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3 years ago

14

ABCD∼EFGH AD=45 in. , EH=75 in. , and AB=30 in. What is EF ? Enter your answer in the box. EF = in.

1 answer:

Helen [10]3 years ago

6 0

The ~ symbol means that the 2 shapes are similar, meaning the same shape but not the same size.

To determine the measure of side length EF, you must find the relationship between the 2 side lengths.

You should set these up as ratios of the length to the width. So, for ABCD, it is 30/45. Make this equal to x/75. If you simplify the first ratio to 2:3, you can then multiply by both numerator and denominator by 25 to get x = 25 in/ 75 in.

Another way is to use cross products with you proportion. 75 x 30 = 45x. Solve for x to get 50 inches.

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LMN is a right-angle triangle. Angle NLM=90. PQ is parallel to LM. The area of triangle PNQ is 8cm^2. The area of triangle LPQ i

likoan [24]

Answer:

The area of LQM is $48cm^2$

Step-by-step explanation:

Given

Area of PNQ = 8

Area of LPQ = 16

See attachment for triangles

The area of PNQ is calculated as:

$Area = \frac{1}{2} * PQ * PN$

Substitute 8 for Area

$8 = \frac{1}{2} * PQ * PN$

$PQ * PN = 16$

The area of LPQ is calculated as:

$Area = \frac{1}{2} * PQ * LP$

Substitute 16 for Area

$16= \frac{1}{2} * PQ * LP$

From the attachment:

$PN + LP =LN$

Make LP the subject

$LP = LN -PN$

So:

$16= \frac{1}{2} * PQ * (LN -PN)$

We have:

$16= \frac{1}{2} * PQ * (LN -PN)$ and $PQ * PN = 16$

Equate both expressions:

$\frac{1}{2} * PQ *(LN - PN) = PQ * PN$

Divide both sides by PQ

$\frac{1}{2} (LN - PN) = PN$

Multiply both sides by 2

$LN - PN = 2PN$

$LN= 3PN$

Since PNQ is similar to LNM, the following equivalent ratios exist:

$\frac{LM}{PQ} = \frac{LN}{PN}$

Substitute $LN= 3PN$

$\frac{LM}{PQ} = \frac{3PN}{PN}$

$\frac{LM}{PQ} = 3$

$LM = 3PQ$

Area of LQM is:

$Area = \frac{1}{2} * LM * LP$

This gives:

$Area = \frac{1}{2} * 3PQ * LP$

$Area = 3 *\frac{1}{2} *PQ * LP$

Recall that:

$16= \frac{1}{2} * PQ * LP$

So:

$Area = 3 *16$

$Area = 48cm^2$

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