Parallelogram Perpendiculars

Since $AC$ is the diagonal, Areas of the triangles $ABD$ and $ACD$ are same. By the formula "Area of a triangle= $\frac {1}{2} \times$ Base $\times$ Height", Area of $ABD= \frac {1}{2} \times BD \times 7$ and Area of $ACD= \frac {1}{2} \times CD \times 24$ . These two expressions are equal,so,by combinng them, $\frac {BD}{CD}= \frac {24}{7} (i)$ . Also the semi perimeter of the parallelogram is $62$ , that is $BD+CD=62 (ii)$ . Solving $(i),(ii)$ we have $BD=48,CD=12$ and then the area of the parallelogram=area of $ABD +$ area of $ACD$ = $\frac {1}{2} \times 48 \times 7+\frac {1}{2} \times 12 \times 24=336$

Let $AB=x$ and $AD=y$ . Note that since $ABCD$ is a parallelogram, $CD=AB=x$ and $BC=AD=y$ . Thus, the perimeter is $2x+2y=124$ or $x+y=62$ . Thus, $y=62-x$ . Also, note that the area of $ABCD$ is $BC\cdot AE=(62-x)\cdot 7$ .

But the area of $ABCD$ is also $CD\cdot AF=x\cdot 24$ . Thus, $(62-x)\cdot 7=x\cdot 24$ . This simplifies to $31x=62\cdot 7$ which implies that $x=14$ . Thus, the area of $ABCD$ is $14\cdot 24=\boxed{336}.$

Let $\angle ADC=\angle ABC=x$ .

Suppose that ABCD is a rectangle. Then, it's perimeter would have been $2(24+7)=62 \neq 124$ . So, ABCD cannot be a rectangle.

Now, $AD=\frac{24}{sinx}$ and $AB=\frac{7}{sinx}$ ( $sinx\neq0$ as ABCD is not a rectangle).

According to question, $AB+AD=62$ $\implies \frac{31}{sinx}=62$ $\implies sinx=\frac{1}{2}$

So, $AD=BC=48$ and $AB=CD=14$

Thus, area of parallelogram ABCD=area of triangle ACD+area of triangle ABC = $24\times7+7\times24=336$

Note:The above solution applies for both cases, i.e. one in which E,F lie between (B and C) and (C and D) respectively and also for the case where D,B lie between (C and F) and (C and E) respectively. In the latter case we have to use that $sinx=sin(\pi-x)$

Let $DC=a$ and $BC=b$ . We know diagonal of a parallelogram divides it into 2 congruent triangle, which have the same area. By considering the perimeter, we have $2a+2b=124$ . By considering the area of $ABC$ and $CDA$ , we get that $24a=7b$ . Solving this, we get $a = 14, b = 48$ . Hence, the area of the parallelogram is $48 \times 7 = 336$ .

[Edits for clarity - Calvin]

Since ABCD is a parallelogram, thus the area of triangle ABC is equal to the area of triangle ADC. Thus $\frac{1}{2}AE\cdot AC=\frac{1}{2}AF\cdot DC$ . Thus $7(BC)=24(DC)$ . Therefore $\frac{BC}{BC+DC}=\frac{24}{24+7}=\frac{24}{31}$ .

But since AD=BC and AB=CD, thus $BC+DC=\frac{1}{2}\cdot 124 = 62$ . Thus $BC=\frac{24}{31}\cdot 62 = 48$ .

Therefore the area of the parallelogram is $BC\cdot AE = 48\cdot 7 = 336$ .

Area of parallelogram = $AE*BC = AF*CD$ . Since $\frac{AE}{AF} = \frac {7}{24}$ , we must have $\frac {BC}{CD} = \frac {1}{\frac {7}{24}} = \frac {24}{7}$ .

But $BC + CD = \frac 12 (Perimeter of ABCD) = 62$

So $BC = \frac {24}{24+7} * 62 = 48$ and the area of ABCD is $48*7 = 336$ .

As opposites sides equal in case of parallelogram perimeter = 124 let say AB = DC = x cm and AD = BC = y cm

2x + 2y = 124 (perimeter) => x + y = 62

so , DC = x and BC = (62 - x)

now we have given two heights i.e. AE = 7 cm and AF = 24 cm but area should be equal by taking any of height area = base*height

so , BC(base) AE(height) = DC(base) AF(height) 24 a = 7 (62 - a) => a = 14

area of parallelogram = 14*26 = 336

Let AB and CD be x, BC and AD be y, angle ABC be θ.

From triangle ABE, sinθ = 7/x.

From triangle ADF, sinθ = 24/y.

Therefore, 7/x = 24/y.

y = 24x/7.

2x + 2y = 124

x + y = 62

x + 24x/7 = 62

x = 14

Area of ABCD = x * 24 = 14 * 24 = 336

Finding the area of the parallelogram requires the unknown AD, and the known AE. Let the 2 triangles formed by the feet of the perpendicular be ABE and ADF. Since ABCD is a parallelogram, angles ABE and ADF are identical, making triangles ABE and ADF similar triangles. Therefore, AE/AF= AB/AD = (AE+AF)/(AB+AD) Since the perimeter of ABCD is 124, AB + AD is 62. AB/AD = 7/24. AD is then equal to 62 x (24/(24+7)) = 62 x 24/31 = 2 x 24 = 48. The area of the parallelogram is then simply AD x AE = 48 x 7 = 336.

Perimeter is 124 so AB + BC = 62. Now BC \times AE = area = AB \times AF. So BC \times 7 = AB \times 24. AB + BC = 62 \rightarrow 7 \times AB + 7 \times BC = 434. From previous equation we substitute value of 7 \times BC. We get 31 \times AB = 434 \Rightarrow AB = 14. As area is AB \times AF \Rightarrow 14 \times 24 = 336 = area.

let the bases be x and y. that is the parallels be x and y.now the area of the parallelogram with base x is ( x )( 7 )........(1) and the area with base y is ( y )( 24 )........(2) equating we get y = 7 ( x ) / ( 24 ). now we know that x + y = 124...so putting values of x there we get x =48....now area will be 7 x 48 = 336

In parallelogram ABCD take its diagonal AC which divide area of parallelogram into two equal area of triangles i-e ACB=ACD . 1/2 * AE * BC = 1/2 * AF * CD it gives 7 *BC = 24 * CD ......(i)

Then Given that: 2(BC + CD)=124 .........(ii) from equ i and ii we get BC = 48 and CD = 14. area of parallelogram ABCD=2 * ar(ACB) $\Rightarrow$ 2 * 168 = 336.

CD=14

[Manual revision]

Let x = AB and y = CD. Using the given information, the area of triangle ABC is 3.5x, and the area of triangle ACD is 12y.

Since ABCD is a parallelogram, triangle ABC is congruent to triangle ACD, and so they have equal area. So 3.5x = 12y.

Again, since ABCD is a parallelogram, x+y is half the perimeter of the parallelogram, so x+y = 62. Solve these two equations simultaneously for x and y to get x = 48 and y = 14. So the areas of triangle ABC and ACD are both 168, and so the area of the entire parallelogram is 336.

We all know that perimeter is equal to the sum of all sides of a polygon so we have,

Perimeter = AB + BC + CD + AD = 124

We have Area of Parallelogram = base * height = CD * AF = CD * 24 = 24CD -> let's make that as Eq. 1

We also have another Area of Parallelogram = base * height = BC * AE (because AE is perpendicular to BC) = BC * 7 = 7BC -> make this as Eq.2

We also know that AB=CD and BC=AD.

To make things clear, let x = AB = CD and let y = BC = AD

From the Perimeter Formula, 2x + 2y = 124 -> x+y=62 -> Eq.3

From Eq. 1, Area = 24CD = 24x

From Eq. 2, Area = 7BC = 7y

Equating those two equations, 24x = 7y -> x = 7/24 y

From Eq.3: x+y=62 -> 7/24y + y = 62 -> 31/24y = 62 -> y = 48

From Eq.3 again, we can solve for x: x + y = 62 -> x = 62 - 48 -> x = 14

Now, we can compute for the area: x = AB = CD and y = BC = AD

Area = base * height = CD * AF =14 * 24

Area = 336

or

Area = base * height = BC * AE = 48 * 7

Area = 336

We know that area of a parallelogram equals base * height Taking the side AB=x BC=y,we get x+y=62 Now taking BC as the base and height equals to AE, and applying the above formula, we get area = 7y Now taking CD as the base and height equal to AF we get area equal to 24x So area in both the case must be equal,so further simplification of the two equations ,we get our required answer......

Area of a parallelogram = height * base, hence the area of parallegram ABCD = AE BC and it is also equal to AF CD. Hence AE BC = AF CD. Hence (AF/AE) CD = BC or (24/7) CD = BC(1). We know that perimeter = 124. Therefore AB + BC + CD + AD = 124. But since the opposite sides of a parallelogram are equal, CD + CD + BC + BC = 124. Hence BC + CD = 62. Replacing BC by (1), (24/7) CD + CD = 62. Hence (31 CD)/7 = 62. We get CD = 14. Multiplying CD by AF, we get the area which is euqal to 14*24 = 336.

Let $[PQRS]$ denote the area of figure $PQRS$ . Let $S=[ABCD]$ . We have $S = [ABCD]= 2[ABC] = BC\times AE$ $\Rightarrow BC=\frac {S}{7}$ . Similarly, $S = [ABCD]= 2[ACD] = CD\times AF$ $\Rightarrow CD=\frac {S}{24}$ .

From the question we have that $\frac {124}{2} = BC+CD$ $= S \left(\frac {1}{7} + \frac {1}{24}\right)$ , thus $S = 336$ .

Note: We should also check that the such a parallelogram can exist, which it can.

CD×24=BC×7And 2(BC+CD)=124, BC+CD=62, 7BC+7CD=434, 24CD+7CD=434, CD=14. Area=14×24=336.