In the diagram, PQRS is a quadrilateral. < PQR = < PRS = 90°, /PQ/ = 3 cm , /QR/ = 4 cm and /PS/ = 13 cm, Find the area of the quadrilateral.

36cm²

Step-by-step explanation:

From the sketched diagram of the Quadrilateral above, we have two triangles that are joined together, ∆PQR and ∆PSR

Both triangles are right angled triangles.

Step 1

We solve for ∆PQR

When solving for the sides of a right angled triangle, if two sides are given, we solve for the third sides using Pythagoras Theorem

Pythagoras Theorem =

c² = a² + b²

Where c = Longest side

For ∆PQR,

/PQ/ = 3 cm = a

/QR/ = 4 cm = b

/PR/ = c ??

c² = 3² + 4²

c² = 9 + 16

c² = 25

c = √25

c = 5cm

/PS/ = 5cm

We find the Area of this triangle using Heron's formula

= √s(s - a)(s - b)(s - c)

Where s = a + b + c/2

s = 3 + 4 + 5/2

= 12/2

s = 6

Area of the ∆PQR = √6(6 - 3)(6 - 4)(6 - 5)

= √6 × 3 × 2 × 1

= √36

= 6cm²

Area of the ∆PQR

Step 2

We solve for ∆PSR. Just like in Step 1, we solve for third side not given using Pythagoras Theorem

/PS/ = 13cm = c

/PR/ = 5cm = a

/SR/ = ?? = b

c² = a² + b²

13² = 5² + b²

b² = 13² - 5²

b² = 169 - 25

b² = 144

b = √144

b = 12cm

/SR/ = 12cm

We find the Area of this triangle using Heron's formula

= √s(s - a)(s - b)(s - c)

Where s = a + b + c/2

s = 5 + 12 + 13/2

= 30/2

s = 15

Area of the ∆PSR = √15(15 - 5)(15 - 12)(15 - 13)

= √15 × 10 × 3 × 2

= √900

= 30cm²

Area of the ∆PSR = 30 cm²

Step 3

We find the Area of the Quadrilateral

Area of the Quadrilateral (PQRS)= Area of ∆PQR + Area of ∆PSR

= 6cm² + 30cm²

= 36cm²

Therefore the Area of the Quadrilateral PQRS = 36cm²

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step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.