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pychu [463]
2 years ago
7

(Bro I'm so confused) In the diagram, PQRS is a square. Points T, U, V, and W are on sides PQ, QR, RS, and ST, respectively, for

ming square TUVW. If PT=QU=RV=SW=4 m and PQRS has area 256 m squared determine the area of TUVW. (Please I'll mark brainliest and heart if you can help me out ty)

Mathematics
1 answer:
Gnesinka [82]2 years ago
8 0

The total area of square TUVW is 160 meters.

<h3>Find out the sides of PQRS:</h3>

Total area: side x side

256 m = side x side

side =  √ 256

side= 16 m

This means PQ, QR, RS, SP = 16 m

<h3>Find out the area of one triangle:</h3>

Area: 1/2 x base x height

Let's find out the area of the triangle UVR although all triangles are the same.

Base: 4 meters

Height:

RQ= 16

QU= 4

16-4 = 12

Area: 1/2 x 4 x 12 = 24

<h3>Find the total area of the triangles:</h3>

24 x 4 = 96

Finally, find out the total area of the grey square:

Total area of the grey square = Total area of the white square - area of the triangles

Total area = 256 - 96

Total area= 160

Learn more about squares in: brainly.com/question/13747846

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In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some poin
Mademuasel [1]

Answer:  7\frac{5}{7} unit and 18 unit

Step-by-step explanation:

Let ABCD is a trapezoid where AB and CD are the bases. ( In which AB is greatest base which shown in below figure)

AD and BC are the legs of the trapezoid ABCD.

Now, we have ( According to the question ),

AB = 18 unit, BC = 7 unit, AD = 3 unit and DC = 11 unit.

Here the leg AD extends from point D.

Similarly leg BC extends from point C.

Let they meet at point P ( shown in below diagram)

Since In triangles PAB and PDC,

∠PDC ≅ ∠PAB ( because DC ║ AB )

And, ∠ PAB ≅ ∠ PBA

∠DPC ≅ ∠ APB ( reflexive)

Therefore, By AAA similarity postulate,

\triangle PDC \sim \triangle PAB

Thus, By the definition of similarity,

\frac{PD}{PA} = \frac{DC}{AB}

\frac{PD}{PD+3} = \frac{11}{18} ( because PA = PD+DA)

⇒ 18 PD = 11 PD +33

⇒7PD = 33

⇒ PD = 33/7

Again by the definition of similarity,

\frac{PC}{PB} = \frac{DC}{AB}

\frac{PC}{PC+7} = \frac{11}{18} ( because PB = PC + CB)

⇒ 18 PC = 11 PD +77

⇒7PC = 77

⇒ PC = 11

Thus, PA =  PD+DA = 33/7 + 3 = 7\frac{5}{7}

And, PB = PC + CB = 11 + 7 = 18


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