Question

In A PQR, QX I PR and PY IQR. If PQ =
10 cm, QR = 8 cm and PR = 5.6 cm, find the
length of PY, where QX = 7.2 cm.
P
X Х
Q
R
Y
Fig. 15.24​

Answer 1

Answer:

There is a problem with your question:

If pq = 10 cm, qr = 8 cm and pr = 5.6 cm then if qx is perpendicular to pr through q it does NOT equal 7.2 cm; it is approx 8.0 cm:

Let X be the distance px, then xr = 5.6 - X

Using Pythagoras:

In pxq: 10² = qx² + X² → qx² = 10² - X²

In rxq: 8² = qx² + (5.6 - X)² → qx² = 8² - (5.6 - X)²

→ 10² - X² = 8² - (5.6 - X)²

→ 10² - X² = 8² - 5.6² + 2×5.6×X - X²

→ 2×5.6×X = 10² - 8² + 5.6²

→ X = (10² - 8² + 5.6²)/(2×5.6)

→ qx² = 10² - ( (10² - 8² + 5.6²)/(2×5.6) )²

→ qx = √(10² - ( (10² - 8² + 5.6²)/(2×5.6) )²) ≈ 7.989265 cm ≈ 8.0 cm

Similarly, for py:

py = √(10² - ( (10² - 5.6² + 8²)/(2×8) )²) ≈ 5.59248 cm ≈ 5.6 cm

The obtuse triangle has angles: qpr ≈ 53°, prq ≈ 93°, rqp ≈ 34°; the perpendiculars qx and py lie outside the triangle; angle prq ≈ 93° which is not far off a right angle making sides pr and qr approximately perpendicular, and shows that the perpendiculars to the sides next to it (ie the perpendiculars to pr and qr) will be approximately equal to the lengths of the other side (next to it, ie length of perpendicular to pr will be approx qr, and the length of the perpendicular to qr will be approx pr).

Answer 2

Answer:

Step-by-step explanation:

Hi bro, so there we go with your answer-->

if you take PR (5.6cm)as the base and QX(7.2cm)as the height, then area of triangle = bh/2.

That means our triangles area will be 5.6 × 7.2/2 = 20.16 sq.cm.

So now we got our Area = 20.16 sq.cm

                                Height = h

                                Base = 8

now lets simply the equation bh/2 = 20.16

8 × h/2 = 20.16

h = 20.16 × 2/8 = 5.04cm

∴ h = 5.04cm