Math, asked by Dumitrescunarcis19, 2 months ago

Probleme combinate
254
Diferenta
doilea număr este cu 72 mai mare decat
a doua numere este 140. Af
dublul primului număn Care sunt​

Answers

Answered by HolyGirl
4

Given:

Vertices of Parallelogram – A(-2,-1), B(4,-1), C(0,2) and D(6,2).

The smaller coungruent sides of Parallelogram is having length of 3.6 units.

To Find:

Area of Parallelogram & Perimeter of Parallelogram.

Solution:

Diagonals of Parallelogram divides it into two congruent triangles, which are ∆ ABC and ∆ ACD.

Therefore,

Area of Parallelogram = Area of ∆ ABC + Area of ∆ ACD.

 \\

Formula for calculating Area of Triangle

 = \frac{1}{2} | x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \\ \\

\bf \underline{ For \: \triangle ABC :}

 (x_1 , y_1) = A(-2,-1),  (x_2,y_2)=B(4,-1),  (x_3,y_3)=C(0,2).

→ Area of ∆ ABC

 = \frac{1}{2} | x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \\

 = \frac{1}{2} \small { | -2 (-1 - 2) + 4(2 - (-1)) + 0(-1 - (-1)) |} \\

 = \frac{1}{2} | -2 (-3) + 4(2+1) + 0(-1 +1) | \\

 = \frac{1}{2} | 6 + 4(3) + 0(0) | \\

 = \frac{1}{2} | 6 + 12 + 0 | \\

 = \frac{1}{2} | 18 | \\

 = \frac{18}{2} \\

 = \cancel{\frac{18}{2}} \\

 \therefore \fbox {Area \: of \: ∆ ABC = 9 \: units.} \\ \\

 \bf \underline {For \: \triangle ACD :}

 (x_1 , y_1) = A(-2,-1),  (x_2,y_2)=D(6,2),  (x_3,y_3)=C(0,2).

→ Area of ∆ ACD

 = \frac{1}{2} | x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \\

 = \frac{1}{2} \small { | -2 (2 - 2) + 6(2 - (-1)) + 0(-1 - 2) |} \\

 = \frac{1}{2} | -2 (0) + 6(2+1) + 0(-3) | \\

 = \frac{1}{2} | 0 + 6(3) + 0 | \\

 = \frac{1}{2} | 0 + 18 + 0 | \\

 = \frac{1}{2} | 18 | \\

 = \frac{18}{2} \\

 = \cancel{\frac{18}{2}} \\

 \therefore \fbox {Area \: of \: ∆ ACD = 9 \: units.} \\

 \\

 \Rightarrow Area of Parallelogram = Area of ∆ ABC + Area of ∆ ACD.

 \therefore Area of Parallelogram = 9 units + 9 units

 \therefore \fbox {Area \: of \: Parallelogram = 18 \: units.} \\ \\

______________________

Perimeter of Parallelogram = 2(a + b)

where a = side and b = base of Parallelogram.

a = AD = 3.6 units (given), b = AB = ?

We can find base by using distance formula:

\bf \underline{ For \: line \: segment \: AB :}

 (x_1 , y_1) = A(-2,-1),  (x_2,y_2)=B(4,-1),

 AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

 AB = \sqrt{(4-(-2))^2 + (-1-(-1))^2}

 AB = \sqrt{(4+2)^2 + (-1+1)^2}

 AB = \sqrt{(6)^2 + (0)^2}

 AB = \sqrt{36+0}

 \therefore \fbox{AB = 6 units = b} \\ \\

 \Rightarrow Perimeter of Parallelogram = 2(a + b)

= 2(3.6 + 6) units

= 2(9.6) units

 \therefore \fbox{ Perimeter \: of \:  Parallelogram = 19.2 \: units} \\ \\

Hence, the area of Parallelogram is 18 units and Perimeter of Parallelogram is 19.2 units.

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