Question

find the area of right angled triangle of side 8cm and 10cm find the base ​

Answer 1

$\texttt{\green{ Height = 8cm }}$

$\texttt{\green{ Hypotenuse = 10cm }}$

$\texttt{\green{ Base = x }}$

$\tt{\green{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: }}$

We know that Hypotenuse ² = height ² + base²

$\tt{\green{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: }}$

$\tt{\red{ {Hypotenuse}^{2} = {height}^{2} + {base}^{2} }}$

$\tt{\red{\implies {10cm}^{2} = {8cm}^{2} + {x}^{2} }}$

$\tt{\red{\implies (10cm \times 10cm) = (8cm \times 8cm) + (x \times x) }}$

$\tt{\red{\implies {100cm}^{2} = {64cm}^{2} + {x}^{2} }}$

$\tt{\red{\implies {100cm}^{2} - {64cm}^{2} = {x}^{2} }}$

$\tt{\red{\implies {36cm}^{2} = {x}^{2} }}$

$\tt{\red{\implies \sqrt{{36cm}^{2}} = x }}$

$\tt{\red{\implies 6cm = x}}$

$\tt{\green{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: }}$

Height is 6cm

$\tt{\green{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: }}$

$\tt{\blue{ Area \:of\:triangel = \dfrac{ base \times height}{2}}}$

$\tt{\blue{\implies Area \:of\:triangel = \dfrac{ 6cm \times 8cm}{2}}}$

$\tt{\blue{\implies Area \:of\:triangel = \dfrac{ {48cm}^{2}}{2}}}$

$\tt{\blue{\implies Area \:of\:triangel = \dfrac{ \cancel{{48cm}^{2}}}{\cancel{2}}}}$

$\tt{\blue{\implies Area \:of\:triangel = {24cm}^{2}}}$

$\tt{\green{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: }}$

Area is 24cm²

Answer 2

ANSWER:

Given

• Height = 8cm
• Hypotenuse = 10cm

TO FIND:

• Area of triangle
• Base

SOLUTION:

Using Pythagoras theorem

Pythagoras theorem : (Hypotenuse)² = (Base)² + (Height)²

Substituting the values of base and height

Let base be x

10² = x² + 8²

→ 100 - 64 = x²

→ x = √36

→ x = 6

$\small \purple{\therefore} \rm{\red{Base }\green{=}} \pink{6 \: cm}$

_______________________________

Area of triangle = $\rm{\bold{ \frac{1}{2} \times Base \times Height}}$

Substituting the values of base & height

$\to \dfrac{1}{ \cancel2} \times 6 \times \cancel 8 \\ \to6 \times 4 \\ \to24$

Hence area of triangle = 24cm²

$\large\bf{\boxed{ Base = 6cm \: , Area = 24cm^2 }}$