Question

Find the area of a right triangle with hypotenuse 10 cm and base 6cm.​

Answer 1

Solution..

by using Pythagoras theorem

$\tt \pink{ {CB}^{2} = {AB}^{2} + {AC}^{2}}$

AB = 6 cm

CB = 10 cm

AC = ?

$\tt{ ={CB}^{2} = {AB}^{2} + {AC}^{2}}$

$\tt {= {10}^{2} = {6}^{2} + {AC}^{2}}$

$\tt{= 100 = 36 + {AC}^{2} }$

$\tt {= {AC}^{2} = 100 - 36 }$

$\tt {= {AC}^{2} = 64}$

$\tt {= AC = \sqrt{64}}$

$\tt {= AC = 8 }$

Area of triangle =$\tt { \frac{1}{2} × Base × height }$

$\tt {= \frac{1}{2} × 6 × 8}$

$\tt {= \frac {48}{2}}$

$\tt {= {24cm}^{2}}$

Answer 2

Solution..

by using Pythagoras theorem

$\tt \pink{ {CB}^{2} = {AB}^{2} + {AC}^{2}}$

AB = 6 cm

CB = 10 cm

AC = ?

$\tt{ ={CB}^{2} = {AB}^{2} + {AC}^{2}}$

$\tt {= {10}^{2} = {6}^{2} + {AC}^{2}}$

$\tt{= 100 = 36 + {AC}^{2} }$

$\tt {= {AC}^{2} = 100 - 36 }$

$\tt {= {AC}^{2} = 64}$

$\tt {= AC = \sqrt{64}}$

$\tt {= AC = 8 }$

Area of triangle =$\tt { \frac{1}{2} × Base × height }$

$\tt {= \frac{1}{2} × 6 × 8}$

$\tt {= \frac {48}{2}}$

$\tt {= {24cm}^{2}}$