Question

t ule JoLOLULZg:
1. The length of a rectangle is 16 cm and the length of its diagonal is 20 cm. The area of the
rectangle is
(a) 320 cm (b) 160 cm
(c) 192 cm
(d) 156 cm​

Answer 1

Length of the given rectangle = 16 cm

Length of its diagonal = 20 cm

Therefore, it's breadth = $\sqrt{ {20}^{2} - {16}^{2} }</strong><strong> </strong><strong>$ cm = 12 cm

Therefore, it's area = (16×12) cm² = 192 cm²

Answer 2

$\huge\sf\pink{Answer}$

Area = 192 cm²

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ Diagonal of a rectangle(AC) = 20 cm

✭ Length of the rectangle(AB) = 16 cm

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ Area of the rectangle?

$\rule{110}1$

$\huge\sf\purple{Steps}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{D}}\put(9.6,1.7){\sf{\large{20 m}}}\put(7.7,1){\large{A}}\put(9.1,0.7){\sf{\large{10 m}}}\put(11.1,1){\large{B}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(8,1){\line(3,2){3}}\put(11.1,3){\large{C}}\put(10.8,1){\line(0,2){0.2}}\put(10.8,1.2){\line(2,0){0.2}}\end{picture}$

Finding base(BC) of the rectangle

Using, Pythagoras Theorem

$\underline{\boxed{\sf (Hypotenuse)^2 = (base)^2 + (Perpendicular)^2 }}$

On Substituting the given values,

➝ $\sf{(20)^2 = (b)^2 + (16)^2}$

➝ $\sf{ 400 = b^2 + 256 }$

➝ $\sf{ 400 - 256 = b^2}$

➝ $\sf{ 144 = b^2 }$

➝ $\sf{ \sqrt{144} = b }$

➝ $\sf{\red{ 12 \: cm = b }}$

Area of a rectangle is given by,

$\underline{\boxed{\sf Area = Length \times Breadth }}$

Substituting the values,

≫ $\sf{ Area = 16 \times 12 }$

≫ $\sf{\orange{ Area = 192 \: cm^2}}$

$\rule{170}3$