Warm up problem 8: parallelogram, not enough info?

Let $a$ be the top and bottom parallel sides, and $b$ be the left and right parallel sides. Then we know that

$2(a+b)=176$
$10a=12b$ .

A little algebra gets us $a=48$ , so that the area is $480$ square units. $\square$

Edit: Here's the actual parallelogram, drawn to accurate scale

Funny Parallelogram

where $10$ and $12$ represent the distances between the $2$ pairs of parallel lines.

There are two ways to find the area (A), one for each height. Let a = base for 10, and b = base for 12.

A = 10a = 12b.

a = A/10

b = A/12

Perimeter: 2a + 2b = 176

Divide by 2: a + b = 88

Substitution gives: (A/10) + (A/12) = 88

Multiply by 120: 12A + 10A = 10560

Add: 22A = 10560

Divide by 22: A = 480

No need to find either base length.

Let a and b be the sides. The area =10 * a =12 * b. Now dividing these two we get $\frac ab=\frac{12}{10}$ .

Therefore $\frac{(a + b)}{b} = \frac{22}{10} \text{ and } a + b = \frac{176}{2} \Rightarrow b = 40$

Hence Area is $40 * 12 = 480.\square$

Let $\theta$ be the angle opposite to 12 and 10.

The perimeter is 2a + 2b = 176, so a + b = 88.

Since $sin \theta$ = 10/a and $sin \theta$ = 12/b

12/b = 10/a

12a = 10b

a = 5b/6

Thus, a + b = 88

5b/6 + b = 88

11b/6 = 88

b/6 = 8

b = 48

10 is the altitude since it's perpendicular to a base.

Therfore A = (10)(48) = 480 sq. units

This is not the most mathematical way to do it but hey, its easier :D

Let the sides be x and y

The perimeter equation reads 2*(x+y) = 176; x+y = 88 Keep that in mind

Now the two heights are given as 12 and 10 Logically, this means that X>Y Keep that in mind too

Now the area of the parallelogram equation gives x*10 which can be 240 or 360 or 480 or 420 so x can be 24, 36, 48 or 42 Substituting the above values in x+y=88 gives only one possible solution where x>y, when x is 48

So, the area has to be 10x which is 480

$let\quad one\quad side\quad with\quad 10\quad as\quad perpendicular\quad be\quad x\quad and\quad other\quad be\quad y,\\ now\quad we\quad know\quad ,\\ 2(x+y)\quad =\quad 176\\ (x+y)\quad =\quad 88\\ \\ Now\quad join\quad the\quad one\quad daigonal\quad (\quad in\quad between\quad the\quad \bot \quad s\quad )\\ \\ so\quad ,\quad \Lambda \quad of\quad triangle\quad 1\quad =\quad \Lambda \quad of\quad triangle\quad 2\quad (\quad made\quad by\quad daigonal)\\ \quad \quad \quad \quad 1/2\quad *\quad 10\quad *\quad x\quad =\quad 1/2\quad *\quad 12\quad *\quad y\\ \quad \quad \quad \quad \quad x/y\quad =\quad 1.2\quad \\ \quad \quad \quad \quad \quad \quad x\quad =\quad 1.2\quad y\quad -------\quad eq.\quad 1\\ \\ Now,\quad given\quad ,\quad x+y\quad =\quad 88\\ put\quad eq\quad 1\quad in\quad it\quad ,\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 1.2y\quad +\quad y\quad =88\quad \Rightarrow \quad y\quad =\quad 40\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad so\quad ,\quad x\quad =\quad 48\\ Now\quad we\quad know\quad ,\\ \Lambda \quad of\quad parallelogram\quad =\quad sum\quad of\quad those\quad 2\quad triangles\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 1/2\quad *\quad 10\quad *\quad 48\quad +\quad 1/2\quad *\quad 12\quad *\quad 40\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 240\quad +\quad 240\quad \Rightarrow \quad 480\quad \\ SIMPLE\quad ????$

Let's start off by labeling the sides. The shorter side I will label as "x" and the longer side I will label "y." This means that with the perimeter constraint we have the following equation:

$2x+2y=176$

Which reduces to

$x+y=88$

Next, you should draw the diagonal through the parallelogram. Once you do this, you should notice that you have split the parallelogram into two triangles. These triangles are congruent due to the side-side-side proof. You will also notice that the two heights given in the parallelogram are also two different heights within these triangles. This means that:

$\frac { 1 }{ 2 } \left( 10 \right) \left( y \right) =\frac { 1 }{ 2 } \left( 12 \right) \left( x \right)$

Which, when solved for y, becomes:

$y=\frac { 6 }{ 5 } x$

Once you substitute back into the perimeter equation you get:

$x+\frac { 6 }{ 5 } x=88$

This gives the solution for x:

$x=40$

and y:

$y=48$

From here, pick a side and height and perform the area equation:

$base \times height=Area$

$48 \times 10=Area$

$Area=480$

Call top member AB&bottom (CD),m is intersection between Line DC&line =10, at∆Bmc ,,sin(mcb)=10\bc,call intersection between( line=12 &BC ) f, at∆ dcf ,sin(dcf)=12\Dc,angle (mcb)&(dcf) is the same, so 10\bc=12\Dc ,5dc=6bc, from given 2dc+2bc=176,solve 2 expressions ,dc=AB=48,BC=AD=40,area =2∆bcm+rectangle, mc=√(1600-100)=10√5,area=2×.5×10√5×10+10×(48-10√5)=100√5+480-100√5=480###

Let x be the side having 12 altitude.since opposite sides are equal,sum of adjacent sides=176/2=88.Adjacent side of x=88-x . Area=12x=10(88-x). 12x=880-10x 22x=880 x=880/22=40 Area=12*40=480

Let bottom side length is x, and so right side length is 176/2-x = 88-x. Then we get area A = 10x or A = 12(88-x). So, 10x = 12(88-x) => 22x = 1056 => x = 48. So area A = 10x = 480.

Let: x be the hypotenuse of the right triangle formed by the altitude w/ a length of 10. y be the hypotenuse of the right triangle formed by the altitude w/ a length of 12.

This 2 triangles are similar because there angles are congruent.

Then we can say that:

x/10 = y/12 (Eq. 1)

Perimeter of Parallelogram = Sum of all sides = 176

2x + 2y = 176 or x + y = 88 (Eq. 2)

x = 88 - y (from Eq. 2) substituting this to Eq. 1 gives us

(88 - y)/10 = y/12 ; 1056 - 12y = 10y; 1056 = 22y ; y = 48

Area of Parallelogram = Height * Base

where Height is 10 and Base us the just solved y which is 48

Area = 10 * 48 = 480

Simplify the ratio --> 10 : 12 = 5 : 6

*Add the elements * ---> 5+6 = 11

Divide by perimeter ---> 176/11= 16

width (divide it by two bcz its of both sides) = (16 * 5) / 2= 40 ,

base = (16 * 6 )/2 = 48

*Area * = base * height = 48 * 10 = 480

2(a+b)=176, a+b=88, where a and b are the sides of the parallelogram, draw a diagonal which divides the parallelogram into two triangles of equal area, hence 5a=6b, but the area we require is 5a+6b=10a, solving we get a=48, hence 10a= 480

Let the upper side be 'a' and the adjacent side be 'b' . Divide the remaining area of the //gm into 2 . Consider the base of the first triangle as x. Consider the base of the second triangle shown in figure as y.

Area of first triangle shown = 1/2 * x* 10

Area of second triangle shown in figure = 1/2 y 12

Area of third triangle (which we have formed)= 1/2 * (a-x) * 10

Area of fourth triangle (which we have formed)= 1/2 (b-y) 12

Add these up and equate it to 10a(area of //gm).

You'll get two linear equations . a+b = 88 6b -5a = 0

Solve them .

You"ll get the area as 480

A simple solution .... Let the area of the parrelelogram be A,

Then the breadth (right and left sides )=A/12 and The length ( top and bottom sides) = A/10 Now the perimeter =176, We get the eq, 2(A/12+ A/10)= 176, Solving,we get, A=480.

A=12a=10b

a = 5n, b = 6n, n ϵ Z

2 * (5n+6n)=176

n=8, a=40 b=48

12 * 40=48 * 10=480

Since the two triangles formed are similar triangles, the hypotenuse of the smaller triangle is 10/12 the length of the hypotenuse of the larger triangle. Setting the hypotenuse of the larger triangle as x, the hypotenuse of the smaller triangle would be 10/12 x. Since the perimeter is 176, we can say that x plus 10/12 x equals 88. Solve for x and you get x=48. If x is 48, than 10/12 x is 40. (48+40=88). So the area is 48 (the base I used) x 10 (the height).

in the figure, two triangle are similar.Let a be perpendicular 10 and let b be 12. then we know that a/10=b/12 so 10b=12a now 2(a+b)=176 a=88-b we get 10b=12(88-b) b=48 now the area=b10=480

let one side be x(normal to 10 unit length perpendicular) and the other side be y(normal to 12 unit length perpendicular).

it is given that the perimeter is 176, that is, 2(x+y)=176

therefore, x+y=88 and x=(88-y)

by equating areas we have,

12 y=10 (88-y)

12y=880-10y

22y=880

y=40

also, it gives x=48 ...[x=(88-y)]

therefore, the area of the parallelogram is base height= 12 y= 12*40= 480

let the two sides of the parallelogram be l and b as the perimeter of the parallelogram is 176cm 2(l+b)=176 ------------(1) l+b=88 10l=12b 5/6l=b now put the value in eq.1, 5/6l+l=88 11/6l=88 l=48cm b=40cm,area of the parallelogram is 40x12=480cmsq.

If we denote with $h_1=10,\,h_2=12,\,L_1,\,L_2,\,$ the lenghts of the heights and the sides of the parallelogram, $P$ the perimeter, $A\,$ the area of its surface, it can be observed that:

$\frac {L_1}{L_2}=\frac {h_1}{h_2};\,L_1={\frac{12}{10}}{L_2};$

$P=2(L_1+L_2)=176=2(L2+\frac{12}{10}L_2);\,L2=40,\,L_1=48$

$A={L_1}{h_1}={L_2}{h_2}=48\times{10}=\boxed{480}$

$2\quad (x\quad +\quad y)\quad =\quad 176\quad (1)\\ 10x\quad =\quad area\quad \quad \times \quad 12\\ 12y\quad =\quad area\quad \quad \times \quad 10\\ \\ \quad \quad 120x\quad =\quad 12area\\ +\\ \quad \quad 120y\quad =\quad 10area\\ ------------------\\ 120\quad (x\quad +\quad y)\quad =\quad 22area\\ (x\quad +\quad y)\quad =\quad \frac { 22area }{ 120 } \quad \quad (2)\\ \\ substituting\quad (2)\quad into\quad (1)\\ 2(\frac { 22area }{ 120 } )\quad =\quad 176\\ area\quad =\quad 480$

let X the top rib & Y is the closing one 2X+2Y=176 :: X+Y=88 ::: Y=88-X :: :A=area
A= 10(88-X) =12X ::X= 40 ::A= 40*12 =480

I solved this one like Michel using a system of equations. However, it also looks like the two right triangles are similar. Is anyone able to solve it by proving the 2 right triangles are similar and then finding a and b since the two hypotenuses would be proportional?

Perimeter is 176,so two sides of parallegram sum will be 176/2 = 88 Cm,Let a and b are longer and smaller sides of parallelogram so, 2(a +b ) = 176,since height is 10 cm and a+b =88, ratio of a to b will be 12/10 and accordingly a side length is 48 therefore area is 480 Ans K.K.GARG ,India