Question

If the base of a right angled triangle is 8 cm and the hypotenuse is 17 cm, find its area.

​

Answer 1

Question :

If the base of a right angled triangle is 8 cm and the hypotenuse is 17 cm, find its area.

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

⠀

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.1,2){\sf{\large{?cm}}}\put(9,0.7){\sf{\large{8cm}}}\put(9.4,1.9){\sf{\large{17 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

⠀

$\underline{\bigstar\:\boldsymbol{By\: Using\; Pythagoras\: Theorem :}}$

⠀⠀⠀⠀

$\begin{gathered}:\implies\sf (AB)^2 + (BC)^2 = (AC)^2 \\\\\\:\implies\sf (AB)^2 = (AC)^2 - (BC)^2 \\\\\\:\implies\sf (AB)^2 = (17)^2 - (8)^2 \\\\\\:\implies\sf (AB)^2 = 289 - 64 \\\\\\:\implies\sf AB^2 = 225 \\\\\\:\implies\sf AB = \sqrt{225} \\\\\\:\implies{\underline{\boxed{\sf{AB = 15\;cm}}}}\end{gathered}$

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

◗ To find out the Area of a right angle triangle formula is given by :

⠀⠀

$\star\;\boxed{\sf{\pink{Area_{\:(triangle)} = \dfrac{1}{2} \times Base \times Height}}}$

$\underline{\bf{\dag} \:\mathfrak{Putting\;values\: :}}$

⠀⠀⠀⠀

$\begin{gathered}:\implies\sf Area_{\;(triangle)} = \dfrac{1}{\cancel{\;2}} \times \: \cancel{\;8} \; \times 15 \\\\\\:\implies\sf Area_{\;(triangle)} = 4 \times 15 \\\\\\:\implies{\underline{\boxed{\frak{\pink{Area_{\:(triangle)} = 60\;cm^2}}}}}\;\bigstar\end{gathered}$

⠀⠀⠀⠀

$\therefore{\underline{\sf{Hence,\;area\;of\;right\;angle\; \triangle\;is\;\bf{ 60\;cm^2}.}}}$

Answer 2

Here,

h²= p²+b²

(17cm)²= p²+(8cm)²

p²= (17cm)²-(8cm)²

p= √(17+8)cm (17-8)cm

= √25cm×9cm

= √225cm²

=15cm

Area= 1/2 (b×h)

= 1/2 (8cm×15cm)

= 4cm×15cm

= 60cm²