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Maths-

General

Easy

Question

In the figure above, △ACD is a right triangle and BE is parallel to CD. What is the perimeter of △ACD to the nearest tenth of a unit?

29.7
36.0
41.5
50.9

The correct answer is: 41.5

Step by step explanation:
Given:
In $straight triangle ACD$ ,
BE ॥ CD, CD = 9 cm, BE = 3 cm, AB = 5 cm.
Step 1:
Consider two triangles, $straight triangle ACD text and end text straight triangle AEB$
$straight angle ACD equals straight angle ABE equals 90 to the power of ring operator$ (As, BE ॥ CD and $straight angle ACD equals 90 to the power of ring operator$ is given)
$straight angle CAD equals straight angle BAE$ (Common angle)
So,
According to AA rule
$straight triangle A C D tilde straight triangle A E B$
Step 2:
We know that, when two triangles are similar their slides are in same ratio.
∴ $AC over AB equals CD over BE equals AD over AE$
$not stretchy rightwards double arrow fraction numerator A C over denominator 5 end fraction equals 9 over 3$
$not stretchy rightwards double arrow AC equals fraction numerator 9 cross times 5 over denominator 3 end fraction$
$not stretchy rightwards double arrow AC equals 45 over 3$
⇒ AC = 15
∴ AC = 15 cm
Step 3:
Find Side AD of $straight triangle A C D$
in $straight triangle A C D$ ,
(AD)² = (AC)² + (CD)²
⇒ (AD)² = (15)² + (9)²
⇒ (AD)² = 225 + 81
⇒ (AD)² = 306
$not stretchy rightwards double arrow left parenthesis AD right parenthesis equals square root of 306$
⇒ AD = 17.49 cm
AD = 17.49 cm
Step 3:
Find perimeter of $straight triangle ACD$
Perimeter of $straight triangle ACD$ = AC + CD + AD
⇒ Perimeter of $straight triangle ACD$ = 15 + 9 + 17.5
⇒ Perimeter of $straight triangle ACD$ = 41.5
Hence, the perimeter of $straight triangle ACD$ is 41.5 cm.

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