Prove that the area of a rhombus is one-half the product of the lengths of the diagonals. Suppose that Delta ABCis equilateral triangle and that P is a point in the interior of this triangle. Prove that the sum of the perpendicular distances from P to each of the sides of the triangle is equal to the height of the triangle. Prove that the area of a triangle is equal to one-half the product of its perimeter and the length of the radius of a circle inscribed within the triangle. Prove that the diagonal of a kite connecting the vertices where the congruent sides intersect bisects the angles at these vertices and is the perpendicular bisector of the other diagonal. (Use the strict definition of a kite which is a quadrilateral with two distinct pairs of congruent adjacent sides.)

Mariyah Bell 2022-10-14 Answered
1. Prove that the area of a rhombus is one-half the product of the lengths of the diagonals.
2. Suppose that Δ A B C is equilateral triangle and that P is a point in the interior of this triangle. Prove that the sum of the perpendicular distances from P to each of the sides of the triangle is equal to the height of the triangle.
3. Prove that the area of a triangle is equal to one-half the product of its perimeter and the length of the radius of a circle inscribed within the triangle.
4. Prove that the diagonal of a kite connecting the vertices where the congruent sides intersect bisects the angles at these vertices and is the perpendicular bisector of the other diagonal. (Use the strict definition of a kite which is a quadrilateral with two distinct pairs of congruent adjacent sides.)
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Answers (1)

t5an1izqp
Answered 2022-10-15 Author has 13 answers
Given ABCD is a rhombus the diagonal AC and BD cut at point O
Then A O D = A O B = C O D = B O C = 900
The area of rhombus ABCD divided diagonal in four parts
So area of rhombus ABCD =area of triangle AOD+area of triangle AOB+area of triangle BOC+area of triangle COD
= 21 × A O × O D + 21 × A O × O B + 21 × B O × O C + 21 × O D × O C
= 21 × A O ( O D + O B ) + 21 O C ( B O + O D )
= 21 × A O × B D + 21 × O C × B D
= 21 B D ( A O + O C ) = 21 × B D × A C
So area of rhombus is equal to half of the product of diagonals....
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