Find the blue shaded area | Geometry Problem | Important Geometry Skills Explained

Find the blue shaded area | Geometry Problem | Important Geometry Skills Explained

Join this channel to get access to perks:

MEMBERS OF THIS CHANNEL
••••••••••••••••••••••••••••••••••••••••
Mr. Gnome
Sambasivam Sathyamoorthy
ஆத்தங்கரை சண்முகம்
Εκπαιδευτήρια Καντά
AR Knowledge
堤修一
Sidnei Medeiros Vicente
Mark Ludington
Grace Cooper
Nicholas Garcia
Ben Stone
Karlo Karbs
Debashis Kesh

[miguelgnievesl]

There is a shorter way. We draw a parallel to BC in A and mark it with E. So DE=10 and CE=4. Therefore AD =2√41 and consequently PD =√41. Since the triangles ADE and PDR are similar, then PR=4 and DR=5 and we also obtain CR=9. In conclusion Blue Area=1/2•PR•CR=1/2•4•9=18

[اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

[spacer]

The blue area is actually 1/4 of the area of the trapezoid. This is because the blue area has the same area as triangle PRC, while the trapezoid can be reconfigured as a rectangle of RCxBC by rotating triangle PRD over point P so that point D lines up to point A. The area of trapezoid is (4+14)*8/2 = 18*4. The blue area is 1/4 of that which is 18.

[santanuganguly]

6:31 we can directly extablish it by mpt that PM = 1/2(AB + DC)

[marioalb]

Area trapezoid =
Area of rectangle with
height = 8 cm and
base = (14+4)/2 = 9 cm

Area rectangle and trapezoid:
Ar = b. h = 8 . 9 = 72 cm²

Area triangle :
At = ¼ Ar = 72 / 4
At = 18 cm² (Solved √ )

[ROCCOANDROXY]

Draw a perpendicular from A to DC intersecting at T and let y = DR and z = PR implies Triangle(ADT) ~ Triangle(DPR) implies 1/2 = y/10 = z/8 implies y = 5 and z = 4 implies RT = 5 implies
RC = 9.
Draw a perpendicular from P to BC intersecting at S and let QS = w implies A(Triangle(PQR)) = A(RPQC) - A(Triangle(RQC)) = 1/2 * 9 * (8 + w) - 1/2 * 9 * (4 + w) = 9/2 * (4) = 18.

In General, letting PB = a, BC = b and DC = c implies A(Triangle(PQR)) = 1/8 * b * (a + c).

[StefanNr]

It is easier to solve for the area of the trapezoid în 2 ways (we need this to find the segment BQ, because if we have BQ then we have the segment QC and we can easly solve for RC fro. The triangle DPR similar With the triangle forme with the trapezoids altitudine. S-o like this we have the segment RQ, we can easly solve again for PR from the similarities between the 2 triangles from before and PQ we can solve from the trapezoid RCQP. Everything will be în function of BQ. I canotaj Wright the whole solution because it will Take too long on a keyboard.

[dileepmv]

Why so much calculations? It can even be calculated in mind

[marioalb]

Base of triangle:
b = 8 / 2 = 4 cm

Height of triangle:
h = (14 - 4)/2 + 4
h = 9 cm

Area = b. h / 2
Area = 18 cm² ( Solved √ )