The area of a parallelogram is 86 square meters, which is the same as the area of a triangle with the same base and height______ ．

The area of a parallelogram is 86 square meters, which is the same as the area of a triangle with the same base and height______ ．

86 △ 2 = 43 (square meters) a: the area of a triangle with the same base and height is 43 square meters. So the answer is: 43 square meters

The area of a parallelogram is 2.5 square centimeters larger than that of a triangle of equal height. What is the area of a parallelogram?
A primary school question

If the base of a parallelogram is a, the height is B, and the area is x, then:
ab=x
The area of the triangle is AB / 2
ab/2+2.5=x
Then AB / 2 + 2.5 = ab
2.5=ab/2
ab=x=5

The area of a triangle is ()% of the area of a parallelogram with its equal base and height

50%
Area of parallelogram = base * height;
The area of triangle = base * height / 2;

What is the area ratio of a parallelogram with equal base and height to a triangle?
Which should a (1:2) B (1:1) C (2:1) choose?

C(2:1)
Because the area of parallelogram = base * height
Triangle area = 1 / 2 * bottom * height

A parallelogram and a triangle have the same area and the same base. The ratio of their height is ()

A parallelogram and a triangle have the same area and the same base. The ratio of their height is (1:2)

A parallelogram and a triangle have the same area and height. The bottom of the triangle is 3.6 decimeters, and the bottom of the parallelogram is () decimeters

A parallelogram and a triangle have the same area and height. The bottom of the triangle is 3.6 decimeters, and the bottom of the parallelogram is (7.2) decimeters
Area of parallelogram = base × height
Triangle area = bottom × height △ 2
They have the same area and the same height
The bottom of the triangle is 2 times that of the parallelogram, which is 7.2 decimeters
What don't understand can ask!

A triangle has a base of 12 decimeters and a height of 6 decimeters. A parallelogram with the same area has a base of 9 decimeters and a height of () decimeters

4

As shown in the figure e, f are two points on the edge BC of the triangle ABCD, and be = EF = DC = AE = AF, prove ∠ BAC = 2 (∠ B + ∠ C)

Because AE = EF = AF, triangle AEF is equilateral triangle, so angle AEF = angle AFE = angle EAF = 60 degrees, because be = AE, so angle B = angle BAE, because angle AEF = angle B + angle BAE = 60 degrees, so angle B = angle BAE = 30 degrees, because AF = CF, so angle c = angle CAF, because angle AFE = angle c +

As shown in the figure, in trapezoidal ABCD, AD / / BC, e is the midpoint of BC, EM ⊥ AB is in M, en ⊥ CD is in M, and EM = Mn

EM = en?
prove:
∵EM⊥AB,EN⊥CD
∴∠BME=∠CNE=90°
∵ e is the midpoint of BC
∴EB=EC
∵EM=EN
∴△EBM≌△ECN
∴∠B=∠C
ABCD is trapezoidal
The quadrilateral ABCD is an isosceles trapezoid

As shown in the figure, in the inscribed quadrilateral ABCD, AE bisects ∠ bad and intersects the circumscribed circle at point e. the distances from point e to BC and DC are em, en respectively. Verify EM = en

Certification:
Connect be, de
∵ AE bisection ∠ bad
∴∠BAE=∠DAE
Be = de
∵EM⊥BC,EN⊥DC
∴∠BME=∠DNE=90º
And ∵ ∠ EBM = ∠ EDN [equal to the circumference angle of arc CE]
∴⊿BME≌⊿DNE（AAS）
∴EM=EN