9514 1404 393
Answer:
34.8
Step-by-step explanation:
The only polygon in the figure is the triangle outlined in red. Its perimeter is the sum of its side lengths: 15 +6.1 +13.7 = 34.8 units.
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The two halves of the chord are the same length.
The area of the composite figure from the image attached is 112.5 units²
<h3>What is the area of a composite figure?</h3>
The area of a composite figure refers to the sum total of all the areas of the shapes in that composite figure. This can be done by first identifying the shapes in that composite figure, then finding each area, followed by the addition of all the areas to determine the area of the composite figure.
From the given figure; we can break it down into:
- A parallelogram
- A rectangle
- A triangle
The area of a parallelogram = b × h
where;
- b & h refers to the length of the two opposite diagonal lines.
The area of a parallelogram = 9 × 6
The area of a parallelogram = 54 units²
The area of a rectangle = Length × breadth
The area of a rectangle = (9 × 3) units²
The area of a rectangle = 27 units²
The area of the triangle 
where;
The area of the triangle 
The area of the triangle 
The area of the triangle 
Therefore, the area of the composite figure is:
= 54 + 27 + 31.5
= 112.5 units²
Learn more about composite figures here:
brainly.com/question/15981553
Answer:
80 men
Step-by-step explanation:
4 : 5 10 : 9
40 : 50 : 45
40 x 2 = 80
It’s a right triangle so the answer would be 40
Answer:
The answer is ΔJMK ≈ ΔMLK ≈ ΔJLM ⇒ answer (A)
Step-by-step explanation:
* Lets start with the equal angles i the three triangles
- In ΔJMK
∵ m∠JKM = 90°
∴ m∠KJM + m∠KMJ = 90 ⇒ (1)
- IN ΔMLK
∵ m∠MKL = 90°
∴ m∠KML + m∠KLM = 90° ⇒ (2)
∵ m∠KMJ + m∠KML = 90° ⇒ (3)
- From (1) , (2) , (3)
∴ m∠KJM = m∠KML
∴ m∠KMJ = m∠KLM
* Now lets check the condition of similarity in the 3 triangles
- At first ΔJMK and ΔMLK
- In triangles JMK , MLK
∵ m∠KJM = m∠KML
∵ m∠KMJ = m∠KLM
∵ m∠JKM = m∠MKL
∴ ΔJMK ≈ ΔMLK ⇒ (4)
- At second ΔJMK and ΔJLM
∵ m∠KJM = m∠MJL
∵ m∠KMJ = m∠MLJ
∵ m∠JKM = m∠JML
∴ ΔJMK ≈ ΔJLM ⇒ (5)
* If two triangles are similar to one triangle, then they are
similar to each other
- From (4) and(5)
∴ ΔJMK ≈ ΔMLK ≈ ΔJLM