Question

In figure (ii) given below, angle BAC = 90°, angle ADC = 90°, AD = 6 cm, CD = 8 cm and
BC = 26 cm. Find
(i) AC (ii) AB (iii) area of the shaded region.​

Answer 1

Given:

∠BAC = 90°

∠ADC = 90°

AD = 6 cm

CD = 8 cm

BC = 26 cm

To find:

(i) AC

(ii) AB

(iii) Area of the shaded region​

Solution:

Formulas to be used:

$\boxed{\bold{Pythagoras\: theorem \rightarrow Hypotenuse^2 = Perpendicular^2 + Base^2}}\\\\\boxed{\bold{Area \:of\: a\: triangle =\frac{1}{2}\times base\times height }}$

(i). Finding the length of AC:

In Δ ADC, using the Pythagoras theorem, we get

$AC^2 = AD^2 + CD^2$

substituting the given values

$\implies AC^2 = 6^2 + 8^2$

$\implies AC = \sqrt{36 + 64}$

$\implies AC = \sqrt{100}$

$\implies AC = 10\:cm$

Thus, $\boxed{\boxed{\bold{AC = 10\:cm}}}$.

(ii). Finding the length of AB:

In Δ ABC, using the Pythagoras theorem, we get

$BC^2 = AC^2 + AB^2$

substituting the given values

$\implies 26^2 = 10^2 + AB^2$

$\implies AB^2 = 26^2 - 10^2$

$\implies AB = \sqrt{676 - 100}$

$\implies AB = \sqrt{576}$

$\implies AB = 24\:cm$

Thus, $\boxed{\boxed{\bold{AB = 24\:cm}}}$.

(iii). Area of the shaded region:

The area of the shaded region is,

= [Area of Δ ABC] - [Area of Δ ADC]

= $[\frac{1}{2}\times AB\times AC ] - [\frac{1}{2}\times AD\times CD]$

substituting the values, we get

= $[\frac{1}{2}\times 24\times 10 ] - [\frac{1}{2}\times 6\times 8]$

= $[12\times 10 ] - [3\times 8]$

= $120 - 24$

= $96\:cm^2$

Thus, $\boxed{\boxed{\bold{Area\:of\:the\:shaded\:region = 96\:cm^2}}}$

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Answer 2

Given:

∠BAC = 90°

∠ADC = 90°

AD = 6 cm

CD = 8 cm

BC = 26 cm

To find:

(i) AC

(ii) AB

(iii) Area of the shaded region

Solution:

Formulas to be used:

\begin{gathered}\boxed{\bold{Pythagoras\: theorem \rightarrow Hypotenuse^2 = Perpendicular^2 + Base^2}}\\\\\boxed{\bold{Area \:of\: a\: triangle =\frac{1}{2}\times base\times height }}\end{gathered}Pythagorastheorem→Hypotenuse2=Perpendicular2+Base2Areaofatriangle=21×base×height

(i). Finding the length of AC:

In Δ ADC, using the Pythagoras theorem, we get

AC^2 = AD^2 + CD^2AC2=AD2+CD2

substituting the given values

\implies AC^2 = 6^2 + 8^2⟹AC2=62+82

\implies AC = \sqrt{36 + 64}⟹AC=36+64

\implies AC = \sqrt{100}⟹AC=100

\implies AC = 10\:cm⟹AC=10cm

Thus, \boxed{\boxed{\bold{AC = 10\:cm}}}AC=10cm .

(ii). Finding the length of AB:

In Δ ABC, using the Pythagoras theorem, we get

BC^2 = AC^2 + AB^2BC2=AC2+AB2

substituting the given values

\implies 26^2 = 10^2 + AB^2⟹262=102+AB2

\implies AB^2 = 26^2 - 10^2⟹AB2=262−102

\implies AB = \sqrt{676 - 100}⟹AB=676−100

\implies AB = \sqrt{576}⟹AB=576

\implies AB = 24\:cm⟹AB=24cm

Thus, \boxed{\boxed{\bold{AB = 24\:cm}}}AB=24cm .

(iii). Area of the shaded region:

The area of the shaded region is,

= [Area of Δ ABC] - [Area of Δ ADC]

= [\frac{1}{2}\times AB\times AC ] - [\frac{1}{2}\times AD\times CD][21×AB×AC]−[21×AD×CD]

substituting the values, we get

= [\frac{1}{2}\times 24\times 10 ] - [\frac{1}{2}\times 6\times 8][21×24×10]−[21×6×8]

= [12\times 10 ] - [3\times 8][12×10]−[3×8]

= 120 - 24120−24

= 96\:cm^296cm2

Thus, \boxed{\boxed{\bold{Area\:of\:the\:shaded\:region = 96\:cm^2}}}Areaoftheshadedregion=96cm2