Deltoid Diagonal Problem: Finding Secondary Diagonal Length Using 5:1 Ratio

Question

The length of the main diagonal in the deltoid is equal to 42 cm.

The secondary diagonal divides the main diagonal in the ratio of 5:1

The area of the small isosceles triangle, whose secondary diagonal forms its base, is equal to 12 cm².

Find the length of the secondary diagonal.

Video Solution

Solution Steps

00:00 Find the length of the secondary diagonal

00:03 The diagonal equals the sum of its parts, ratio of parts according to given

00:08 Substitute appropriate values according to given, and solve for X

00:16 This is value X

00:19 Substitute value X to find diagonal parts

00:26 These are the parts of diagonal AC

00:34 Use the formula for calculating triangle area

00:38 (height multiplied by base) divided by 2

00:50 Substitute appropriate values according to given, and solve for BD

00:58 Isolate diagonal BD

01:10 And this is the solution to the question

Step-by-Step Solution

To solve this problem, we'll employ the formula for the area of a triangle and the information about the main diagonal's division:

Step 1: Determine segment lengths using the ratio 5:1 and calculate $x$ where $6x = 42$ . Thus, $x = 7$ . Therefore, $AE = 5x = 35 \, \text{cm}$ and $EC = x = 7 \, \text{cm}$ .
Step 2: Using the area of triangle $BDE$ , given $EC$ as the base, we use the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ where base = $EC = 7$ .
Step 3: Calculate $\text{BD}$ by relating it to height: Given the area of $BDE$ is 12 cm²:
$\frac{1}{2} \times 7 \times \frac{\text{BD}}{2} = 12$ .
Solve:
$\frac{7 \times \text{BD}}{4} = 12$
Multiply through by 4 to clear fraction: $28 \times \text{BD} = 48$
$\text{BD} = \frac{48}{28} = \frac{24}{14} = \frac{12}{7} \times 2 = 4 \, \text{cm}$ .

The secondary diagonal $BD$ has a length of 4 cm.

The correct choice from the given options is $\boxed{4}$ .