Question

ABC is a triangle, right angled at C. If BC = 24 cm and AC = 7 cm.

Find AB​

Answer 1

GIVEN :

• ABC is right angled at C

$\qquad\quad {:}\longmapsto\sf \angle {C}=90°$

• BC=24 cm
• AC=7 cm

TO FIND :

• AB =?

S O L U T I O N :-

Here , in triangle ABC :

• Perpendicular(P)=AC=7 cm
• Base (B)=BC=24 cm
• We need to find Hypontenuse(H)= AB

According to Pythagorean theorem,

• ${\boxed{\bf H^2=P^2+B^2}}$

$\qquad\quad {:}\longmapsto\sf h=\sqrt {P^2+B^2}$

• Substitute the values :

$\qquad\quad {:}\longmapsto\sf h=\sqrt {(7)^2+(24)^2}$

$\qquad\quad {:}\longmapsto\sf h=\sqrt{49+576}$

$\qquad\quad {:}\longmapsto\sf h=\sqrt {625}$

$\qquad\quad {:}\longmapsto\tt h=25\:cm$

$\therefore{\underline{\boxed{\bf {\overline{AB}=25\:cm.}}}}$

Answer 2

GIVEN :

ABC is right angled at C

$\qquad\quad {:}\longmapsto\sf \angle {C}=90°$

BC=24 cm

AC=7 cm

TO FIND :

AB =?

S O L U T I O N :-

Here , in triangle ABC :

Perpendicular(P)=AC=7 cm

Base (B)=BC=24 cm

We need to find Hypontenuse(H)= AB

According to Pythagorean theorem,

${\boxed{\bf H^2=P^2+B^2}}$

$\qquad\quad {:}\longmapsto\sf h=\sqrt {P^2+B^2}$

Substitute the values :

$\qquad\quad {:}\longmapsto\sf h=\sqrt {(7)^2+(24)^2}$

$\qquad\quad {:}\longmapsto\sf h=\sqrt{49+576}$

$\qquad\quad {:}\longmapsto\sf h=\sqrt {625}$

$\qquad\quad {:}\longmapsto\tt h=25\:cm$

$\therefore{\underline{\boxed{\bf {\overline{AB}=25\:cm.}}}}$