What is the perimeter of the figure below?
2 answers:
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Answer:
UV=29
Step-by-step explanation:
In right triangles AQB and AVB,
∠AQB = ∠AVB ...(i) {Right angles}
∠QBA = ∠VBA ...(ii) {Given that they are equal}
We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle
∠AQB+∠QBA+∠BAQ=180 ...(iii)
∠AVB+∠VBA+∠BAV=180 ...(iv)
using (iii) and (iv)
∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV
∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV (using (i) and (ii))
∠BAQ=∠BAV...(v)
Now consider triangles AQB and AVB;
∠BAQ=∠BAV {from (v)}
∠QBA = ∠VBA {from (ii)}
AB=AB {common side}
So using ASA, triangles AQB and AVB are congruent.
We know that corresponding sides of congruent triangles are equal.
Hence
AQ=AV
5x+9=7x+1
9-1=7x-5x
8=2x
divide both sides by 2
4=x
Now plug value of x=4 into UV=7x+1
UV=7*4+1=28+1=29
<u>Hence UV=29 is final answer.</u>
Solution :
Let be the unit vector in the direction parallel to the plane and let
be the component of F in the direction of
and
be the component normal to
.
Since,
Therefore,
From figure,
We know that the direction of is opposite of the direction of
, so we have
The unit vector in the direction normal to the plane, has components :
Therefore,
From figure,
∴
Therefore,
