find the area of a right triangle whose hypotenuse measure 17 cm and o

1.	find the area of a right triangle whose hypotenuse measure 17 cm and one of the other two sides is 8 cm
Answer» Answer: As this is RIGHT Angle Triangle, so Either we can USE Pythagoras Theorem or PYTHAGOREAN Triplet to Find the THIRD Side. $\begin{tabular}{\|c \|c \| c\|}\cline{1-3}p/b & b/p & h \\\cline{1-3}3 & 4 & 5 \\5 & 12 &13\\8 & 15&17 \\7 & 24&25\\\cline{1-3}\end{tabular}$ So we can SEE that [8 , 15 , 17] is the Triplet, therefore Other Side will be 15 cm. ⋆ DIAGRAM : $\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\thicklines\put(7.7,2.9){\large{A}}\put(7.7,1){\large{B}}\put(10.6,1){\large{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.2,2){\sf{\large{15 cm}}}\put(9,0.7){\sf{\large{8 cm}}}\put(9.4,1.9){\sf{\large{17 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$ $\rule{160}{1}$ $\underline{\bigstar\:\:\textsf{Area of Right Angle Triangle ABC :}}$ $\dashrightarrow\tt\:\:Area=\dfrac{1}{2} \times Base \times Height\\\\\\\dashrightarrow\tt\:\:Area=\dfrac{1}{2} \times BC \times AB\\\\\\\dashrightarrow\tt\:\:Area=\dfrac{1}{2} \times 8\:cm \times 15\:cm\\\\\\\dashrightarrow\tt\:\:Area=4\:cm \times 15\:cm\\\\\\\dashrightarrow\:\:\underline{\boxed{\textsf{\textbf{Area = 60 cm$^{\textbf2}$}}}}$

find the area of a right triangle whose hypotenuse measure 17 cm and one of the other two sides is 8 cm

Answer»

Answer:

As this is RIGHT Angle Triangle, so Either we can USE Pythagoras Theorem or PYTHAGOREAN Triplet to Find the THIRD Side.

$\begin{tabular}{|c |c | c|}\cline{1-3}p/b & b/p & h \\\cline{1-3}3 & 4 & 5 \\5 & 12 &13\\8 & 15&17 \\7 & 24&25\\\cline{1-3}\end{tabular}$

So we can SEE that [8 , 15 , 17] is the Triplet, therefore Other Side will be 15 cm.

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\thicklines\put(7.7,2.9){\large{A}}\put(7.7,1){\large{B}}\put(10.6,1){\large{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.2,2){\sf{\large{15 cm}}}\put(9,0.7){\sf{\large{8 cm}}}\put(9.4,1.9){\sf{\large{17 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

$\rule{160}{1}$

$\underline{\bigstar\:\:\textsf{Area of Right Angle Triangle ABC :}}$

$\dashrightarrow\tt\:\:Area=\dfrac{1}{2} \times Base \times Height\\\\\\\dashrightarrow\tt\:\:Area=\dfrac{1}{2} \times BC \times AB\\\\\\\dashrightarrow\tt\:\:Area=\dfrac{1}{2} \times 8\:cm \times 15\:cm\\\\\\\dashrightarrow\tt\:\:Area=4\:cm \times 15\:cm\\\\\\\dashrightarrow\:\:\underline{\boxed{\textsf{\textbf{Area = 60 cm$^{\textbf2}$}}}}$