Question

Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,3)\qbezier(0,0)(0,0)(6,0)\qbezier(0,0)(0,0)(6,2)\qbezier(6,2)(6,2)(6,0)\qbezier(6,0)(6,0)(1,3)\qbezier(3.84,1.3)(3.8,0)(3.8,0)\put(-0.2,-0.2){\sf B}\put(0.5,3){\sf A}\put(6.2,-0.2){\sf F}\put(3.8,-0.4){\sf D}\put(3.8,1.6){\sf C}\put(6.2,2.2){\sf E}\end{picture}$

in the diagram $\sf \overline{AB} || \overline{CD}||\overline{EF}$

EF=8cm
BD=6cm
DF=4cm

Find AB=?



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Go to brainly.in and see the diagram .

Answer only if you know

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Answer 1

$\Large{\underbrace{\underline{\sf{Understanding\; the\; Question}}}}$

Here in this question, concept of similarity of triangle as well as theorem of similarity of triangle is used. We are given a figure and we have to find a particular side. We can solve this question by using similarity criteria for triangles.

So let's do it!!

$\rule{290}{1.5}$

Consider, in ∆ABF and ∆CDF

∠DFC=∠DFC (Common in both triangles)

∠CDF= ∠ABF (Pair of corresponding angles)

$\sf So,\triangle ABF\sim\triangle CDF \:(By\;AA)$

Note→ A.A. is a criteria for similarity of triangle. We have several criteria for proving 2 triangles similar.

Now using theorem:

Theorem 1: The ratio of sides of similar triangle is always equal.

As ∆ABF  and ∆CDF are similar:

$\sf \longmapsto\dfrac{AB}{CD}=\dfrac{BF}{DF}=\dfrac{AF}{CF}$

So,

$\sf\longmapsto \dfrac{AB}{CD}=\dfrac{BF}{DF}$

Now putting given values:

$\sf\longmapsto \dfrac{AB}{CD}=\dfrac{BD+DF}{4cm}$

$\sf\longmapsto \dfrac{AB}{CD}=\dfrac{6cm+4cm}{4cm}$

$\sf\longmapsto\dfrac{AB}{CD}=\dfrac{10cm}{4cm}---(1.)$

$\sf \longmapsto AB=\dfrac{10cm\times CD}{4cm}---(2.)$

$\rule{290}{1.5}$

Now consider triangles, In ∆EFB and ∆CDB

∠CBD=∠CBD (Common in both triangles)

∠CDB=∠CFB (Pair of corresponding angles)

$\sf So, \triangle EFB \sim \triangle EDB\:(By\:A.A.)$

According to theorem 1:

$\sf\longmapsto\displaystyle\frac{EF}{CD} =\frac{FB}{DB} =\frac{EB}{CB}$

Also,

$\sf \longmapsto\displaystyle\frac{EF}{CD} =\frac{FB}{DB}$

By putting given values:

$\sf \longmapsto\displaystyle\frac{8cm}{CD} =\frac{BD+DF}{6cm}$

$\sf \longmapsto\displaystyle\frac{8cm}{CD} =\frac{6cm+4cm}{6cm}$

$\sf\longmapsto\displaystyle\frac{8cm}{CD} =\frac{10cm}{6cm}---(3.)$

$\rule{290}{1.5}$

Now add equation (1) and (3)

$\sf \longmapsto\dfrac{AB}{CD}+\dfrac{8cm}{CD}=\dfrac{10cm}{4cm}+\dfrac{10cm}{6cm}$

Now solve it!

$\sf\longmapsto \dfrac{AB+8cm}{CD}=\dfrac{30cm+20cm}{12cm}$

$\sf \longmapsto\dfrac{AB+8cm}{CD}=\dfrac{50cm}{12cm}$

By transporting 8cm from LHS to RHS

$\sf\longmapsto AB+8cm=\dfrac{50cm\times CD}{12cm}$

Now put value of AB from equation (2.)

$\sf\longmapsto\bigg[ \dfrac{10\times CD}{4cm}\bigg]+8cm=\dfrac{50cm\times CD}{12cm}$

$\sf \longmapsto\dfrac{ 10CD}{4cm}+8cm=\dfrac{50cm\times CD}{12cm}$

$\sf \longmapsto 8cm=\dfrac{50CD}{12cm}-\dfrac{10CD}{4cm}$

$\sf\longmapsto 8cm=\dfrac{50CD-30CD}{12cm}$

$\sf\longmapsto 8cm=\dfrac{20CD}{12cm}$

$\sf\longmapsto 8cm\times\dfrac{12cm}{20cm}=CD$

$\sf\longmapsto \dfrac{24cm}{5cm}=CD$

Put this value in equation (1)

$\rule{290}{1.5}$

$\sf\longmapsto\dfrac{AB}{CD}=\dfrac{10cm}{4cm}$

$\sf\longmapsto\dfrac{AB}{\dfrac{24cm}{5cm}}=\dfrac{10cm}{4cm}$

$\sf\longmapsto\dfrac{AB\times 5cm}{24cm}=\dfrac{10cm}{4cm}$

$\sf\longmapsto AB=\dfrac{10cm}{4cm}\times\dfrac{24cm}{5cm}$

Now, by solving it we get:

$\sf\longmapsto AB=12cm$

So the value of AB is 12cm.

$\rule{290}{1.5}$

Answer 2

Answer:

Diagram:-

\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,3)\qbezier(0,0)(0,0)(6,0)\qbezier(0,0)(0,0)(6,2)\qbezier(6,2)(6,2)(6,0)\qbezier(6,0)(6,0)(1,3)\qbezier(3.84,1.3)(3.8,0)(3.8,0)\put(-0.2,-0.2){\sf B}\put(0.5,3){\sf A}\put(6.2,-0.2){\sf F}\put(3.8,-0.4){\sf D}\put(3.8,1.6){\sf C}\put(6.2,2.2){\sf E}\end{picture}

in the diagram \sf \overline{AB} || \overline{CD}||\overline{EF}

AB

∣∣

CD

∣∣

EF

EF=8cm

BD=6cm

DF=4cm

Find AB=?

\begin{gathered}\\ \end{gathered}

Note:-

Go to brainly.in and see the diagram .

Answer only if you know

spam=strict actions will be taken.

Step-by-step explanation:

\Large{\underbrace{\underline{\sf{Understanding\; the\; Question}}}}

UnderstandingtheQuestion

Here in this question, concept of similarity of triangle as well as theorem of similarity of triangle is used. We are given a figure and we have to find a particular side. We can solve this question by using similarity criteria for triangles.

So let's do it!!

\rule{290}{1.5}

Consider, in ∆ABF and ∆CDF

∠DFC=∠DFC (Common in both triangles)

∠CDF= ∠ABF (Pair of corresponding angles)

\sf So,\triangle ABF\sim\triangle CDF \:(By\;AA)So,△ABF∼△CDF(ByAA)

Note→ A.A. is a criteria for similarity of triangle. We have several criteria for proving 2 triangles similar.

Now using theorem:

Theorem 1: The ratio of sides of similar triangle is always equal.

As ∆ABF and ∆CDF are similar:

\sf \longmapsto\dfrac{AB}{CD}=\dfrac{BF}{DF}=\dfrac{AF}{CF}⟼

CD

AB

=

DF

BF

=

CF

AF

So,

\sf\longmapsto \dfrac{AB}{CD}=\dfrac{BF}{DF}⟼

CD

AB

=

DF

BF

Now putting given values:

\sf\longmapsto \dfrac{AB}{CD}=\dfrac{BD+DF}{4cm}⟼

CD

AB

=

4cm

BD+DF

\sf\longmapsto \dfrac{AB}{CD}=\dfrac{6cm+4cm}{4cm}⟼

CD

AB

=

4cm

6cm+4cm

\sf\longmapsto\dfrac{AB}{CD}=\dfrac{10cm}{4cm}---(1.)⟼

CD

AB

=

4cm

10cm

−−−(1.)

\sf \longmapsto AB=\dfrac{10cm\times CD}{4cm}---(2.)⟼AB=

4cm

10cm×CD

−−−(2.)

\rule{290}{1.5}

Now consider triangles, In ∆EFB and ∆CDB

∠CBD=∠CBD (Common in both triangles)

∠CDB=∠CFB (Pair of corresponding angles)

\sf So, \triangle EFB \sim \triangle EDB\:(By\:A.A.)So,△EFB∼△EDB(ByA.A.)

According to theorem 1:

\sf\longmapsto\displaystyle\frac{EF}{CD} =\frac{FB}{DB} =\frac{EB}{CB}⟼

CD

EF

=

DB

FB

=

CB

EB

Also,

\sf \longmapsto\displaystyle\frac{EF}{CD} =\frac{FB}{DB}⟼

CD

EF

=

DB

FB

By putting given values:

\sf \longmapsto\displaystyle\frac{8cm}{CD} =\frac{BD+DF}{6cm}⟼

CD

8cm

=

6cm

BD+DF

\sf \longmapsto\displaystyle\frac{8cm}{CD} =\frac{6cm+4cm}{6cm}⟼

CD

8cm

=

6cm

6cm+4cm

\sf\longmapsto\displaystyle\frac{8cm}{CD} =\frac{10cm}{6cm}---(3.)⟼

CD

8cm

=

6cm

10cm

−−−(3.)

\rule{290}{1.5}

Now add equation (1) and (3)

\sf \longmapsto\dfrac{AB}{CD}+\dfrac{8cm}{CD}=\dfrac{10cm}{4cm}+\dfrac{10cm}{6cm}⟼

CD

AB

+

CD

8cm

=

4cm

10cm

+

6cm

10cm

Now solve it!

\sf\longmapsto \dfrac{AB+8cm}{CD}=\dfrac{30cm+20cm}{12cm}⟼

CD

AB+8cm

=

12cm

30cm+20cm

\sf \longmapsto\dfrac{AB+8cm}{CD}=\dfrac{50cm}{12cm}⟼

CD

AB+8cm

=

12cm

50cm

By transporting 8cm from LHS to RHS

\sf\longmapsto AB+8cm=\dfrac{50cm\times CD}{12cm}⟼AB+8cm=

12cm

50cm×CD

Now put value of AB from equation (2.)

\sf\longmapsto\bigg[ \dfrac{10\times CD}{4cm}\bigg]+8cm=\dfrac{50cm\times CD}{12cm}⟼[

4cm

10×CD

]+8cm=

12cm

50cm×CD

\sf \longmapsto\dfrac{ 10CD}{4cm}+8cm=\dfrac{50cm\times CD}{12cm}⟼

4cm

10CD

+8cm=

12cm

50cm×CD

\sf \longmapsto 8cm=\dfrac{50CD}{12cm}-\dfrac{10CD}{4cm}⟼8cm=

12cm

50CD

−

4cm

10CD

\sf\longmapsto 8cm=\dfrac{50CD-30CD}{12cm}⟼8cm=

12cm

50CD−30CD

\sf\longmapsto 8cm=\dfrac{20CD}{12cm}⟼8cm=

12cm

20CD

\sf\longmapsto 8cm\times\dfrac{12cm}{20cm}=CD⟼8cm×

20cm

12cm

=CD

\sf\longmapsto \dfrac{24cm}{5cm}=CD⟼

5cm

24cm

=CD

Put this value in equation (1)

\rule{290}{1.5}

\sf\longmapsto\dfrac{AB}{CD}=\dfrac{10cm}{4cm}⟼

CD

AB

=

4cm

10cm

\sf\longmapsto\dfrac{AB}{\dfrac{24cm}{5cm}}=\dfrac{10cm}{4cm}⟼

5cm

24cm

AB

=

4cm

10cm

\sf\longmapsto\dfrac{AB\times 5cm}{24cm}=\dfrac{10cm}{4cm}⟼

24cm

AB×5cm

=

4cm

10cm

\sf\longmapsto AB=\dfrac{10cm}{4cm}\times\dfrac{24cm}{5cm}⟼AB=

4cm

10cm

×

5cm

24cm

Now, by solving it we get:

\sf\longmapsto AB=12cm⟼AB=12cm

So the value of AB is 12cm.

\rule{290}{1.5}