The sum of two numbers is 15, and the sum of their square is 137. What are the numbers?
1 answer:
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Answer: 24 unit
Step-by-step explanation:
Here, MNOK is a trapezoid, MN=OK, m∠M=60°, NK ⊥ MN , MK=16cm
Since, a mid segment is the line segment which joins the mid points of the equal sides of the isosceles trapezoid.
Let LO is the mid segment of trapezoid MNOK
Where, Let J is the intersection point of KN and LO.
Therefore, LO= LJ+JO --------(1)
And, KL=LO , MO=ON
Since, In triangle KLJ,
∠LKJ=90° ( given), ∠KLJ=60° ( because, LO║ON and it is given m∠M=60°)
Thus, ∠KJL=30°
Therefore, sin 30°=LK/LJ=8/LJ ( Because ΔMKN is a isoceleus triangle where ∠MKN=∠MNK=30°⇒KM=MN)
⇒LJ=8×2=16
Now, In ΔNOJ,
∠ONJ=∠OJN=30°
Therefore, ON=OJ ( by the property of isosceles triangle)
⇒OJ=8
Thus, By putting these values in equation 1) we get,
LO=16+8=24 cm
Answer: The hook would be 2.2 inches (approximately) above the top of the frame
Step-by-step explanation: Please refer to the picture attached for further details.
The top of the picture frame has been given as 9 inches and a 10 inch ribbon has been attached in order to hang it on a wall. The ribbon at the point of being hung up would be divided into 5 inches on either side (as shown in the picture). The line from the tip/hook down to the frame would divide the length of the frame into two equal lengths, that is 4.5 inches on either side of the hook. This would effectively give us two similar right angled triangles with sides 5 inches, 4.5 inches and a third side yet unknown. That third side is the distance from the hook to the top of the frame. The distance is calculated by using the Pythagoras theorem which states as follows;
AC^2 = AB^2 + BC^2
Where AC is the hypotenuse (longest side) and AB and BC are the other two sides
5^2 = 4.5^2 + BC^2
25 = 20.25 + BC^2
Subtract 20.25 from both sides of the equation
4.75 = BC^2
Add the square root sign to both sides of the equation
2.1794 = BC
Rounded up to the nearest tenth, the distance from the hook to the top of the frame will be 2.2 inches
