Charlene puts together two isosceles triangles so that they share a base, creating a kite. The legs of the triangles are 10 inches and 17 inches, respectively. If the length of the base for both triangles is 16 inches, what is the length of the kite's other diagonal?

A. 6 inches
B. [tex]$4 \sqrt{21}$[/tex] inches
C. [tex]$16 \sqrt{21}$[/tex] inches
D. 21 inches



Answer :

Let's solve the problem step-by-step.

Charlene creates a kite by putting together two isosceles triangles, which share a base that is 16 inches long. The legs of the first triangle are each 10 inches, and the legs of the second triangle are each 17 inches. We need to find the length of the kite's other diagonal.

Step 1: Find the height of the first triangle.

For an isosceles triangle, we can split it into two right triangles by drawing a perpendicular line from the vertex opposite the base to the midpoint of the base.
- Base = 16 inches, so each half of the base is [tex]\( \frac{16}{2} = 8 \)[/tex] inches.
- Legs = 10 inches.

Using the Pythagorean theorem in one of the right triangles:
[tex]\[ (\text{leg})^2 - (\text{base}/2)^2 = (\text{height})^2 \][/tex]
[tex]\[ 10^2 - 8^2 = (\text{height})^2 \][/tex]
[tex]\[ 100 - 64 = (\text{height})^2 \][/tex]
[tex]\[ 36 = (\text{height})^2 \][/tex]
[tex]\[ \text{height} = \sqrt{36} = 6 \text{ inches} \][/tex]

So, the height of the first triangle is 6 inches.

Step 2: Find the height of the second triangle.

- Base = 16 inches, so each half of the base is [tex]\( \frac{16}{2} = 8 \)[/tex] inches.
- Legs = 17 inches.

Again, using the Pythagorean theorem in the right triangle:
[tex]\[ 17^2 - 8^2 = (\text{height})^2 \][/tex]
[tex]\[ 289 - 64 = (\text{height})^2 \][/tex]
[tex]\[ 225 = (\text{height})^2 \][/tex]
[tex]\[ \text{height} = \sqrt{225} = 15 \text{ inches} \][/tex]

So, the height of the second triangle is 15 inches.

Step 3: Find the length of the kite's other diagonal.

The other diagonal is the sum of the heights of the two triangles:
[tex]\[ \text{other diagonal} = 6 \text{ inches} + 15 \text{ inches} = 21 \text{ inches} \][/tex]

Thus, the length of the kite's other diagonal is [tex]\( 21 \)[/tex] inches.

Among the options, the correct answer is:
[tex]\[ \boxed{21 \text{ inches}} \][/tex]

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