Question

Find the area of a rhombus whose side is 5 cm and its altitude is 4 cm. If one of its diagonal is 8 cm long, find the length of the other diagonal........ Plz do it on a paper drawing a rhombus and plz label it with altitude, diagonal, sides etc. U will get 35 points. And I will mark u the brainliest

Answer 1

*Reference of rhombus is in the attachment .

AnswEr:-

Area of rhombus = 24 cm²

Other diagonal of rhombus = 6 cm

Step-by-step explanation:

According to attachment,the diagonal divided the rhombus into two same triangles.

Here,the same triangles are ∆ABD & ∆BDC

These two triangles are isosceles triangles because both the triangles have same two sides i.e. AB = AD = 5 cm & BC = BD = 5 cm.

As we know,

$\bf{\dag}\:{\boxed{\sf{Area \: of \: rhombus = \dfrac{1}{2}\times Diagonal_1 \times Diagonal_2 }}}$

Then,we also know,

$\bf{\dag}\:{\boxed{\sf{Area \: of \: isosceles\: triangle = \dfrac{b}{4}\sqrt{4a^2 - b^2} }}}$

Here,b = inequal side. & a = equal side.

$\dashrightarrow\sf Area =\cancel{ \dfrac{8}{4}} \sqrt{4(5)^2 - 8^2} \\\\\dashrightarrow\sf Area = 2 \sqrt{4 \times 25 - 64} \\\\\dashrightarrow\sf Area = 2 \sqrt{100 - 64}\\\\\dashrightarrow\sf Area = 2 \sqrt{36} \\\\\dashrightarrow\sf Area = 2 \times 6 \\\\\dashrightarrow\large{\underline{\boxed{\sf\blue{Area = 12 \: cm^2 }}}}$

Therefore,

Area of ∆ABD or ∆BCD = 12 cm²

$\rule{200}{1}$

Now,

Area of rhombus = Area of ∆ABD + ∆BDC

➩ Area of rhombus = 12 + 12

➩ Area of rhombus = 24 cm²

Now according to question,

$\nrightarrow\sf \dfrac{1}{2}\times Diagonal_1 \times Diagonal_2 = 24 \\\\\nrightarrow\sf \dfrac{1}{2}\times 8 \times Diagonal_2 = 24 \\\\\nrightarrow\sf 4 \times Diagonal_2 = 24 \\\\\nrightarrow\sf Diagonal_2 =\cancel{ \dfrac{24}{4}} \\\\\nrightarrow\large{\underline{\boxed{\sf\blue{Diagonal_2 = 6 \: cm }}}}$

Therefore,

Area of rhombus = 24 cm² & Length of other diagonal = 6 cm

Answer 2

AnswEr :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

A rhombus whose side is 5 cm and it's altitude is 4 cm. If one of its diagonal is 8 cm long.

$\setlength{\unitlength}{1.2cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(7.7,0.9){\large{B}}\put(9.2,0.7){\large{\sf(5\:cm)}}\put(11.1,0.9){\large{C}}\put(9.9,2.1){\large{O\:(8\:cm)}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(11.7,2){\large{\sf(4\:cm)}}\put(11,1){\line(-1,1){2}}\put(8,1){\line(2,1){4}}\put(8,1){\line(2,1){4}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\end{picture}$

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The area of rhombus and the length of the other diagonal.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

$\bf{We\:have}\begin{cases}\sf{Altitude\:of\:\triangle BCD=4\:cm}\\ \sf{Base\:of\:the\:\triangle BCD=5\:cm}\\ \sf{Diagonal\:of\:rhombus\:(d_{1})=8\:cm}\end{cases}}$

formula use :

$\bf{\boxed{\sf{Ar.\:of\:triangle\:=\:\frac{1}{2} \times base \times height}}}}}$

$\mapsto\sf{\pink{Area\:of\:rhombus\:=\:2\times Area\:of\: \triangle BCD }}$

$\mapsto\sf{Area\:of\:rhombus=2\times \dfrac{1}{\cancel{2}} \times 5\times\cancel{ 4}}\\\\\\\mapsto\sf{Area\:of\:rhombus=2\times 5\times 2}\\\\\\\mapsto\sf{\red{Area\:of\:rhombus=20\:cm^{2} }}$

Now;

$\bf{\boxed{\sf{Ar.\:of\:rhombus\:=\:\frac{1}{2} \times d_1 \times d_2}}}}}$

$\mapsto\sf{20\:cm^{2} =\dfrac{1}{\cancel{2}}\times \cancel{8cm}\times d_2}\\\\ \\ \mapsto\sf{20cm^{2} =4cm\times d_2}\\\\\\\mapsto\sf{d_2=\cancel{\dfrac{20cm^{2} }{4cm} }}\\\\\\\mapsto\sf{\red{d_2=5\:cm}}$

Thus;

The area of rhombus = 20 cm² and other diagonal is 5 cm.