A geometry problem by Arpit Sah

Let the side be $a$ and diagonals $p,q$ , then $4a = 48$ , so $a = 12$ and $p^2 + q^2 = 4a^2 = 576$ $p + q = 26$ Thus $pq = \frac{(p + q)^2 - (p^2 + q^2)}{2} = \frac{100}{2} = 50$ So the area is $A = \frac{pq}{2} = \boxed{25}$

In a rhombus, the diagonals are perpendicular bisectors of each other, dividing the rhombus into 4 congruent right-angled triangles of hypotenuse 12. Let us call the legs of these triangles $a$ and $b$ .

From the Pythagorean theorem, we know that:

$a^{2}+b^{2} = 12^{2} = 144$

From the given sum of the diagonals, we also know that:

$2a + 2b = 26 \Longrightarrow a+b = 13$

Squaring the second equation:

$a^{2}+b^{2}+2ab = 13^{2} = 169$

But $a^{2}+b{2} = 144$ , from the first equation. Hence:

$144 + 2ab = 169 \Longrightarrow 2ab = 25$

This does not seem useful until we try to calculate the area of the rhombus, which is given by half the product of the length of its diagonals, $2a$ and $2b$ .

$Area_{rhombus} = \dfrac{1}{2} \times 2a \times 2b = 2ab = \boxed{25}$

Let one diagonal $x$

And another $26-x$

The all sides of rhombus is equal. So one side length $\frac{48}{4} \implies 12$

We know when two diagonals across each other in rhombus then it makes 4 right angle and equally divide each other.

So, we get a equation in any right triangle that $(\frac{x}{2})^2 + ( \frac{(26-x)}{2})^2 = 12^2$

Solving this equation we get two values of $x = (13 - \sqrt{119})$ and $(13 + \sqrt{119})$ .

So, the two diagonals are $(13 - \sqrt{119})$ and $( 13 + \sqrt{119})$ .

And the area of the rhombus $\implies \frac{1}{2} \times (13-\sqrt{119}) \times (13+\sqrt{119}) \implies \frac{1}{2} \times (13^2 - (\sqrt{119})^2) \implies \frac{1}{2} \times 169 - 119 \implies \frac{1}{2} \times 50 \implies \boxed{25}$

Once the perimeter is 48, we know the value od the sides is 48/4 = 12

Calling the diagonals D and d , and using Pythagoras theorem, we got:

(D/2)^2 + (d/2)^2 = 12^2

D²/4 + d²/4 = 144

D² + d² = 144*4 = 576

Is given D+d = 26 , so:

(D+d)² = (26)² = D²+d²+2Dd , but we have alredy found D²+d²

26² = (576)+2Dd

676 = 576 + 2Dd

100 = 2Dd , Dd = 50

How the area of a rhombus is given by Dd/2 ...

50/2 = 25 u² = area.

Since the length of the sides of a rhombus are all equal, the length of one side is $12$ . Look at a right triangle formed by each of the diagonals and a side. The hypotenuse of this right triangle is $12$ , and the sum of the other two sides $a$ and $b$ is $\frac{26}{2} = 13$ . Using the pythagorean theorem, we the two equations:

$a^2 + b^2 = 144$

$a+b = 13$

Squaring the second one, we get $a^2 + b^2 = 169 - 2ab$

Substituting it into the above equation, we get $2ab = 25$ .

Note that the area of a rhombus is given by $\frac{1}{2} d_1 d_2$ , where $d_1$ and $d_2$ represent the length of each diagonal. Since $a$ and $b$ are each half the length of a diagonal, we get $A = \frac{1}{2} (2a) (2b) = 2ab$

Thus, $A = 25$ .

Lets call $d_{1}$ and $d_{2}$ the bigger and smaller diagonal, respectively. There are two equations given:

$1)$ $d_{1} + d_{2} = 26 \rightarrow d_{2} = 26 - d_{1}$ .

$2)$ $\frac {d_{1}^{2}}{4} + \frac {d_{2}^{2}}{4} = 144$ .

$d_{1}^{2} + d_{2}^{2} = 576$ .

$d_{1}^{2} + (26 - d_{1})^{2} = 576$ .

$d_{1}^{2} - 26d_{1} + 50 = 0$ .

As $d_{1}$ is the bigger, then we need to say:

$d_{1} = 13 + \sqrt {119}$

$d_{2} = 13 - \sqrt {119}$

We know the rhombus area is $\frac {d_{1}\cdot d_{2}}{2}$ . Hence, our answer is $\frac {(13 + \sqrt {119})(13 - \sqrt {119})}{2} = \frac {50}{2} = \boxed {25 u^{2}}$ .

Perimeter of rhombus is 48 so the length of side is 12. Let the half diagonals be x and y. Since the half diagonals are at right angle, therefore $Area=4\left ( \frac{1}{2} \right )xy$ . Using Pythagora's Theorem: $12^{2}=x^{2}+y^{2}$ $12^{2}=\left ( x+y \right )^{2}-2xy$ $12^{2}=\left ( x+y \right )^{2}-2xy...{\color{#D61F06} 1}$ Now, the sum of diagonals of the rhombus is 26. Since x and y are half diagonals, thus $2x+2y=26$ which makes $x+y=13$ . Plug in $x+y=13$ into equation ${\color{#D61F06} 1}$ . $Area=2xy=\left ( 13 \right )^{2}-12^{2}$ $Area=\boxed{25}$

let a + b = 26

obviously, measure of each side will be 12

so (a+b)^{2} = 26^2

a^2 + b^2 + 2ab = 676

in the rhombus diagonLS INTERESRCT at 90. so (a/2)^2 +( b/2)2 = 12^2

solving the two equation we get a*b= 50

so area of rhombus is 1/2 *50=25

as diagonals are at right angles. and length of a side is 12 cm,

hence (d1/2)^2 + (d2/2)^2 = 12^2

solving further :

d1^2 + d2^2 = 576 _(1)

second condition : sum of diagonals is 26.

d1 + d2 = 26

squaring

d1^2 + 2d1d2 + d2^2 = 676

from (1)

2d1d2 + 576 = 676

2d1d2 = 100

d1d2 = 50

area of rhombus = d1d2/2 = 50/2 = 25

Firstly, we know that the sides of a rhombus are all congruent. So, if the perimeter is 48, then each side must be 48/4 or 12. We also know that the diagonals bisect each other perpendicularly, so the area can be found by dividing the rhombus into four congruent right triangles. Let's take one of these triangles and find all the side lengths so we can find the total area. We already know the hypotenuse length-12, so we can use a system of equations to solve-a^2 + b^2=12^2 and 2a + 2b=26. The second equation is like shown because the sum of the diagonal lengths of the rhombus is 26 and the two unknown triangle leg lengths are each half of their respective diagonal. (Like I said earlier, the diagonals of a rhombus are perpendicular bisectors of each other.) We should simplify the second equation by dividing by 2 on both sides to make life easier. By substitution, the equations can be simplified down to a polynomial that can be solved using the quadratic formula. Although this equation has two solutions, when each solution is plugged back into the original equation it gives the other. Once we have the two leg lengths, simply apply the formula for the area of a triangle, then multiply by 4 (because a rhombus is made up of 4 congruent triangles) to obtain an area and answer of 25.

Rhombus has 4 equal sides, hence the side equals $\frac {48}{4}=12$ . Let $d_1$ and $d_2$ denote the diagonals of the rhombus. Then it is stated in the problem that

$d_1+d_2=26...(1)$

But if we observe the rhombus, we could easily verify that

$(\frac {1}{2}d_{1})^{2}+(\frac {1}{2}d_{2})^{2}=12^{2}...(2)$

Now we have two equations to solve.

From (2) we have

$(\frac {1}{2}d_{1})^{2}+(\frac {1}{2}d_{2})^{2}=\frac {1}{4}(d_{1}^{2}+d_{2}^{2})$

$\Leftrightarrow \frac {1}{4}[(d_{1}+d_{2})^2-2d_{1}d_{2}]=144...(3)$

Substitute (1) to (3), and we can solve for $\frac {1}{2}d_{1}d_{2}$ which is the area of the rhombus.

$\frac {1}{4}[(d_{1}+d_{2})^2-2d_{1}d_{2}]=144$

$\Leftrightarrow \frac {1}{4}(26^2)-\frac{1}{2}d_{1}d_{2}=144$

$\Leftrightarrow \frac{1}{2}d_{1}d_{2}=\frac {1}{4}(26^2)-144=169-144=\boxed{25}$

Let the diagonals of the rhombus be a and b.

Thus, a + b ₌ 26 ….. (1)

The diagonals intersect at right-angles dividing the rhombus into four equal right-angled triangles.

From the information given about the perimeter of the rhombus, we can deduce that the length of each side is 12 units (the 4 sides of a rhombus are equal).

Considering one of the right-triangles, let

the hypotenuse side be 12 units

the other two sides that form the right angle be a/2 and b /2 (when the diagonals intersect, the sides that form the right angle are 1/2 the length of each diagonal).

By Pythagoras theorem,

(a/2)^2 + (b/2)^2 = (12)^2

(a/4)^2 + (b/4)^2 = 144

a^2 + b^2 = 576 …..(2)

Solving the two equations gives,

a ₌ 23.90871211uints and b ₌ 2.091287885 units

Applying the formula for the area of a rhombus,

A ₌ ab/2

A ₌ (23.90871211 ×2.091287885)/2

A ₌ 25 sq. units.

Perimeter = 48

Therefore, each side = $\frac{48}{4} = 12$

Also, the diagonals intersect at right angles giving four congruent right triangles.

Lets focus on one of them.

The hypotenuse is of course 12 .

Since the sum of the diagonals is 26, perpendicular+base = $\frac{26}{2} = 13$

If the perpendicular is x then the base has to be (13 - x)

Applying Pythagoras' theorem: $x^{2} + (13 - x)^{2} = 12^{2}$ ⇒ $2x^{2} - 26x + 169 = 144$ ⇒ $x^{2} - 13x = \frac{25}{2}$ ⇒ $x(x-13) = \frac{25}{2}$ ⇒ $\frac{1}{2} \times x(x-13) = \frac{25}{4}$

But of course, this is nothing but the area of the triangle-

We simply multiply it by 4 to find the area if the rhombus.

Area of rhombus $= \frac{25}{4} \times 4 = \boxed { 25 sq. units}$

We can write down two equations describing the lengths of the diagonals: $a+b=26$ $12^{2}=(\frac{a}{2})^{2}+(\frac{b}{2})^{2} \Rightarrow 576=a^{2}+b^{2}$ We can now solve for $a$ in terms of $b$ : $a=26-b$ We now plug this into the other equation and get: $576=(26-b)^{2}+b^{2} \Rightarrow 0=b^{2}-26b+50 \Rightarrow b=13 \pm \sqrt{119}$ Since the area is equal to $\frac{1}{2}ab$ , we get that: $A=\frac{13^{2}-119}{2}=\frac{50}{2}=\boxed{25}$

let the diagonals be x and y units in length, then according to the problem: x + y = 26 (i) also, the diagonals of a rhombus bisect each other perpendicularly, therefore four right angled triangles would be formed, applying pythagoras theorem to any of the triangles we get : (x/2)^2 + (y/2)^2 = (48/4)^2 ( as side of rhombus = perimeter/4) x^2 + y^2 = 576 (x+y)^2 - 2xy = 576 26^2 -2xy = 576 676 - 2xy = 576 xy = 50 now, area of rhombus = xy/2 = 50/2 = 25 sorry to give quite a long method.

it is given that side length=12 (4a=48) and d1 +d2=26 by using the formula (side length) =1/2√(d1)²+(d2)² weget (d1)(d2)=50 then area =1/2(d1)(d2)=1/2(50)=25

Let one of the diagonals be ' d ' , then the other diagonal will be ' 26 - d '. Since perimeter is 48 , side of a rhombus = 12 Since , diagonals of rhombus bisect at right angles , we get , by pythagoras ,

12^ 2 = (d/2) ^ 2 + ( 13-d/2 ) ^ 2 Solving for d , we get d = 13+119 ^ 1/ 2, other diagonal becomes 26 - d = 13 - 119 ^ 1/2

Area of rhombus = 1/2 * product of diagonals , Therefore area = 25 ( by multiplication )

First split the rhombus into four equal triangles and label the perpendicular lengths $a$ and $b$ .

The area of the rhombus will be 4 times the area of each triangle: $\text{Area} = 4\times \frac{1}{2}ab = 2ab$

The perimeter of the rhombus will be 4 times the length of the hypotenuse of each triangle. By Pythagoras:

$\text{Perimeter} = 4\times \sqrt{a^2+b^2} = 48 \Rightarrow \sqrt{a^2 + b^2} = 48\div 4 = 12 \Rightarrow a^2 + b^2 = 12^2 = 144$

The sum of the lengths of the diagonals is expressed as: $2a + 2b = 26 \Rightarrow a + b = 13$

$\Rightarrow (a+b)^2 = 13^2 = a^2 + 2ab + b^2$

$\Rightarrow 2ab = 13^2 - (a^2 + b^2) = 169 - 144 = 25$

So the perimeter of the rhombus is $\boxed{25}$

1. let one half of the larger diagonal be X
2. let one half of the smaller diagonal be Y
3. This means X^2 + Y^2 = 12^2 = 144
4. X+Y = 26/2 = 13
5. (X+Y)^2 = X^2 + 2XY + Y^2 =169
6. The diagonals cut the rhombus into 4 right angle triangle each with base = X and height = Y
7. The are of the rhombus = 4 * 1/2 * X * Y = 2XY
8. Subtract the equations in step 5 and 3, you get 2XY = 169 - 144 = 25 which is the are of the rhombus

Let 2x, 2y are the lengths of the rhombus' diagonals. So its area is (2x)(2y)/2 = 2xy.

Sum of the lengths of its diagonals is 26, then 2x + 2y = 26 ==> x + y = 13.

The perimeter of the rhombus is equal to 48 ==> length of 1 side = 48/4 = 12.

Apply Pythagore's theorem: x^2 + y^2 = 12^2

(x+y)^2 - 2xy = 12^2

13^2 - 2xy = 12^2

2xy = 13^2 - 12^2

2xy = (13-12)(13+12)

2xy = 25

The area of the rhombus is 2xy = 25

In order to solve this question, we begin by drawing a picture, like this one : we draw a rhombus and its diagonals and then move two of the resulting (congruent) triangles. We now have a picture containing a rectangle and a rhombus with equal area. If we call the half-diagonals of the rhombus $a$ and $b$ , we now notice that this area is $2ab$ .

We know that all sides of a rhombus are of equal length, so the given perimeter tells us that each side has length $12$ . Using our picture, Pythagoras now tells us that $a^2+b^2=12^2=144$ . We also know $a+b$ , which is exactly half the sum of the lengths of the diagonals, so $13$ .

Using these facts we are done: $A=2ab=(a+b)^2-(a^2+b^2)=13^2-144=169-144=25$ .

perimeter of rhombus ABCD is 48 cm(given). all the sides of rhombus are equal. so each side of rhombus will be 12 cm. BD and AC are its diagonals. BD + AC = 26 cm (given) (BD/2)^2 + (AC/2)^2 = (12)^2 (by using pythagoras theorem) BD^2 + AC^2=576 (BD + AC)^2=BD^2 + AC^2 +2.BD.AC (USING IDENTITY) BD.AC=50 area of rhombus = 1/2 * d1* d2 (where d1 and d2 are diagonals of rhombus) area of rhombus= 1/2 BD AC=1/2 * 50=25 cm.sq.

Since all side lengths in a rhombus are equal, each side length is equal to $\dfrac{48}{4}=12$ . Also, since the diagonals of a rhombus are perpendicular to each other, we can use the Pythagorean Theorem on one of the four congruent right triangles when the two diagonals are drawn. Let $a$ and $b$ be half the length of each diagonal.

By Pythagoras, we have $a^2+b^2=12^2=144$

We also know what $a+b$ is. Since the sum of the diagonals is 26, $a+b$ is simply half of that, or 13.

Since $(a+b)^2-2ab=a^2+b^2$ , we can then find $2ab$ . We have

$13^2-2ab=12^2$

$\implies 2ab=25$

Since the formula for the area of a rhombus is equal to $d_1d_2$ where $d_1$ and $d_2$ are the two diagonals, here we have $d_1=2a$ and $d_2=2b$ , so the area is equal to $\dfrac{2a\times 2b}{2} = \dfrac{4ab}{2}=2ab$

Since we have already found what $2ab$ is, we are done, and our desired answer is equal to $2ab=\boxed{25}\blacksquare$

We know lengths of diagonals of a rhombus are $2a \sin \theta$ and $2a \cos \theta$ where $a$ is the given side. Also area of a rhombus is $2a^2 \sin \theta \cos\theta$ Here perimeter given is $48$ .

$\therefore\ 4a= 48$

$\Rightarrow\ a=12$

$2a \sin \theta + 2a \cos \theta$ = $26$ Squaring we get $4a^2(1+2\sin \theta \cos\theta) = 676$

Hence we need to find $2a^2 \sin \theta \cos\theta$ which is easy once we plug in the value of $a$ .

That gives us the area to be $\boxed{25}$