Question

ABCD is a rectangle of dimension 12 cm and ac is a rectangle drawn in such a way that the diagonal AC of the first rectangle is one of its sides and the side opposite to it is touching the first rectangle at D at show in the figure what is the ratio of area of the rectangle ABCD 2 AEFC


A. 3:1
B.2:3
C.1:1
D.5:4​

Answer 1

Therefore, The ratio of Area of rectangle $ABCD$ to $AEFC$ is $1:1$

Step-by-step explanation:

Given,

Rectangle $ABCD$ where $AB=12cm$ and $BC=5cm$

∴ Area of rectangle $ABCD=(AB\times BC)$

$=(12\times 5)$

$=60cm^{2}$

∴ Area of $\triangle ADC=\frac{1}{2}$ Area of Rectangle $ABCD$

⇒Area of $\triangle ADC=\frac{1}{2}\times 60$

∴ Area of $\triangle ADC=30cm^{2}$

And,

$AC=\sqrt{AB^{2}+BC^{2} }$

⇒$AC=\sqrt{12^{2}+5^{2} }$

∴$AC=\sqrt{169}=13cm$

From $\triangle ABC$,

$\frac{1}{2}\times AC\times Height= 30$

⇒$Height=\frac{60}{13}$

∴$Height=AE=4.615cm$

∴Area of Rectangle $AEFC=(AC\times AE)$

$=(13\times 4.615)$

$=59.995$

≅$60cm^{2}$

∴ The ratio of Area of rectangle $ABCD$ to $AEFC$$=\frac{60}{60} =\frac{1}{1}$

∴ The ratio of Area of rectangle $ABCD$ to $AEFC$ is $1:1$