Question

in figure BD = 10 cm Ac = 12 cm and bc = 26 cm A) find area of abc b) Calculate the height AEin figure BD = 10 cm Ac = 12 cm and bc = 26 cm A) find area of abc b) Calculate the height AE ( Figure given below
)

Answer 1

Given:

In figure BD = 10 cm AC = 12 cm and BC = 26 cm

To find:

A) Area of ABC

B) The height AE

Solution:

Finding the value of AD:

In Δ BDC, using Pythagoras theorem, we get

$BC^2 = CD^2 + BD^2$

$\implies 26^2 = (12 + AD)^2 + 10^2$

$\implies 676 = (12 + AD)^2 + 100$

$\implies 576 = (12 + AD)^2$

$\implies 24 = 12 + AD$

$\implies \bold{AD = 12 \:cm}$

Finding the value of AB:

In Δ ABD, using Pythagoras theorem, we get

$AB^2 = AD^2 + BD^2$

$\implies AB = \sqrt{AD^2 + AD^2}$

$\implies AB = \sqrt{12^2 + 10^2}$

$\implies \bold{ AB = 15.62\:cm}$

Finding the area of Δ ABC:

Using Heron's formula in Δ ABC, we have

a = 15.62 cm, b = 12 cm & c = 26 cm

Semi-perimeter, S = $\frac{a+b+c}{2} = \frac{15.62 + 12 + 26}{2} = \frac{53.62}{2} = 26.81 \:cm$

∴ Area of Δ ABC,

= $\sqrt{s (s - a) (s -b)(s-c)}$

= $\sqrt{26.81 (26.81 - 15.62) (26.81 - 12)(26.81-26)}$

= $\sqrt{26.81 \times 11.19 \times 14.81 \times 0.81}$

= $59.99$

≈ $\boxed{\bold {60 \: cm}}$

Finding the height AE:

We know,

$\boxed{\bold{Area \:of\:a\:triangle = \frac{1}{2} \times base \times height}}$

∴ Area of Δ ABC = $\frac{1}{2} \times BC \times AE$

$\implies 60 = \frac{1}{2} \times 26 \times AE$

$\implies 60 = 13 \times AE$

$\implies AE = \frac{60}{13}$

$\implies \boxed{\bold{AE =4.6\: cm}}$

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Answer 2

Answer:

In the given figure, BD = 10 cm, AC = 12 cm, and BC = 26 cm

(a) Find the area of AABC. D

(b) Calculate the height AE. А 10 cm 12 cm E В. С 26 cm