Question

Given two sticks AB and CD of length AB = 16 cm and CD = 12 cm. Using these sticks, make a kite. Find the perimeter of the kite whose shape is a rhombus.

Answer 1

Hint: To solve the above question, we will take the sticks as the diagonals of the given kite. Then we will use the property that if a kite is a rhombus, then the diagonals will bisect each other and the angle between these diagonals will be \[{{90}^{o}}.\] Then we will consider one out of the four triangles which are formed due to construction of diagonals. We will then apply Pythagoras theorem in this triangle to get one side of the rhombus. Then to find the perimeter of the kite, we will multiply the side length into four.

Complete step by step solution:
We are given that we have to make a kite from two sticks whose shape is a rhombus. Now, we will take these sticks as diagonals of the rhombus. Before proceeding further, we need to know what a rhombus is and some of its properties. A rhombus is a special type of parallelogram whose opposite sides are parallel and all the sides are equal in length. One property of the diagonals of a rhombus is that the diagonals of a rhombus bisect each other and the angle between these diagonals is \[{{90}^{o}}.\] A rough sketch of the kite is given to understand things better.

Now, AB and CD are diagonals which intersect each other at O. Now, the diagonals AB and CD bisect each other. So, we have,
\[AO=BO=\dfrac{16cm}{2}=8cm\]
\[CO=DO=\dfrac{12cm}{2}=6cm\]
Now, AOD is a right-angled triangle, so we will apply the Pythagoras theorem to find the length of side AD. The Pythagoras theorem states that if H is the hypotenuse, B is the base and P is the perpendicular, then,
\[{{H}^{2}}={{P}^{2}}+{{B}^{2}}\]
In our case, H = AD, P = AO, B = OD. So, we have,
\[{{\left( AD \right)}^{2}}={{\left( AO \right)}^{2}}+{{\left( OD \right)}^{2}}\]
\[\Rightarrow {{\left( AD \right)}^{2}}={{\left( 8cm \right)}^{2}}+{{\left( 6cm \right)}^{2}}\]
\[\Rightarrow {{\left( AD \right)}^{2}}=64c{{m}^{2}}+36c{{m}^{2}}\]
\[\Rightarrow {{\left( AD \right)}^{2}}=100c{{m}^{2}}\]
\[\Rightarrow AD=10cm\]
Now, we know that all the sides are equal in length. So,
\[\text{Perimeter }=4\times side\]
\[\Rightarrow \text{Perimeter }=4\times AD\]
\[\Rightarrow \text{Perimeter }=4\times 10cm\]
\[\Rightarrow \text{Perimeter }=40cm\]

Note: The side length of the rhombus can also be calculated in an alternate way. We will consider triangle AOD now.

We know that \[\tan \theta \] is a ratio of the perpendicular and the base. So, we will get,
\[\tan \theta =\dfrac{8}{6}\]
\[\Rightarrow \tan \theta =\dfrac{4}{3}\]
\[\Rightarrow \theta ={{53}^{o}}\]
Now, we know that \[\sin \theta \] is a ratio of the perpendicular and the hypotenuse. So, we will get,
\[\sin \theta =\sin {{53}^{o}}=\dfrac{8}{x}\]
\[\Rightarrow \dfrac{4}{5}=\dfrac{8}{x}\]
\[\Rightarrow x=10cm\]
\[\Rightarrow AD=10cm\]