Question

8. ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of ΔABC. What will be the height from C to AB i.e., CE?

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Answer 1

Answer:

The area is 27 sq. cm.

Step-by-step explanation:

hight =6 cm

base=9 cm

Area= 1/2 ×6cm×9cm

= 27 sq. cm

Answer 2

SOLUTION-;

Finding Area of ∆ ABC

using BC as base and AD as a height

: In ΔABC, AD= 6cm and BC= 9cm

$Area \: of \: triangle= \frac{1}{2} \times base \times height</p><p> </p><p> </p><p>$

$= \frac{1}{2} \times BC×AD$

$= \frac{1}{2} \times 9 \times 6 = 27 {cm}^{2}$

Now,

we have to find EC which is height corresponding to base AB

So, finding Area using EC as a height AB as a base

Base = AB = 7.5

height = EC = ?

$Area \: of \: triangle = \frac{1}{2} \times base \times height$

$= \frac{1}{2} \times AB×EC$

$⇒27 = \frac{1}{2} \times 7.5 \times EC$

$⇒ EC = \frac{27 \times 2}{7.5}$

$⇒ \frac{27 \times 2 \times 10}{75}$

$⇒ \frac{9 \times 2 \times 10}{25}$

$⇒ \frac{9 \times 2 \times 2}{5}$

$⇒ \frac{36}{5}$

$⇒E C =7.2cm$

since

Area of ∆ABC = 27cm²

and Height CE = 7.2 cm