Question

А.
From the adjoining figure, find
a) the area of AABC
6 cm
b) length of AC
C
8 cm
B
c) the length of an
altitude from B to AC​

Answer 1

(a) Area of ∆ABC :

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

$\frac{1}{2} \times 6 \times 8$

Area = 24cm²

(b) Notice that ∆ABC is right angled

hence,

(AB)² + (BC)² = (AC)² [Pythagoras theorem]

6² + 8² = (AC)²

36 + 64 = (AC)²

100 = (AC)²

√100 = AC

10cm = AC

(c) Let AC be the base now and the altitude be the height

as we know,

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

$\frac{1}{2} \times 10 \times x = 24$

(as we have already calculated the area before and let us take the altitude as x)

$10x = \frac{24}{2}$

$10x = 12$

$x = \frac{12}{10}$

therefore the altitude's length is 1.2cm

Hope it helps

have a great day