Question

find the area of triangle ABC,where AB=5CM,BC=8CM,AC=9CM.Find the length of perpendicular from A to BC.​

Answer 1

Step-by-step explanation:

Given that: Triangle ABC,where AB=5CM,BC=8CM,AC=9CM.

To find:Area of ∆ ABC and length of perpendicular from A to BC.

Solution:

Tip: When all sides are given Heron's formula can be applied.

$\bold{Heron \: formula = \sqrt{S(S - a)(S - b)(S - c)}} \\ \\ here \\ \\ S = semi-primeter \: of \: triangle \\ \\ a,b,c \: are \: side \: lengths$

Let

a=5 cm

b= 8 cm

c=9 cm

$S = \frac{a + b + c}{2} \\ \\ S= \frac{5 +8 + 9 }{2} \\ \\ S = \frac{22}{2} \\ \\ S = 11 \: cm$

apply the values in formula

$AREA= \sqrt{11(11 - 5)(11 - 8)(11 - 9)} \\ \\ = \sqrt{11 \times 6 \times 3 \times 2} \\ \\ area \: of \: triangle= 6 \sqrt{11} \: {cm}^{2}$

Thus,

Area of ∆ABC = 6√11 cm².

2. To find length of perpendicular from A to BC:

In this case,CB is base of triangle and height(perpendicular) is to recalculated

Area of Triangle= 1/2× Base× height

But we know that area of ∆ ABC is 6√11 cm²

Thus,

$6 \sqrt{11} = \frac{1}{2} \times 8 \times h \\ \\ h = \frac{12 \sqrt{11} }{8} \\ \\ h = 4.97 \: cm$

Thus,

Length of perpendicular from A to BC is 4.97 cm ≈ 5 cm

Hope it helps you.

Answer 2

Answer:

Let

a=5 cm

b= 8 cm

c=9 cm

\begin{gathered}S = \frac{a + b + c}{2} \\ \\ S= \frac{5 +8 + 9 }{2} \\ \\ S = \frac{22}{2} \\ \\ S = 11 \: cm \\ \\ \end{gathered}

S=

2

a+b+c

S=

2

5+8+9

S=

2

22

S=11cm

apply the values in formula

\begin{gathered} AREA= \sqrt{11(11 - 5)(11 - 8)(11 - 9)} \\ \\ = \sqrt{11 \times 6 \times 3 \times 2} \\ \\ area \: of \: triangle= 6 \sqrt{11} \: {cm}^{2} \end{gathered}

AREA=

11(11−5)(11−8)(11−9)

=

11×6×3×2

areaoftriangle=6

11

cm

2

Thus,

Area of ∆ABC = 6√11 cm².

2. To find length of perpendicular from A to BC:

In this case,CB is base of triangle and height(perpendicular) is to recalculated

Area of Triangle= 1/2× Base× height

But we know that area of ∆ ABC is 6√11 cm²

Thus,

\begin{gathered}6 \sqrt{11} = \frac{1}{2} \times 8 \times h \\ \\ h = \frac{12 \sqrt{11} }{8} \\ \\ h = 4.97 \: cm \\ \\ \end{gathered}

6

11

=

2

1

×8×h

h=

8

12

11

h=4.97cm

Thus,

Length of perpendicular from A to BC is 4.97 cm ≈ 5 cm

Step-by-step explanation:

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