In rectangle ABCD, e is the midpoint of BC, and AE ⊥ ed, if AE = 2, root 2, calculate the area of rectangle

In rectangle ABCD, e is the midpoint of BC, and AE ⊥ ed, if AE = 2, root 2, calculate the area of rectangle

According to △ Abe ≌ △ DCE, △ AED is an isosceles right triangle
From Pythagorean theorem, we get ad = 4, ab = 2
The area of the rectangle is 8

In rectangle ABCD, BC = 2Ab, e is the midpoint of BC, AE = 2, radical 2, find the area and perimeter of rectangle ABCD

Because AE = 2, radical 2
ab=be
According to Pythagorean theorem
The square of AB + the square of be = the square of AE
So AB = be = 2
Because e is the midpoint
So BC = 2be = 4
So s quadrilateral = AB * BC = 2 * 4 = 8
C quadrilateral ABCD = AB + BC + CD + Da = 2 + 2 + 4 + 4 = 12

In rectangular ABCD, BC = 2Ab, e is the midpoint of BC, AE = 2 radical cm, find the perimeter and area of rectangular ABCD

It is known that be = AB, and because AE = √ 2, ab = 1, BC = 2
So perimeter = 2 + 2 + 1 + 1 = 6, area = 2 * 1 = 2

As shown in the figure, the quadrilateral ABCD is a parallelogram, be: EC = 1:2, f is the midpoint of DC, the area of triangle Abe is 12cm2, then the area of triangle ADF is______ cm2．

The area of triangle ADF is 12 times of the area of triangle ACD, and the area of triangle ADF is equal to 12 × 12 = 14 of the area of parallelogram. The area ratio of triangle Abe and triangle AEC is 1:2, so the area of triangle ABC is 12 × (2 + 1) = 36

In the parallelogram ABCD, BC: EC = 1:2, f is the midpoint of DC, the area of triangle Abe is 6cm & # 178;, then the area of triangle ADF is () square centimeter
The original title is: in the parallelogram ABCD, be: EC = 1: F is the midpoint of DC, the area of triangle Abe is 6cm & # 178;, then the area of triangle ADF is () square centimeter

Is e on BC?
Connect AC
A quadrilateral ABCD is a parallelogram
∴S△ABC=S△ADC
BC:EC=1:2
E is the midpoint of BC
∴S△ABE=1/2S△ABC=6
∴S△ABC=12
∴S△ADC=12
∵ f is the midpoint of CD
S △ ADF = 1 / 2S △ ADC = 6 square centimeter

As shown in the figure is a parallelogram, CE = 2be, f is the midpoint of DC, the area of triangle Abe is 6cm2, then the area of triangle ADF is______ cm2．

Because CE = 2be, so BC = 3bE, and because the area of triangle Abe is 6 square centimeters, so the area of triangle ABC is 6 × 3 = 18 (square centimeters), then the area of triangle ACD is 18 square centimeters; because f is the midpoint of CD, so the area of triangle ADF is 18 △ 2 = 9 (square centimeters), answer: the area of triangle ADF is 9 square centimeters. So the answer is: 9

A parallelogram be: EC = 1:2, f is the midpoint of DC, the area of Abe is 10 square centimeter, find the area of triangle ADF

Link AC
∵ E on BC, be: EC = 1:2
∴S△ABE：S△ABC = BE：BC = 1：3
∴S△ABC = 30cm²
∵ AC is the diagonal of the parallelogram ABCD
∴S△ABC = S△ACD
∵ f is the midpoint of CD
∴S△AFD：S△ADC = DF：CD = 1：2
∴S△AFD = 15cm²

As shown in the figure, in the parallelogram ABCD, AC is a diagonal, ∠ B = 60 °, ab = 2, BC = 4. Prove that AC is perpendicular to ab

Proof
CE ⊥ AB over C, AB over e
∵∠B=60°,BC＝4
∴BE＝BC×cos∠B=4×cos60=4×1/2=2
∵AB=2
The point e coincides with the point a
∴AC⊥AB

In the parallelogram ABCD, we know AB = 12, BC = 10, a = 60 ° to find the length of two diagonals of the parallelogram

AC=√(AB^2+AD^2+2AB*AD*cos60°)=√(12^2+10^2+2*12*10*1/2)=2√91
BD=√(AB^2+AD^2-2AB*AD*cos60°)=√(12^2+10^2-2*12*10*1/2)=2√31

As shown in the figure, in the parallelogram ABCD, m n is the point on ad AB and BM = DN, and its intersection point is p. try to compare the size of ∠ CPB and ∠ cpd
RT

: equal
The area connecting MC, NC, Δ BCM and Δ CDN is equal to half of the area of quadrilateral ABCD
The area of Δ BCM = the area of Δ CDN
Do CE ⊥ BM, CF ⊥ dn
∴1/2BM•CE=1/2DN•CF,∵BM=DN
∴CE=CF,∴∠CPB=∠CPD