Question

the area of an isosceles triangle is 60 CM square and the length of each one of its equal sides is 13 cm find its base

Answer 1

GIVEN:

• Area of an isosceles triangle = 60 cm²
• Equal sides of an isosceles triangle = 13 cm

TO FIND:

• What is the length of the base ?

SOLUTION:

Let the length of the base be 'a' cm

and height be 'h' cm

In triangle ABD, we use the formula:-

❰ AREA = $\bf{\dfrac{1}{2}}$ $\times$ B $\times$ H ❱

According to question:-

➺ 60 = $\sf{\dfrac{1}{2}}$$\times$ 2a $\times$ h

➺ 60 $\times$ 2 = 2ah

➺ 120 = 2ah....1)

Applying Pythagoras theorem in triangle ABD:-

➺ (AB)² = (BD)² + (AD)²

➺ (13)² = a² + h²

➺ 169 = a² + h²....2)

Adding Equation 1) and 2), we get

➺ 120 + 169 = 2ah + a² + h²

➺ 289 = (a + h)²

Taking square root from both sides

➺ 17 = a + h

➺ 17 –a = h....3)

Put the value of h in equation 1)

➺ 120 = 2 $\times$ a $\times$ (17–a)

➺ $\sf{\cancel\dfrac{120}{2}}$ = 17a –a²

➺ 60 = 17a –a²

➺ a² –17a + 60 = 0

➺ a² –(12+5)a + 60 = 0

➺ a² –12a –5a + 60 = 0

➺ a(a –12) –5(a –12) = 0

➺ (a –5)(a–12) = 0

❮ a = 5 ❯

❮ a = 12 ❯

➺ BASE = 2a = 2(5) = 10 cm

⠀⠀⠀⠀⠀⠀⠀ or

➺ BASE = 2a = 2(12) = 24 cm

❝ Hence, the base of an isosceles triangle is 10 cm or 24 cm ❞

______________________

Answer 2

Solution :

Let the base be 2x and height be h.

$\sf In \: \triangle ABD \\\\\\:\implies\sf AB^2 = AD^2 + BD^2\:\:\:\:\:\Bigg\lgroup \textbf{By Phythagoras theorem}\Bigg\rgroup\\\\\\:\implies\sf 13^2 = h^2 + x^2\:\:\:\:\:\ \Bigg\{\textbf{equation (1)}\Bigg\}$

$\begin{lgathered}\dashrightarrow\sf\:\:Area_{{\tiny\triangle ABC}}=\dfrac{1}{2} \times Base \times Height\\\\\\\dashrightarrow\sf\:\:60 = \dfrac{1}{2} \times 2x \times h \\\\\\\dashrightarrow\sf\:\:120 = 2xh\:\:\:\:\:\ \Bigg\{\textbf{equation (2)}\Bigg\}\end{lgathered}$

☯ $\underline{\boldsymbol{Adding\: equation\: (1) \:and\: 2 :}}$

$\begin{lgathered}:\implies\sf 17 = x + h\:\:\:\:\:\ \Bigg\{\textbf{equation (3)}\Bigg\}\\\\\\:\implies\sf h = 17 - x \end{lgathered}$

$\underline{\boldsymbol{Put\: the\:value\:of\:h\:in\: equation\: (2) \:we\: get :}}$

$\begin{lgathered}:\implies\sf x^2 - 17x + 60 = 0\\\\\\:\implies\sf x^2 - 12x+ 5x + 60 = 0\\\\\\:\implies\sf x(x - 12) - 5(x - 12) = 0 \\\\\\:\implies\sf (x - 12) (x - 5) = 0 \\\\\\:\implies\sf x = 5\: or\: x = 12\end{lgathered}$

$\begin{lgathered}\bullet\:\:\textsf{If x = 5 ,BC = 2x = 2(5) = \textbf{10 cm}}\\\bullet\:\:\textsf{If x = 12 ,BC = 2x = 2(12) = \textbf{24 cm}}\end{lgathered}$