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3 years ago

Which congruence theorems can be used to prove △EFG ≅ △JHG? Check all that apply.

Mathematics

2 answers:

sasho [114]3 years ago

4 0

Answer:

ASA and AAS

Step-by-step explanation:

We do not know if these are right triangles; therefore we cannot use HL to prove congruence.

We do not have 2 or 3 sides marked congruent; therefore we cannot use SSS or SAS to prove congruence.

We are given that EF is parallel to HJ. This makes EJ a transversal. This also means that ∠HJG and ∠GEF are alternate interior angles and are therefore congruent. We also know that ∠EGF and ∠HGJ are vertical angles and are congruent. This gives us two angles and a non-included side, which is the AAS congruence theorem.

Since EF and HJ are parallel and EJ is a transversal, ∠JHG and ∠EFG are alternate interior angles and are congruent. Again we have that ∠EGF and ∠HGJ are vertical angles and are congruent; this gives us two angles and an included side, which is the ASA congruence theorem.

UkoKoshka [18]3 years ago

3 0

Answer: ASA and AAS

Step-by-step explanation:

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Alenkasestr [34]

Answer:

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Step-by-step explanation:

4 0

3 years ago

A merchant bought 1000 electric bulbs at the rate of Rs.150 each. 50 of them were broken on the way and the remaining he sold at

neonofarm [45]

Answer:

₹165.79

Step-by-step explanation:

Given:-

No. of electric bulbs = 1000

cost of each electric bulb = ₹ 150

No. of bulbs broken = 50

Selling price of each bulb = x

Profit percentage = 5%

To Find:-

The selling price of each bulb.

Solution:-

Cost price of 1000 electric bulbs,

= 1000 × ₹150

= ₹1,50,000

5% profit on the total cost price,

= {5}/{100}× ₹150000

= ₹7500

Total selling price = ₹157500

No. of bulbs remaining = 950

Therefore, selling price of each bulb,

= {₹157500}/{950}

= ₹165.79

Therefore,

Selling price of each bulb = ₹165.79

5 0

1 year ago

Find the area of the circle to the nearest whole number, if necessary!

astra-53 [7]

Answer:

The circle has an area of about 1385 square mm.

Step-by-step explanation:

Let's recall that circles have an area that can be found with the following formula:

$A=\pi r^{2}$

where r is the radius of the circle.

Now, focus your eyes on the circle. We are shown that the diameter of this circle is 42 mm, but we only want the radius. Since the radius is half the diameter, the radius is 21 mm. Now, we can solve for the area of the circle.

$A=\pi (21)^2\\A=441\pi \\A=1385.44236...$