A Rhombus whose one side is 10 cm has one diagonal length is 12 cm. Fi

1.	A Rhombus whose one side is 10 cm has one diagonal length is 12 cm. Find the measurement of other diagonals.
Answer» » To Find : The length of other diagonal . » We Know : Pythagoras theorem : $\sf{\underline{\boxed{h^{2} = p^{2} + b^{2}}}}$ Where , h = hypotenuse b = base p = height » Concept : We Know ,that the in rhombus the diagonals intersect each other equally and at 90°. So the length AO will be Equal to length OC. Hence , we get the length as : $\sf{\underline{\boxed{AO = OC = 6 cm}}}$ And the given length is 10 cm . According to the Diagram , if we look at the FIGURE ADO ,we get a right-angled triangle having 90° at point O . So by this information ,we can find the length OD , which is the CORRESPONDING height ,by using the Pythagoras theorem . » Solution : Given Information : Hypotenuse = 10 cm Base = 6 cm Taken : Let the Corresponding height be x Formula : $\sf{\underline{\boxed{h^{2} = p^{2} + b^{2}}}}$ By substituting the values in it ,we get : $\sf{\Rightarrow 10^{2} = x^{2} + 6^{2}}$ Now, Subtracting 6² from both the sides ,we get : $\sf{\Rightarrow 10^{2} - 6^{2} = x^{2} + 6^{2} - 6^{2}}$ $\sf{\Rightarrow 10^{2} - 6^{2} = x^{2}}$ By Square rooting on both the sides ,we get : $\sf{\Rightarrow \sqrt{10^{2} - 6^{2}} = \sqrt{x^{2}}}$ $\sf{\Rightarrow \sqrt{10^{2} - 6^{2}} = x^{2}}$ $\sf{\Rightarrow \sqrt{100 - 36} = x}$ $\sf{\Rightarrow \sqrt{64} = x}$ $\sf{\Rightarrow 8 cm = x}$ Hence , the Corresponding height is 8 cm . We Know, that the Diagonals intersect each other equally. And the Corresponding height is the length DO , which is equal to OB ,so the other diagonal is twice of DO, i.e. $\sf{\Rightarrow D_{2} = 2 \times DO}$ Substituting the value of DO ,in the above expression ,we get : $\sf{\Rightarrow D_{2} = 2 \times 8}$ $\sf{\Rightarrow D_{2} = 16 cm}$ Hence , the other diagonal is 16 cm. » ADDITIONAL information : Volume of a Cylinder = πr²h Surface AREA of a Cylinder = 2πr(h + r) Curved-surface area of a Cylinder = 2πrh Area of triangle = ½ × base × height

A Rhombus whose one side is 10 cm has one diagonal length is 12 cm. Find the measurement of other diagonals.

Answer»

» To Find :

The length of other diagonal .

» We Know :

Pythagoras theorem :

$\sf{\underline{\boxed{h^{2} = p^{2} + b^{2}}}}$

Where ,

h = hypotenuse
b = base
p = height

» Concept :

We Know ,that the in rhombus the diagonals intersect each other equally and at 90°.

So the length AO will be Equal to length OC.

Hence , we get the length as :

$\sf{\underline{\boxed{AO = OC = 6 cm}}}$

And the given length is 10 cm .

According to the Diagram , if we look at the FIGURE ADO ,we get a right-angled triangle having 90° at point O .

So by this information ,we can find the length OD , which is the CORRESPONDING height ,by using the Pythagoras theorem .

» Solution :

Given Information :

Hypotenuse = 10 cm

Base = 6 cm

Taken :

Let the Corresponding height be x

Formula :

$\sf{\underline{\boxed{h^{2} = p^{2} + b^{2}}}}$

By substituting the values in it ,we get :

$\sf{\Rightarrow 10^{2} = x^{2} + 6^{2}}$

Now, Subtracting 6² from both the sides ,we get :

$\sf{\Rightarrow 10^{2} - 6^{2} = x^{2} + 6^{2} - 6^{2}}$

$\sf{\Rightarrow 10^{2} - 6^{2} = x^{2}}$

By Square rooting on both the sides ,we get :

$\sf{\Rightarrow \sqrt{10^{2} - 6^{2}} = \sqrt{x^{2}}}$

$\sf{\Rightarrow \sqrt{10^{2} - 6^{2}} = x^{2}}$

$\sf{\Rightarrow \sqrt{100 - 36} = x}$

$\sf{\Rightarrow \sqrt{64} = x}$

$\sf{\Rightarrow 8 cm = x}$

Hence , the Corresponding height is 8 cm .

We Know, that the Diagonals intersect each other equally.

And the Corresponding height is the length DO , which is equal to OB ,so the other diagonal is twice of DO, i.e.

$\sf{\Rightarrow D_{2} = 2 \times DO}$

Substituting the value of DO ,in the above expression ,we get :

$\sf{\Rightarrow D_{2} = 2 \times 8}$

$\sf{\Rightarrow D_{2} = 16 cm}$

Hence , the other diagonal is 16 cm.

» ADDITIONAL information :

Volume of a Cylinder = πr²h

Surface AREA of a Cylinder = 2πr(h + r)

Curved-surface area of a Cylinder = 2πrh

Area of triangle = ½ × base × height