

$\tt{\implies{\underbrace\pink{Answer\implies 212.26cm^2}}}$

$\sf\small\underline\purple{Let:-}$

$\tt{\implies The\: sides\:of\:triangle=x\:and\:y}$

$\sf\small\underline\purple{Given:-}$

$\tt{\implies Sum\:_{(two\: sides)}=52cm}$

$\tt{\implies Difference\:_{(two\: sides)}=28cm}$

$\tt{\implies Third\:side\:_{(triangle)}=36cm}$

$\sf\small\underline\purple{To\:Find:-}$

$\tt{\implies Area\:_{(triangle)}=?}$

$\sf\small\underline\purple{Solution:-}$

To calculate the area of triangle at first we have find out the sides of triangle. To find the sides of this triangle we have to set up equation. Then find out area of triangle. As given in the question that sum and difference are given so, it's a clue by which you can set up equation. you can easily find out the two sides of triangle by solving them. After that you have to calculate semiperimeter then calculate it's area by applying heron's formula:

$\tt{\implies Sum\:_{(two\: sides)}=52cm}$

$\tt{\implies x+y=52----(i)}$

$\tt{\implies Difference\:_{(two\: sides)}=28cm}$

$\tt{\implies x-y=28-----(ii)}$

• Here adding eq (i) and (ii)

$\tt{\implies x+y=52}$

$\tt{\implies x-y=28}$

• By solving we get here,

$\tt{\implies 2x=80}$

$\tt{\implies x=40}$

• putting the value of x=40 in eq (i)

$\tt{\implies x+y=52}$

$\tt{\implies 40+y=52}$

$\tt{\implies y=52-40}$

$\tt{\implies y=12cm}$

$\tt{\implies Here\: sides\:of\: triangle=12cm,\:36cm,\:40cm}$

• Now calculate semiperimeter here:-]

$\tt{\implies S=\dfrac{a+b+c}{2}}$

$\tt{\implies a=12,\:b=36,\:c=40}$

$\tt{\implies S=\dfrac{12+36+40}{2}}$

$\tt{\implies S=\dfrac{88}{2}}$

$\tt{\implies S=44cm}$

• Now calculate area of triangle:-]

$\tt{\implies A\:_{(triangle)}=\sqrt{s(s-a)(s-b)(s-c)}}$

$\tt{\implies A\:_{(triangle)}=\sqrt{44(44-12)(44-36)(44-4)}}$

$\tt{\implies A\:_{(triangle)}=\sqrt{44*32*8*4}}$

$\tt{\implies A\:_{(triangle)}=\sqrt{(11*4)(8*4)*8*4}}$

$\tt{\implies A\:_{(triangle)}=8*4*2\sqrt{11}}$

$\tt{\implies A\:_{(triangle)}=64\sqrt{11}}$

$\sf\large{Hence,}$

$\tt{\implies Area\:_{(triangle)}=64\sqrt{11}cm^2\:or\:212.26cm^2}$

Answer 2

solutions

x+y=52

X-y=28

so, x=40 and y=12 by solving we get

s=a+b+c/2=88/2=44

Area of A=V44(44-12)(44-36) (44-40)=212.26cm

Thus,area of the triangle is 212.26cm