Question

The base of a isosceles triangle is 4/3cm.The perimeter of the triangle is 4 2/15cm.What is the length of either of the remaining equal sides?​

Answer 1

Answer:

• 7/5cm is the required length of remaining sides .

Step-by-step explanation:

According to the Question

It is given that ,

• Base of isosceles triangle = 4/3 cm
• Perimeter of triangle = 64/15 cm

We have to calculate the length of remaining equal sides .

As we know that two sides in Isosceles triangle are equal in length .

Let the equal length of triangle be x cm

Now,

• Perimeter of triangle = Sum of length of all sides

Putting the values we get

➻ x + x + 4/3 = 62/15

➻ 2x + 4/3 = 62/15

➻ 2x = 62/15 - 4/3

➻ 2x = 62-20/15

➻ 2x = 42/15

➻ x = 42/30 cm

➻ x = 14/10

➻ x = 7/5

• Hence, the length of remaining equal sides of triangle will be 7/5cm .

Answer 2

$\rm \large {\underbrace{\underline{✵Elucidation:-}}}$

$\sf \red {\dag{\underline{\underline{Provided\: that:}}}}$

$\tt {Base(b)\: of\: isosceles\: ∆^{le}=\frac{4}{3}cm}$

$\tt {Perimeter(p)\: of\: the\: ∆^{le}=4\frac{2}{15}cm}$

➻Converting it into improper fraction,

$\large \to \tt {P=\frac{15×4+2}{15}}$

$\large \to \tt {P=\frac{60+2}{15}}$

$\large \tt {★Perimeter=\frac{62}{15}cm}$

$\sf \blue {\dag{\underline{\underline{To\: determine:}}}}$

➻Length of the either of the remaining equal sides.

$\sf \pink {\dag{\underline{\underline{Consideration:}}}}$

➻Let the Length of the either of the remaining equal sides be "a" cm.

$\sf \purple {\dag{\underline{\underline{We\: know:}}}}$

★Perimeter of an isoceles traingle=length of two equal sides +base of the traingle.

➻Substituting in the formula,

$\mapsto \tt {P=a+a+b}$

$\mapsto \tt {P=2a+b}$

$\large \mapsto \tt {\frac{62}{15}=2a+\frac{4}{3}}$

$\large \mapsto \tt {2a=\frac{62}{15}-\frac{4}{3}}$

➻Taking L.C.M,

$\large \mapsto \tt {2a=\frac{62-4(5)}{15}}$

$\large \mapsto \tt {2a=\frac{62-20}{15}}$

$\large \mapsto \tt {2a=\frac{42}{15}}$

$\large \mapsto \tt {a=\frac{42}{15×2}}$

$\large \mapsto \tt {a=\frac{\cancel{42}^{21}}{15×{\cancel{2}}}}$

$\large \mapsto \tt {a=\frac{21}{15}}$

$\large \mapsto \tt {a=\frac{\cancel{21}^7}{\cancel{15}^5}}$ [3 table]

$\large \mapsto \tt {a=\frac{7}{5}}$

(OR)

$\large \mapsto \tt {a=1\frac{2}{5}}$

$\sf \orange {\dag{\underline{\underline{Thereupon,}}}}$

➻Length of the either of the remaining equal sides is $\tt {\frac{7}{5}cm}$ respectively.