Question

In the given figure (not drawn to scale) \[ABCD\] and \[WXYZ\] both are rhombus, \[2\] semi-circles and \[2\] identical triangles are drawn on the sides of the rhombus \[WXYZ\] which of the following option is correct?
\[A)\] Area of \[\Delta WPZ\] is \[104c{m^2}\]
\[B)\] circumference of the semicircle is \[20cm\]
\[C)\] area of a rhombus \[ABCD\] is \[24c{m^2}\]
\[D)\] none of these

Answer 1

Hint: Identify the shape of the given region. First list out the known values. Then find out the given option to correct. The first option is to find the area of a triangle. To find height and breadth value. And the second option is to find the circumference value of the semi-circle. To find the radius of the semi-circle. In the third option to find the area of the rhombus so find the diagonal value of the rhombus.

Complete step-by-step answer:
In this question first, we clearly understand the given data. They have given rhombus, \[2\] semi-circles, and \[2\] identical triangles. In this sum, we check which option is correct. So we move on to the option to check the statement or given information. If the statement is correct we choose that option else we move on to the next option.
Now first we check the option \[A\]
The given information is an area of \[\Delta WPZ\] is \[104c{m^2}\]
Now \[\Delta WPZ\] first assume that center point \[WZ\]is named \[T\]
So that \[WZ \bot TP\]
\[\angle WTP = {90^ \circ }\]
So \[\Delta WTP\] is a right angle. We know the length of \[TP = 8cm\] but we find to \[TW\] and \[WP\] values
So we used Pythagoras theorem,
\[
{\left( {WP} \right)^2} = {\left( {TP} \right)^2} + {\left( {TW} \right)^2} \\
{\left( {TP} \right)^2} = {\left( {WP} \right)^2} - {\left( {TW} \right)^2} \\
{\left( 8 \right)^2} = {\left( {WP} \right)^2} - {\left( {TW} \right)^2} \\
64 = {\left( {WP} \right)^2} - {\left( {TW} \right)^2} \\
\]
We have \[64\] now we find \[WP\] and \[TW\] value. If this right angle so we easily find the balance two sides which two numbers squared subtract value is equal to \[64\]
\[
{\left( {10} \right)^2} = 100 \\
{\left( 6 \right)^2} = 36 \\
= 100 - 36 \\
= 64 \\
\]
So that
\[
WP = 10cm \\
TW = 6cm \\
\]
We know triangles area formula now we find the area of \[\Delta WPZ\]
\[TW\] is \[6cm\]
So
\[
WZ = 2TW \\
= 2\left( 6 \right) \\
WZ = 12cm \\
\]
Area \[ = \dfrac{1}{2}bh\]
\[
= \dfrac{1}{2}\left( {WZ} \right)\left( {TP} \right) \\
= \dfrac{1}{2}\left( {12} \right)\left( 8 \right) \\
= 48c{m^2} \\
\]
But the option given Area of \[\Delta WPZ\] is \[104c{m^2}\] so this option is wrong.
The next option \[B\] is given that the circumference of the semicircle is \[20cm\]
\[WXYZ\] is a rhombus. One of the properties of a rhombus is "Rhombus has all its sides equal". So the length of \[WZ\] is equal to the length of \[YZ\] we take the\[YZ\] radius semi-circle.
Circumference for semi-circle is \[\pi r + 2r\]
Here \[r = \] length of \[YZ\]
\[ = 12cm\]
Circumference is,
\[
= \pi \left( {12} \right) + 2\left( {12} \right) \\
= 12\left( {\pi + 2} \right)cm \\
\]
So that option \[B\] is also wrong.
Next, we move on to option \[C\]
Given that, the area of a rhombus \[ABCD\] is \[24c{m^2}\]
The area of a rhombus is \[ = \dfrac{{{d_1} \times {d_2}}}{2}\]
Here
\[
AC = OA + OC \\
= 3 + 3 \\
{d_1} = 6cm \\
BD = OD + OB \\
= 4 + 4 \\
{d_2} = 8cm \\
= \dfrac{{6 \times 8}}{2} \\
= \dfrac{{48}}{2} \\
= 24c{m^2} \\
\]
So that option \[C\] is correct. And the option \[D\] is given that none of these but option \[C\] is valid so the option \[D\] is not valid. Finally, the option \[\left( C \right)\] area of a rhombus \[ABCD\] is \[24c{m^2}\] is the correct answer.

Note: Find the length of the sides of the given shape. Use those concepts like parallel lines that have the same length. And the area of the semi-circle is half of its circle. Use the Pythagoras theorem, In a right-angled triangle, the squared value of the hypotenuse side on the right-angled triangle is equal to the sum of squared values of the other two sides of the right-angled triangle. Find the sides of the triangle.