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Viktor [21]
3 years ago
9

SOMEONE PPS SHOW ME HOW TO DO THIS???

Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0
A(total) = A(big rectangle) + A(triangle) - A(sml rectangle)

At = 15(10) + (1/2)(15)(5) - (5)[10 - (2+5)]
At = 172.5 unit sq.
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The figure below shows a rectangle ABCD having diagonals AC and DB: A rectangle ABCD is shown with diagonals AC and BD. Jimmy wr
yaroslaw [1]

Answer:

C. DC = DC. By reflexive property of equality.

Step-by-step explanation:

We are given that,

In step 4, Jimmy proved ΔADC ≅ ΔBCD by the SAS Congruence Postulate.

For that, the previous steps states,

Step 1: AD = BC, as opposite sides of a rectangle are congruent.

Step 2: ∠ADC = ∠BCD

So, in order to satisfy the SAS Postulate, we must have another pair of sides equal from the respective rectangles.

As, we can see that in ΔADC and ΔBCD, the side DC is a common side.

Thus, in the missing step, we get,

Step 3: DC = DC, By the reflexive property of equality.

Hence, option C is correct.

5 0
2 years ago
Read 2 more answers
30 POINTS please help me!
Stels [109]
To prove a similarity of a triangle, we use angles or sides.

In this case we use angles to prove

∠ACB = ∠AED (Corresponding ∠s)
∠AED = ∠FDE (Alternate ∠s)

∠ABC = ∠ADE (Corresponding ∠s)
∠ADE = ∠FED (Alternate ∠s)

∠BAC = ∠EFD (sum of ∠s in a triangle)

Now we know the similarity in the triangles.

But it is necessary to write the similar triangle according to how the question ask.

The question asks " ∆ABC is similar to ∆____. " So we find ∠ABC in the prove.

∠ABC corressponds to ∠FED as stated above.
∴ ∆ABC is similar to ∆FED

Similarly, if the question asks " ∆ACB is similar to ∆____. "
We answer as ∆ACB is similar to ∆FDE.

Answer is ∆ABC is similar to ∆FED.
7 0
3 years ago
How to find the nth term
Arte-miy333 [17]

Answer:

Write full question

I cannot understand

6 0
2 years ago
Read 2 more answers
Please help if you know the answer
Pachacha [2.7K]

Answer:

I think the answer to this is C

5 0
2 years ago
Suppose a varies directly as b, and a=7 when b=2 find b when a =21
kozerog [31]
Because α varies directley to β
\alpha = k \beta
(k = constant)

α=7 when β=2
7= 2k
k= \frac{7}{2}

so when α=21
21= \frac{7}{2} b
b= 6

6 0
3 years ago
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