Question

in figur,in triangle ABC,BC = 16 cm,AC = 12, AB=10 cm and altitude AD = 6 cm. Find (a) BE (b) altitude on side AB ​

Answer 1

Answer:

The value of BE is 8 cm and altitude on side AB is 9.6 cm.

Step-by-step explanation:

Since we have given that

In ∆ABC,

BC = 16 cm

AC = 12 cm

AB = 10 cm

altitude = AD = 6 cm

Area of triangle with height AD and base BC,

\begin{gathered}\dfrac{1}{2}\times AD\times BC\\\\=\dfrac{1}{2}\times 6\times 16\\\\=8\times 6\\\\=48\ cm^2\end{gathered}

2

1

×AD×BC

=

2

1

×6×16

=8×6

=48 cm

2

Now, area with base AC and height BE,

\begin{gathered}48=\dfrac{1}{2}\times BE\times AC\\\\48=\dfrac{1}{2}\times BE\times 12\\\\48=BE\times 6\\\\\dfrac{48}{6}=BE\\\\BE=8\ cm\end{gathered}

48=

2

1

×BE×AC

48=

2

1

×BE×12

48=BE×6

6

48

=BE

BE=8 cm

Area of triangle with altitude on side AB is given by

\begin{gathered}48=\dfrac{1}{2}\times AB\times h\\\\48=\dfrac{1}{2}\times 10\times h\\\\48=5\times h\\\\h=\dfrac{48}{5}\\\\h=9.6\ cm\end{gathered}

48=

2

1

×AB×h

48=

2

1

×10×h

48=5×h

h=

5

48

h=9.6 cm

Hence, the value of BE is 8 cm and altitude on side AB is 9.6 cm. please mark me as a brainliest