Question

The area of a right angled triangle,whose base is 8cm and hypotenuse is 10 cm,is: (explain in detail)​

Answer 1

Area of triangle = 1/2 × base × height

= 1/2 × 8 × 10

= 4 × 10

= 40 cm sq

Answer 2

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Given:

• Bas.e = 8cm
• Hypotenuse = 10cm

To Find:

• Area of the right angled triangle

Solution:

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(2.8,3.5){\large\bf 10cm}\put(2.8,.3){\large\bf 8cm}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}</p><p>$

$\\\text{We know that, the area of a right angled triangle =>}\\\implies \dfrac{1}{2}*bas.e*height \\\text{But, we dont have the height given. So, we calculate it using Pythagoras Theorem} \\\implies (hypotenuse)^2 = (ba.se)^2 + (height)^2 \\\implies AC^2 = BC^2 + AB^2 \\\implies AB^2 = 10^2 - 8^2 \\\implies AB^2 = 100 - 64 \\\implies AB^2 = 36 \\\implies AB = \sqrt{36} = 6 \:\:\:\:\:(length\: is \:never \:negative). \\\text{So, the height is 6cm}$

$\text{ \implies Area of Triangle ABC = \dfrac{1}{2} * ba.se * height } \\\\\text{ \implies Area of Triangle ABC = \dfrac{1}{2} * 8 * 6 }\\\\\text{ \implies Area of Triangle ABC = 4 * 6} \\\\\bf\text{ \implies Area of Triangle ABC = 48cm^2}}$

Hope it helps!