Question

Ir the base of a right-angled triangle is 6 cm and the hypotenuse is 10 cm, find its area​

Answer 1

Answer:

refer to the attachment

Answer 2

$\huge\pink{Answer:}$

$\sf \ Area \ of \ the \ triangle \ is \ 10 \ cm.$

$\huge\pink{Explanation:}$

$\sf Let \ ABC \ be \ the \ right \ angled \ \triangle.$

$\sf Let \ base \ (AB) = 6 \ cm \ (given)$

$\sf and \ hypotenuse \ (CB) \ = 10 \ cm \ (given)$

$\sf By \ Pythagoras \ theorem \ in \ triangle \ ABC$

$\sf ⇒ (CB)² = (AC)² + (AB)²$

$\sf ⇒ (10)² = (AC)² + 6²$

$\sf ⇒ (AC)² = (10)² \ – \ (6)²$

$\sf ⇒ (AC)² = 64$

$\sf ⇒ AC = \sqrt{64}$

$\sf So, hypotenuse = 8 cm$

$\sf Now \ area \ of \ the \ \ triangle$

$\sf = 1/2 × b × h$

$\sf = 1/2 × 6 × 8$

$\sf = 24 \ cm².$