Question

45.
In the figure given, ABCD and DEFC are two squares.
There are four identical unshaded triangles. If the area
of each of these unshaded triangle is 16 cm?, then find
the area of shaded portion.
16 cm
E
А
4 cm
cm-4'cm
F
→C
B4
8 cm
-12 cm
48 cm​

Answer 1

Step-by-step explanation:

ABCD and AEFG are squares

By definition of square

AB=BC=CD=AD,AE=EF=FG=AG

Angle A=Angle B=Angle C=Angle D=90 degrees

Angle AGF=Angle GFE=Angle AEF=Angle GAE=90 degrees

To prove that

(i)\frac{AF}{AG}=\frac{AC}{AD}

AG

AF

=

AD

AC

(ii)\triangle ACF△ACF \sim \triangle ADG∼△ADG

Proof:

Construction: Draw AC

(i)AC and AF are diagonals of square ABCD and AEFG.

We know that

Diagonal of square=\sqrt{2}\times

2

× side of square

Using the formula

AF=\sqrt{2}AGAF=

2

AG

AC=\sqrt{2}ADAC=

2

AD

\frac{AF}{AC}=\frac{\sqrt{2}AG}{\sqrt{2}AD}

AC

AF

=

2

AD

2

AG

\frac{AF}{AC}=\frac{AG}{AD}

AC

AF

=

AD

AG

\frac{AF}{AG}=\frac{AC}{AD}

AG

AF

=

AD

AC

Hence, proved.

(ii)Angle AGF=90 degree

AG and GF are equal sides

Therefore, Angle GAF=Angle GFA=45 degrees

Because sum of angles of triangle =180 degrees

Diagonal AC divides the angle in to two equal parts

Therefore, Angle CAD=Angle CAB=45 degrees

Let

Angle FAC=x

Angle GAC=45-x

Angle GAD=Angle CAD-Angle CAG=45-(45-x)=45-45+x=x

Angle FAC=Angle GAD=x

\frac{AF}{AG}=\frac{AC}{AD}

AG

AF

=

AD

AC

\triangle ACF\sim \triangle ADG△ACF∼△ADG

By SAS similarity postulate

Hence, proved.