Let $\displaystyle x$ be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by $\displaystyle x+10$.

Recall how to find the area of a kite:

$\displaystyle \text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for $\displaystyle x$.

$\displaystyle 100=\frac{x(x+10)}{2}$

$\displaystyle x^2+10x=200$

$\displaystyle x^2+10x-200=0$

$\displaystyle (x+20)(x-10)=0$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, $\displaystyle x=10$.

The length of the shorter diagonal is $\displaystyle 10$ units long.

Let $\displaystyle x$ be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by $\displaystyle x+5$.

Recall how to find the area of a kite:

$\displaystyle \text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for $\displaystyle x$.

$\displaystyle 88=\frac{x(x+5)}{2}$

$\displaystyle x^2+5x=176$

$\displaystyle x^2+5x-176=0$

$\displaystyle (x+16)(x-11)=0$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, $\displaystyle x=11$.

The length of the shorter diagonal is $\displaystyle 11$ units long.

Let $\displaystyle x$ be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by $\displaystyle x+8$.

Recall how to find the area of a kite:

$\displaystyle \text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for $\displaystyle x$.

$\displaystyle 64=\frac{x(x+8)}{2}$

$\displaystyle x^2+8x=128$

$\displaystyle x^2+8x-128=0$

$\displaystyle (x+16)(x-8)=0$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, $\displaystyle x=8$.

The length of the shorter diagonal is $\displaystyle 8$ units long.

Let $\displaystyle x$ be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by $\displaystyle x+7$.

Recall how to find the area of a kite:

$\displaystyle \text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for $\displaystyle x$.

$\displaystyle 114=\frac{x(x+7)}{2}$

$\displaystyle x^2+7x=228$

$\displaystyle x^2+7x-228=0$

$\displaystyle (x+19)(x-12)=0$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, $\displaystyle x=12$.

The length of the shorter diagonal is $\displaystyle 12$ units long.

Let $\displaystyle x$ be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by $\displaystyle x+3$.

Recall how to find the area of a kite:

$\displaystyle \text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for $\displaystyle x$.

$\displaystyle 104=\frac{x(x+3)}{2}$

$\displaystyle x^2+3x=208$

$\displaystyle x^2+3x-208=0$

$\displaystyle (x+16)(x-13)=0$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, $\displaystyle x=13$.

The length of the shorter diagonal is $\displaystyle 13$ units long.

Let $\displaystyle x$ be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by $\displaystyle x+5$.

Recall how to find the area of a kite:

$\displaystyle \text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for $\displaystyle x$.

$\displaystyle 18=\frac{x(x+5)}{2}$

$\displaystyle x^2+5x=36$

$\displaystyle x^2+5x-36=0$

$\displaystyle (x+9)(x-4)=0$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, $\displaystyle x=4$.

The length of the shorter diagonal is $\displaystyle 4$ units long.

Let $\displaystyle x$ be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by $\displaystyle x+3$.

Recall how to find the area of a kite:

$\displaystyle \text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for $\displaystyle x$.

$\displaystyle 20=\frac{x(x+3)}{2}$

$\displaystyle x^2+3x=40$

$\displaystyle x^2+3x-40=0$

$\displaystyle (x+8)(x-5)=0$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, $\displaystyle x=5$.

To find the longer diagonal, add $\displaystyle 3$.

The length of the longer diagonal is $\displaystyle 8$ units long.

Let $\displaystyle x$ be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by $\displaystyle x+4$.

Recall how to find the area of a kite:

$\displaystyle \text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for $\displaystyle x$.

$\displaystyle 1680=\frac{x(x+4)}{2}$

$\displaystyle x^2+4x=3360$

$\displaystyle x^2+4x-3360=0$

$\displaystyle (x+60)(x-56)=0$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, $\displaystyle x=56$.

The length of the shorter diagonal is $\displaystyle 56$ units long.