Question

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???​

Answer 1

Answer:

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Answer 2

Answer:

⇒CE= 7.2 cm

Step-by-step explanation:

In ΔABC, AD=6cm and BC=9cm

Area of triangle=

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

$= \frac{1}{2} \times bc \times ad$

$\frac{1}{2} \times 9 \times 6$

= 27 cm²

Again, Area of triangle=

$\frac{1}{2} \times base \: \times height$

$\frac{1}{2} \times ab \times ce$

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

$⇒ce = \frac{27 \times 2}{7.5}$

⇒CE= 7.2 cm

Thus, height from C to AB i.e., CE is 7.2cm.

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