Question

Find the base of a triangle whose area is 0.48dm square and altitude is 8cm

Answer 1

Answer :

Given area of triangle = 0.48 dm

{1dm = 10 cm}

0.48 dm = 4.8 cm

Given altitude or height of triangle (h) = 8 cm

As we know that

$area \: of \: triangle \: = \frac{1}{2} \times base \: \times height \\ \\ therefore \\ \\ \frac{1}{2} \times b \times h = 4.8 \\ \\ \frac{1}{2} \times b \times 8 = 4.8 \\ \\ b = \frac{4.8}{8} \times 2 \\ \\ b = 0.6 \times 2 \\ \\ b = 1.2 \\ \\ therefore \: base = 1.2 \: cm$

$verification \\ \\ put \: the \: value \: b \: in \: l.h.s \\ \\ \frac{1}{2} \times 1.2 \times 8 = 4.8 \\ \\ \frac{3}{5} \times 8 = 4.8 \\ \\ \frac{24}{5} = 4.8 \\ \\ 4.8 = 4.8 \\ \\ since \: l.h.s = r.h.s \\ therefore \: base \: found \: is \: true$

Answer 2

Area of triangle= 0.48dm
= 0.48÷100 (1dm=100cm)
= 48cm
Altitude =8cm
Area of triangle= 1/2×base×altitude
48cm = 1/2×b×8cm
48cm= 4×b
48cm/4=b
12cm=base