1 cm = 3 m
so 5 cm = 3x5= 15 m
the actual tree is 15 m tall
Answer:
<h2>231cm²</h2>
Step-by-step explanation:
First, let's find the surface area of both the triangles
5x3=15
So, the surface area of the triangles is 15 sq.cm
Now, let's find the surface area of the base (large rectangle in the middle)
12x8=?
10x8=80
2x8=16
80+16=96
12x8=96
So, the surface area of the base, is 96sq.cm
Now, let's find the surface area of both of the side rectangles
12x5=60
60x2=120
So, the surface area of the two side rectangles is 120sq.cm
Now, let's find the total surface area by adding all of our answers.
120+96=216
216+15=231
<h2>
So hence, the surface area of this net is 231cm²</h2>
Answer: ↓↓↓Btw sorry its not in order
Step-by-step explanation:
m∠62 and ∠g are vertical angles so they're congruent.
∠g = 62°
∠f and m∠62 are linear pairs so they're supplementary
∠f + 62 = 180
∠f = 118°
∠e is a right angle and is supplementary to a 90° angle
∠e = 90°
Adding ∠e, m∠28° and the missing angle should sum up to 180° since its a triangle. x = missing angle
x + ∠e + m∠28 = 180°
x + 90 + 28 = 180
x + 118 = 180
x = 62°
∠h ≅ x
∠h ≅ 62
∠h = 62°
∠b and m∠28 are vertical angles so they're congruent.
∠b ≅ m∠28
∠b = 28°
∠b and ∠a are linear pairs so they are supplementary.
∠a + ∠b = 180
∠a + 28 = 180
∠a = 152°
∠c and ∠a are corresponding angles so they're congruent.
∠c ≅ ∠a
∠c ≅ 152
∠c = 152°
∠c and ∠j are vertical angles so they're congruent.
∠j ≅ ∠c
∠j ≅ 152
∠j = 152°
The corresponding angle of c is 152° as well. Let's call it y. Y corresponds to the angle with a bisector. Call that whole angle z.
z ≅ y
z ≅ 152
z = 152°
Z has a bisector meaning the two angles that formed because of the bisector are congruent.
∠d and the other angle add up to 152°. They're equal.
∠d = 152 ÷ 2
∠d = 76°
Hope I helped!
Answer:
See the paragraph proof below.
Step-by-step explanation:
Quadrilateral JKLM is given as a parallelogram. By a theorem, opposite sides JK and LM are congruent (1). By the definition of parallelogram, opposite sides KJ and ML are parallel. By the theorem on alternate interior angles, angles KJL and MLJ are congruent (2). Segments JN and PL are given as congruent (3). Using the three statements of congruence labeled above (1), (2), and (3), we now prove that triangles JKN and LMP are congruent by SAS. Sides of the triangles KN and PM are congruent by CPCTC. Sides of quadrilateral KNMP are given as parallel. Therefore, quadrilateral KNMP is a parallelogram by the theorem: If two sides of a quadrilateral are both parallel and congruent, then the quadrilateral is a parallelogram.
