Answer:
∠B ≅ ∠F ⇒ proved down
Step-by-step explanation:
<em>In the </em><em>two right triangles</em><em>, if the </em><em>hypotenuse and leg</em><em> of the </em><em>1st right Δ ≅</em><em> the </em><em>hypotenuse and leg</em><em> of the </em><em>2nd right Δ</em><em>, then the </em><em>two triangles are congruent</em>
Let us use this fact to solve the question
→ In Δs BCD and FED
∵ ∠C and ∠E are right angles
∴ Δs BCD and FED are right triangles ⇒ (1)
∵ D is the mid-point of CE
→ That means point D divides CE into 2 equal parts CD and ED
∴ CD = ED ⇒ (2) legs
∵ BD and DF are the opposite sides to the right angles
∴ BD and DF are the hypotenuses of the triangles
∵ BD ≅ FD ⇒ (3) hypotenuses
→ From (1), (2), (3), and the fact above
∴ Δ BCD ≅ ΔFED ⇒ by HL postulate of congruency
→ As a result of congruency
∴ BC ≅ FE
∴ ∠BDC ≅ ∠FDE
∴ ∠B ≅ ∠F ⇒ proved
Answer:
how tho. people I summon y'all lol
Ok so then line p is the equation y=x
is that all you needed?
2 radians = 2 full rotations
pi = 3.14
circumference of wheel = 2(pi)r
= 20 * 3.14
=62.8 cm
= 0.628m
meters rolled = 0.628 * 2
=1.256 m
Answer:
16. Right isosceles
17. D
Step-by-step explanation:
The triangle has a right angle this makes it a right trangle. Also all parallelograms are not squares, all rectangles aren't squares and all parallelograms aren't rhombuses.
