Math, asked by Devil8161, 1 year ago

heya friends plzz this question ......

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Answered by siddhartharao77
9

Ina  triangle ABC,

Let a = 10cm, b = 16cm, c = 14cm.

We know that Semi-perimeter of a triangle s :

= > (a + b + c)/2

= > (10 + 16 + 14)/2

= > 40/2

= > 20cm.

Area of triangle ABC:

 =>\sqrt{s(s - a)(s - b)(s - c)}

 =>\sqrt{20(20 - 10)(20 - 16)(20 - 14)}

 =>\sqrt{20(10)(4)(6)

 =>\sqrt{4800}

 => 40\sqrt{3}cm

We know that diagonal divides a parallelogram into 2 triangles of equal areas.

= > Area of parallelogram ABCD = 2(Area of triangle ABC)

 => 2 * 40\sqrt{3}

 => 80\sqrt{3}

 => 80 * 1.73

=> 138.4cm^2.



Hope this helps!


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Answered by Swayze
74

hy                                

friend                                        

                         here is your answer                      

                                                                                                                                   

      In Δ ABC

Using Heron's Formula,

S = (a + b + c)/2

⇒ S = (10 + 14 +16)/2

⇒ S = 40/2

⇒ S = 20 


Area =  \sqrt[]{s[s-a][s-b][s-c]}  

∴ Area of the triangles =  \sqrt[]{20[20-10[20-14][20-16]}  

⇒ Area of Δ ABC =  \sqrt[]{20[10][6][4]}  

⇒ Area = √4800

⇒ Area = 10√48

⇒ Area = 10√(2 × 2 × 2 × 2 × 3)

⇒ Area = 10 × 2 × 2√3

⇒ Area of Δ ABC = 40√3 cm²

∴ Area of Δ ABC = 69.28 cm².


Area of the Δ ADC = Area of the ΔABC.

 [∵ Both the triangles are congruent)


∴ Area of the Parallelogram = 2 × Area of Δ ABC.

⇒ Area of the Parallelogram = 2 × 69.28 cm²

∴ Area of the Parallelogram = 138.56 cm²

                                                                                                                                                         

  •      the Area of the Parallelogram is 138.56 cm².

                                                                                                                                   

                                     thankyou

                                    brainly user


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