Construct a quadrilateral MINT such that MI= 5cm, IN = 6cm, NT = 7cm, TM=3cm and MN =9cm
1 answer:
From the given dimensions, of MI, IN, NT, TM, and MN, the quadrilateral
MINT can be drawn as shown in the attached image.
<h3>What are the steps for the construction of MINT?</h3>The given dimensions of the quadrilateral MINT are;
MI = 5 cm
IN = 6 cm
NT = 7 cm
TM = 3 cm
MN = 9 cm
The side MN is a diagonal of MINT, therefore;
ΔMIN, and ΔMTN are triangles with a common base = MN
The steps to construct MINT are therefore;
- Step 1; Draw the line MN = 9 cm.
- Step 2; Place the compass at point <em>M</em> and with a radius MI = 5 cm, draw an arc on one side of MN.
- Step 3; Place the compass at <em>N</em> and with radius IN = 6 cm, draw an arc to intersect the arc dawn in step 1 above.
- Step 4; Place an arc at point <em>M</em> and with radius TM = 3 cm draw an arc on the other side of MN.
- Step 5; Place the compass at point <em>N</em> and with radius NT = 7 cm, draw an arc to intersect the arc drawn in step 3.
- Step 6; Join the point of intersection of the arcs to points <em>M</em> and <em>N</em> to complete the quadrilateral MINT.
Please find attached the drawing (showing the construction arcs) of the
quadrilateral MINT created with MS Word.
Learn more about types of geometric construction here:
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Hello from MrBillDoesMath!
Answer:
4/sin(80)
which is approximately 4.06
Discussion:
From the triangle, not drawn too accurately (see attachment)
sin(80) = side opposite /hypotenuse =>
sin (80) = 4/c => multiply both sides by "c"
c sin(80) = 4/c * c =>
c sin(80) = 4 => divide both sides by sin(80)
c = 4/sin(80)
and 4/sin(80) is approximately 4.06
Thank you,
MrB
