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Question

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[a] $\sqrt{204}$ cm

[b] 204 cm

[c] $2\sqrt{204}$ cm

[d] 408 cm

Answer
Verified

Hint: Use the property that the diagonals of a rhombus bisect each other at right angles. Hence, use Pythagoras theorem to find the length of the diagonal. Alternatively, find the area of rhombus using heronâ€™s formula in triangle ECF and using the property that a diagonal of a parallelogram divides the parallelogram into two congruent triangles and find the area again using the area of a rhombus $=\dfrac{1}{2}{{d}_{1}}{{d}_{2}}$. Hence compare the two areas and from a linear equation in ${{d}_{2}}$. Solve for ${{d}_{2}}$. The value of ${{d}_{2}}$ gives the length of the second diagonal.

Complete step-by-step answer:

We know that the diagonal of a rhombus bisect each other at right angles. Hence we have an angle EGF is a right angle.

Now, given that EC = 28 cm.

Since G is the midpoint of EC, we have EG = 14 cm.

Also EF = 20cm.

We know that in a right-angled triangle, the sum of the squares of legs is equal to the square of the hypotenuse {Pythagoras theorem}.

Using Pythagoras theorem in triangle EGF, we have

$\text{E}{{\text{G}}^{2}}+\text{G}{{\text{F}}^{2}}=\text{E}{{\text{F}}^{2}}$

Substituting EG = 14 and EF = 20, we get

$\begin{align}

& {{14}^{2}}+\text{G}{{\text{F}}^{2}}={{20}^{2}} \\

& \Rightarrow 196+\text{G}{{\text{F}}^{2}}=400 \\

\end{align}$

Subtracting 196 from both sides, we get

$\text{G}{{\text{F}}^{2}}=400-196=204$

Taking square root on both sides, we get

$\text{GF = }\sqrt{204}$ cm

Since G is the midpoint of AF, we have

$\text{AF=2GF=2}\sqrt{204}$ cm.

Hence option [c] is correct.

Note: Alternative Solution:

We have in triangle ECF, EC = 28cm, EF = 20cm and CF = 20cm.

Hence we have a =28, b = 20 and c = 20.

Hence, we have s $=\dfrac{a+b+c}{2}=\dfrac{28+20+20}{2}=34$

We know from heronâ€™s formula that area of a triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Using heron's formula for a = 28, b= 20, c= 20 and s =34

We have the area of triangle ECF $=\sqrt{34\left( 34-28 \right)\left( 34-20 \right)\left( 34-20 \right)}=28\sqrt{51}$.

Since the diagonal of a parallelogram divides the parallelogram into two congruent triangles, we have

$ar\left( \Delta \text{ECF} \right)=ar\left( \Delta \text{ECA} \right)$

Hence area of the rhombus ACFE $=ar\left( \Delta \text{ECF} \right)+ar\left( \Delta \text{ECA} \right)=28\sqrt{51}+28\sqrt{51}=56\sqrt{51}$

Also area of a rhombus $=\dfrac{1}{2}{{d}_{1}}{{d}_{2}}$

We have ${{d}_{1}}=28.$

Hence, we have

$\begin{align}

& \dfrac{1}{2}\left( 28 \right){{d}_{2}}=56\sqrt{51} \\

& \Rightarrow {{d}_{2}}=4\sqrt{51}=2\sqrt{204} \\

\end{align}$

Hence the length of the other diagonal is $2\sqrt{204}$

Complete step-by-step answer:

We know that the diagonal of a rhombus bisect each other at right angles. Hence we have an angle EGF is a right angle.

Now, given that EC = 28 cm.

Since G is the midpoint of EC, we have EG = 14 cm.

Also EF = 20cm.

We know that in a right-angled triangle, the sum of the squares of legs is equal to the square of the hypotenuse {Pythagoras theorem}.

Using Pythagoras theorem in triangle EGF, we have

$\text{E}{{\text{G}}^{2}}+\text{G}{{\text{F}}^{2}}=\text{E}{{\text{F}}^{2}}$

Substituting EG = 14 and EF = 20, we get

$\begin{align}

& {{14}^{2}}+\text{G}{{\text{F}}^{2}}={{20}^{2}} \\

& \Rightarrow 196+\text{G}{{\text{F}}^{2}}=400 \\

\end{align}$

Subtracting 196 from both sides, we get

$\text{G}{{\text{F}}^{2}}=400-196=204$

Taking square root on both sides, we get

$\text{GF = }\sqrt{204}$ cm

Since G is the midpoint of AF, we have

$\text{AF=2GF=2}\sqrt{204}$ cm.

Hence option [c] is correct.

Note: Alternative Solution:

We have in triangle ECF, EC = 28cm, EF = 20cm and CF = 20cm.

Hence we have a =28, b = 20 and c = 20.

Hence, we have s $=\dfrac{a+b+c}{2}=\dfrac{28+20+20}{2}=34$

We know from heronâ€™s formula that area of a triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Using heron's formula for a = 28, b= 20, c= 20 and s =34

We have the area of triangle ECF $=\sqrt{34\left( 34-28 \right)\left( 34-20 \right)\left( 34-20 \right)}=28\sqrt{51}$.

Since the diagonal of a parallelogram divides the parallelogram into two congruent triangles, we have

$ar\left( \Delta \text{ECF} \right)=ar\left( \Delta \text{ECA} \right)$

Hence area of the rhombus ACFE $=ar\left( \Delta \text{ECF} \right)+ar\left( \Delta \text{ECA} \right)=28\sqrt{51}+28\sqrt{51}=56\sqrt{51}$

Also area of a rhombus $=\dfrac{1}{2}{{d}_{1}}{{d}_{2}}$

We have ${{d}_{1}}=28.$

Hence, we have

$\begin{align}

& \dfrac{1}{2}\left( 28 \right){{d}_{2}}=56\sqrt{51} \\

& \Rightarrow {{d}_{2}}=4\sqrt{51}=2\sqrt{204} \\

\end{align}$

Hence the length of the other diagonal is $2\sqrt{204}$