Question

ABC is a triangle in which AB equal to AC equal to 4 cm and angle A equal to 90 degree calculate the area of triangle ABC and the length of perpendicular from a to BC​

Answer 1

Step-by-step explanation:

∠A = 90°

AB = AC = 4 cm

Area of triangle = 1/2*Base*Height

                          =1/2*4*4

                          =8 cm^2

Hypotenuse = [4^2 + 4^2]^1/2

                     = 32^1/2

                     =4root2 cm

Area of triangle = 1/2*hypotenuse*perpendicular on hypotenuse

8 = 1/2*4root2*x

x = 2root2 cm

Hope it helps

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Answer 2

Diagram:

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \put(2,2){\line(0, 1){2.5}} \put(2, 2){\line(1, 0){2}} \put(4,2){\line(-4,5){2}} \put(2,2){\line(3,5){1}} \put(2,4.5){\large{B}} \put(2,1.5){\large{A}}\put(4,1.5){\large{C}} \put(3,3.5){\large{O}} \end{picture}$

Given:

• Right angle ABC
• AB = AC = 4cm
• $\angle$ A = 90°

Find:

• Calculate the area of triangle ABC
• Length of Perpendicular from A to BC

Solution:

we, know that

$\boxed{\red{\rm Area \: of \: triangle = \dfrac{1}{2} \times b \times h}}$

where,

• b = AC = 4cm
• h = AB = 4cm

So,

$\dashrightarrow\purple{\rm Area \: of \: triangle = \dfrac{1}{2} \times b \times h}$

$\dashrightarrow\purple{\rm Area \: of \: triangle = \dfrac{1}{2} \times 4 \times 4}$

$\dashrightarrow\purple{\rm Area \: of \: triangle = \dfrac{1}{2} \times 16}$

$\dashrightarrow\purple{\rm Area \: of \: triangle = \dfrac{16}{2}}$

$\dashrightarrow\purple{\rm Area \: of \: triangle = 8 {cm}^{2} }$

So, area of triangle ABC = 8cm²

________________________

In triangle ABC

BY PYTHOGORAS THEOREM

$\longmapsto\orange{\rm H^2 = P^2 + B^2 }$

$\longmapsto\orange{\rm {BC}^2 = {AB}^2 + {AC}^2 }$

$\longmapsto\orange{\rm {BC}^2 = {4}^2 + {4}^2 }$

$\longmapsto\orange{\rm {BC}^2 = 16 + 16}$

$\longmapsto\orange{\rm {BC}^2 = 32}$

$\longmapsto\orange{\rm BC = \sqrt{32} cm}$

$\longmapsto\orange{\rm BC = \sqrt{16 \times 2} cm}$

$\longmapsto\orange{\rm BC = 4\sqrt{2} cm}$

Now, take BC as base and AO as a height

So, again using

$\boxed{\blue{\rm Area \: of \: triangle = \dfrac{1}{2} \times b \times h}}$

where,

• b = BC = 4√2cm
• Area of triangle = 8cm²

So,

$\dashrightarrow\pink{\rm Area \: of \: triangle = \dfrac{1}{2} \times b \times h}$

$\dashrightarrow\pink{\rm 8 = \dfrac{1}{2} \times 4 \sqrt{2} \times AO}$

$\dashrightarrow\pink{\rm 8 = \dfrac{4 \sqrt{2}}{2} \times AO}$

$\dashrightarrow\pink{\rm 8 = 2\sqrt{2} \times AO}$

$\dashrightarrow\pink{\rm \dfrac{8}{2\sqrt{2}} = AO}$

$\dashrightarrow\pink{\rm AO = \dfrac{4}{\sqrt{2}} }$

$\qquad$ Rationalise the denominator

$\dashrightarrow\pink{\rm AO = \dfrac{4}{\sqrt{2}} \times \dfrac{ \sqrt{2} }{ \sqrt{2} } }$

$\dashrightarrow\pink{\rm AO = \dfrac{4 \sqrt{2} }{2}cm}$

$\dashrightarrow\pink{\rm AO = 2 \sqrt{2}cm}$

Hence, length of perpendicular A to BC is 2√2cm