Answer:
∠BAD=20°20'
∠ADB=34°90'
Step-by-step explanation:
AB is tangent to the circle k(O), then AB⊥BO. If the measure of arc BD is 110°20', then central angle ∠BOD=110°20'.
Consider isosceles triangle BOD (BO=OD=radius of the circle). Angles adjacent to the base BD are equal, so ∠DBO=∠BDO. The sum of all triangle's angles is 180°, thus
∠BOD+∠BDO+∠DBO=180°
∠BDO+∠DBO=180°-110°20'=69°80'
∠BDO=∠DBO=34°90'
So ∠ADB=34°90'
Angles BOD and BOA are supplementary (add up to 180°), so
∠BOA=180°-110°20'=69°80'
In right triangle ABO,
∠ABO+∠BOA+∠OAB=180°
90°+69°80'+∠OAB=180°
∠OAB=180°-90°-69°80'
∠OAB=20°20'
So, ∠BAD=20°20'
Answer:
See below.
Step-by-step explanation:
I'm assuming these questions are about the Midline Theorem (segment AL joins the midpoints of the non-parallel sides.
♦ The midline's length is the average of the lengths of the top and bottom parallel sides.
Use this equation and substitute values given in each problem, then solve for the missing information.
1. AL = x, CE = 9, OR = 5
2. AL = <em>m</em> - 4, CE = 15, OR = 17
3. OR = y + 5, AL = 15, CE = 18
Answer:
length= 24cm
Step-by-step explanation:
interpreting the question
L=3w
perimeter=2(l+w)
64=2(3w+w)
64=2(4w)
64=8w, Divide both sides by 8
w= 8cm
since w= 8
L=3w
L=3×8
L=24cm
Answer:
<em>It has been given that Rectangle Q has an area of 2 square units.</em>
<em>Thea Drew a scaled version of Rectangle Q and marked it as R.</em>
<em>As you must keep in mind If we draw scaled copy of pre-image, then the two images i.e Pre-image and Image are similar.</em>
<em>As you have not written what is the scale factor of transformation</em>
<em>Suppose , Let the Scale factor of transformation= k</em>
<em>Rectangle Q = Pre -image, Rectangle R= Image</em>
<em>If, Pre-Image < Image , then scale factor is k >1.</em>
<em>But If, Pre-Image > Image , then Scale factor will be i.e lies between, 0<k<1.</em>
