Question

Triangle ABC is right-angled at B. Given AB = 9cm, AC = 15cm and D, E are the mid-points of AB and AC respectively, calculate 1) The length of BC 2) The area of

Answer 1

By Pythagoras theorem

AC²=AB²+BC²

15²=9²+BC²

225-81=BC²

144=BC²

√144=BC

12=BC

BC=12cm


Area of triangle =base*hieght/2

Area =12*9/2

Area =6*9

Area =54cm²

Answer 2

Answer:

1. BC= 12 cm

2. The area of the triangle ΔADE= 13.5 sq.cm.

Step-by-step explanation:

1. According to Pythagoras theorem, in ΔABC, if ∠ABC is a right angle, then,

$AC^{2} = AB^{2}+ BC^{2}$

given that, AB= 9cm, AC= 15 cm

∴$15^{2}- 9^{2}= BC^{2}$

or, 225- 81= 144= $BC^{2}$

∴BC= 12cm

2. D,E are the mid points of AB and AC respectively.

So, AD= 4.5cm, AE= 7.5cm

now, DE= $\sqrt[]{(7.5^{2} }- 4.5^{2} )$ = $\sqrt{56.25- 20.25}$= 6 cm.

Hence, The area of the triangle ΔADE= $\frac{1}{2}$×4.5×6 = 13.5 sq.cm.

To learn more, please visit:

https://brainly.in/question/13231423

https://brainly.in/question/54178582

#SPJ6