Question

The altitude drawn to the base of an isosceles triangle is of length 8 cm and the perimeter is 32 cm. Find the area of the triangle.

Answer 1

Final Answer : 48 sq. units

Steps :
1)Let AD be the altitude of triangle ABC.
AD = 8 cm
AB = AC = a (say)

In Isosceles triangle, Altitude is median.
=> AD = CD = b (say)

2) In right angled triangle ACD,
$AB^2 = BD^2 + AD^2 \\ => a^2 = b^2 + 8^2 -(1)$

3) Perimeter = 32 cm
=> a + a + 2b = 32
=> a + b = 16 cm ---(2)

4) Using equation (1) & (2) ,
a - b = 4 cm ---(3)

Using equation (2) ,(3) ,
a = 10cm
b = 6 cm


5) Area of Triangle,
$\Delta ABC = \frac{1}{2}*AD*BC \\ => \frac{1}{2}* 8 * 2b \\ => \frac{1}{2}*8*2*6 \\ => 48 \: sq. units$

Answer 2

Let the base of the isosceles triangle be '2y' and the two equal sides be 'x'.

Given perimeter of an isosceles triangle = 32.

= > x + x + 2y = 32

= > 2x + 2y = 32

= > x + y = 32/2

= > x + y = 16. -------- (1)

Now,

Given Altitude h = 8 cm.

By Pythagoras theorem, we get

= > x^2 = 8^2 + y^2

= > x^2 - y^2 = 8^2

= > x^2 - y^2 = 64

= > (x + y)(x - y) = 64

= > (16)(x - y) = 64

= > (x - y) = 64/16

= > (x - y) = 4 ------ (2)

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On solving (1) & (2), we get

= > x + y = 16

= > x - y = 4

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2x = 20

x = 10.

Substitute x = 10 in (1), we get

= > x + y = 16

= > 10 + y = 16

= > y = 6.

Now,

= > Area of triangle = (1/2) * base * height

= > (1/2) * (2y) * (8)

= > (1/2) * 12 * 8

= > 48cm^2.

Therefore, the area of triangle = 48 cm^2.

Hope this helps!