In. What is the area of this trapezoid? b2 = 5 in. h = 4 in. 2 in. 3 in. b1 = 10 in.
1 answer:
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Answer:
Step-by-step explanation:
In the diagram shown, the measure of angle 1 is oppositely directed to angle 2 and oppositely directed angles are equal.
Hence <1 = <3
Given < 1 = 3x-1 and <3 = 2x+9
Hence 3x-1 = 2x+9
collect like terms
3x-2x = 9+1
x = 10°
Since <1 = 3x-1
on substituting x = 10
<1 = 3(10)-1
<1 = 30-1
<1 = 29°
<1+<2 = 180 (angle on a straight line)
29+<2 = 180
<2 = 180-29
<2 = 151°
Similarly, on substituting x = 10 into <3
<3 = 2x+9
<3 = 2(10)+9
<3 = 20+9
<3 = 29°
<3+<4 = 180 (angle on a straight line)
29+<4 = 180
<4 = 180-29
<4 = 151°
Answer:
See explanation
Step-by-step explanation:
In ΔABC, m∠B = m∠C.
BH is angle B bisector, then by definition of angle bisector
∠CBH ≅ ∠HBK
m∠CBH = m∠HBK = 1/2m∠B
CK is angle C bisector, then by definition of angle bisector
∠BCK ≅ ∠KCH
m∠BCK = m∠KCH = 1/2m∠C
Since m∠B = m∠C, then
m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH (*)
Consider triangles CBH and BCK. In these triangles,
- ∠CBH ≅ ∠BCK (from equality (*));
- ∠HCB ≅ ∠KBC, because m∠B = m∠C;
- BC ≅CB by reflexive property.
So, triangles CBH and BCK are congruent by ASA postulate.
Congruent triangles have congruent corresponding sides, hence
BH ≅ CK.
