Question

find the area of an isosceles triangle whose sides are 10 cm and base is 12cm​

Answer 1

Answer:

$\large\bold\red{48\:{cm}^{2}}$

Step-by-step explanation:

Given,

An isosceles ∆ABC,

• Length of each equal sides = AB = AC = 10 cm

• Length of base = BC = 12 cm

Construction:

• Draw an altitude from the vertex A to the base Bc meeting at D.

Note:- Refer to the attachment for diagram.

Now,

We know that,

• Altitude of an isosceles triangle bisects the base .

Therefore,

We get,

• BD = CD = 6 cm

Let's assume AD = h

Now,

In right ∆ABD

By Pythagoras theorem,

We have,

${AB}^{2}={AD}^{2}+{BD}^{2}\\\\=>{10}^{2}={h}^{2}+{6}^{2}\\\\=>100={h}^{2}+36\\\\=>{h}^{2}=100-36\\\\=>{h}^{2}=64\\\\=>{h}^{2}={8}^{2}\\\\=>h=8\;cm$

Therefore,

We have,

In ∆ABC,

• Altitude = 8 cm
• Base = 12 cm

Therefore,

Area of ∆ABC

$=\frac{1}{2}\times Base\times Altitude\\\\=\frac{1}{2}\times 12 \times 8\\\\=12 \times 4\\\\ =\bold{48 \:{cm}^{2}}$

Answer 2

Answer:

Hello mate here is ur Answer

Step-by-step explanation:

Isosceles triangle

solve for Area

a side => 10cm

b side=>12cm

Using the formula

$A = \frac{bhb}{2} \\ {h}^{b} = \sqrt{ {a}^{2} - \frac { {b}^{2} }{4} } \\ solving \: for \: A \\ A = \frac{1}{4} b \sqrt{ {4a}^{2} - {b}^{2} } \\ = \frac{1}{4}.12. \sqrt{ {4.10 }^{2} - {10}^{2} } \\ = {48cm}^{2}$

have a nyc day✌️✌️✌️