If the base of a right angled triangle is 8 cm and the hypotenuse is 17 cm, find its area. About the author Rose
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 Reply
Question : If the base of a right angled triangle is 8 cm and the hypotenuse is 17 cm, find its area. ⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀ ⠀ ⋆ DIAGRAM : [tex]\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}[/tex] ⠀ [tex]\underline{\bigstar\:\boldsymbol{By\: Using\; Pythagoras\: Theorem :}}[/tex] ⠀⠀⠀⠀ [tex]\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}[/tex] ⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀ ◗ To find out the Area of a right angle triangle formula is given by : ⠀⠀ [tex]\star\;\boxed{\sf{\pink{Area_{\:(triangle)} = \dfrac{1}{2} \times Base \times Height}}}[/tex] [tex]\underline{\bf{\dag} \:\mathfrak{Putting\;values\: :}}[/tex] ⠀⠀⠀⠀ [tex]\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}[/tex] ⠀⠀⠀⠀ [tex]\therefore{\underline{\sf{Hence,\;area\;of\;right\;angle\; \triangle\;is\;\bf{ 60\;cm^2}.}}}[/tex] Reply
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
Question :
If the base of a right angled triangle is 8 cm and the hypotenuse is 17 cm, find its area.
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⋆ DIAGRAM :
[tex]\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}[/tex]
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[tex]\underline{\bigstar\:\boldsymbol{By\: Using\; Pythagoras\: Theorem :}}[/tex]
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[tex]\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}[/tex]
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◗ To find out the Area of a right angle triangle formula is given by :
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[tex]\star\;\boxed{\sf{\pink{Area_{\:(triangle)} = \dfrac{1}{2} \times Base \times Height}}}[/tex]
[tex]\underline{\bf{\dag} \:\mathfrak{Putting\;values\: :}}[/tex]
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[tex]\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}[/tex]
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[tex]\therefore{\underline{\sf{Hence,\;area\;of\;right\;angle\; \triangle\;is\;\bf{ 60\;cm^2}.}}}[/tex]