Question

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

Answer 1

Answer:

Solution :-

By using Pythagoras theorem

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

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

Given:-

• hypotenuse of the triangle = 10cm
• base of the triangle = 6cm

To find:-

• Area of this triangle

Formula to be used:-

Pythagoras theorem:-

$\underline{\boxed{\bf H² = B²+P²}}$

here,

• h = hypotenues
• b = base
• p = perpendicular

$\underline{\boxed{\bf area\: of triangle = \dfrac{1}{2}×base × height}}$

Solution:-

Let's find the value of height by Pythagoras theorem:-

$\longrightarrow$ 10² = 6² + P²

$\longrightarrow$ 100 = 36 + P²

$\longrightarrow$ P² = 64

$\longrightarrow$ P = √64

$\longrightarrow$ P = 8cm

Let's find the area of the triangle now,

$\longrightarrow$ Area = $\sf\dfrac{1}{2}$ × 6 × 8

$\longrightarrow$ Area = 3 × 8

$\longrightarrow$ Area = 24cm²